Alta- Ch. 3 Probability Topics Pt. 2

Let A be the event that a person who lives in New York City is married, and note that P(A)=25. Let B be the event that a person who lives in Honolulu is married, and note that P(B)=720. Events A and B are independent. What is the probability that a randomly chosen person from New York City and a random chosen person from Honolulu will be married? Give your answer as a fraction in simplest form.

$\frac{7}{50}$750​​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=25⋅720=750

Two psychologists are studying the IQs of a group of business professionals. Let A be the event that a professional's IQ is greater than 115, with P(A)=16%. Let B be the event that a professional's IQ is between 115 and 130, with P(B)=14%. The IQs of different professionals are independent events. What is the probability that two random professionals will be chosen and one of their IQs is greater than 115 and the other is between 115 and 130? Give your answer as a decimal, rounded to three decimal places.

$0.022$0.022​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=16%⋅14%=0.16⋅0.14≈0.022

The probability that a student will take loans to pay for their undergraduate education is 0.85, and the probability that a student will go to graduate school given that the student took loans to pay for their undergraduate education is 0.13. What is the probability that a student will go to graduate school and take loans to pay for their undergraduate education? Round your answer to three decimal places.

$0.111$0.111​ Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) So if we think of A as the event of a student taking loans to pay for their undergraduate education and B as being the event of a student going to graduate school, then we can plug in the known information to find P(B AND A)=(0.13)(0.85)≈0.111

The probability that an adult visits their dentist annually is 0.58. The probability that an adult has mouth pain given that they visit their dentist annually is 0.34. What is the probability that an adult has mouth pain and visits their dentist annually? Round your answer to three decimal places.

$0.197$0.197​ Let the probability that an adult visits their dentist annually be P(A)=0.58. Let the probability that an adult has mouth pain given that they visit their dentist annually be P(B|A)=0.34. Remember the multiplication rule for conditional probability: P(A AND B)=P(B|A)P(A) Then we can plug in the known information to find:P(A AND B)=(0.34)(0.58)≈0.197

The outcome X of an experiment has only two possible values, X=4 and X=5. Given that the expected outcome E(X)=4.2, find P(X=5). Give your answer as a decimal.

$0.2$0.2​ Let's say P(X=4)=a, and P(X=5)=b. Since the probabilities must add up to 1, we have a+b=1(∗). Now we use the fact that E(X)=4.2. Using the formula for E(X) we can write an equation for a and b as follows:4a+5b=4.2(∗∗).Solving (*) and (**) simultaneously givesa=0.8,b=0.2,so we haveP(X=4)=0.8,andP(X=5)=0.2.

The probability that an adult smokes cigarettes is 0.14. The probability that an adult has a cardiovascular disease and they smoke cigarettes is 0.028. What is the probability that an adult has a cardiovascular disease given that they smoke cigarettes? Round your answer to two decimal places.

$0.20$0.20​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging, we see thatP(A|B)=P(A AND B)P(B) So if we think of A as the event that an adult has a cardiovascular disease and B as the event that an adult smokes cigarettes, then we can plug in the known information to find P(A|B)=0.0280.14=0.20

The probability that an adult is actively looking for a job is 0.18. The probability of an adult being unemployed and actively looking for a job is 0.04. What is the probability of an adult being unemployed given that they are actively looking for a job? Round your answer to two decimal places.

$0.22$0.22​ Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) Rearranging, we see thatP(B|A)=P(B AND A)P(A) So if we think of A as being the event that an adult is actively looking for a job and B as the event that an adult is unemployed, then we can plug in the known information to find P(B|A)=0.040.18≈0.22

An electric company has to track on-time payments and late payments for each customer monthly. It is impossible for a customer to pay on-time and late each month. If the probability that a customer pays on-time each month is 0.55, and the probability that a customer pays late or on-time each month is 0.82, what is the probability that a customer pays late each month?

$0.27$0.27​ Because it is impossible for a customer to pay on-time and late each month, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find thatP(A)=P(A OR B)−P(B)So plugging in the given values, we findP(A)=0.82−0.55=0.27 The probability that a customer pays late each month is 0.27.

A random variable X has the following probability distribution: x12345P(X=x)0.1ab0.10.2 Given that E(X)=3.0, find the value of a. Give your answer as a decimal to the first decimal place.

$0.3$0.3​ First, we note that the probabilities must add up to 1. Therefore 0.1+a+b+0.1+0.2=1, which givesa+b=0.6. This is the first equation. We now use the fact that we know the expected value, so we can write 3=1⋅0.1+2a+3b+4⋅0.1+5⋅0.2, which gives the equation2a+3b=1.5. This is the second equation. If we solve the first and second equations simultaneously we get a=b=0.3.

The following Venn diagram shows the percent of people who own a cat and own a dog. A Venn diagram with an unlabeled universal set contains two intersecting circles labeled Has a dog and Has a cat that divide the universal set into four regions labeled as follows, where the label is given first and the content is given second: Has a dog only, 35 percent; Has a dog and has a cat only, 10 percent; Has a cat only, 15 percent; outside the circles only, 40 percent. Given that a randomly selected person has a cat, what is the probability that the person also has a dog? Give your answer as a decimal without any percent signs. Round to two decimal places.

$0.40$0.40​ We are given that the person has a cat, so we want to look in the red circle. Within that circle, the 10% portion (the portion which is also part of the blue circle) represents the people who also have a dog. Therefore, the probability that a person has a dog, given that the person has a cat, is 10%10%+15%=10%25%=1025=0.40

The probability that a student is a business major and in a statistics course is 0.29. The probability that a randomly chosen student is in a statistics course given that the student is a business major is 0.67. What is the probability a student is a business major? Round your answer to three decimal places.

$0.433$0.433​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging, we find thatP(B)=P(A AND B)P(A|B) So if we think of A as being the event that a student is in a statistics course and B as being the event a student is a business major, then we can plug in the known information to find P(B)=0.290.67≈0.433

A hospital conducts a study of men and women to see if they are satisfied with their current weight. Let A be the event that a randomly selected woman is satisfied with her weight and B be the event that a randomly selected man is satisfied with his weight. The probabilities of these events are as follows: P(A)=56% and P(B)=85% A and B are independent events. What is P(A AND B), the probability that a randomly selected man and woman will both be satisfied with their weight? Give your answer as a decimal, rounded to three decimal places.

$0.476$0.476​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=56%⋅85%=0.56⋅0.85=0.476

If a police officer pulls over someone for speeding, the police officer can either give a ticket or a warning, so it is impossible for a police officer to give a ticket and a warning for speeding. If the probability that a police officer will give a warning for speeding is 0.03, and the probability that a police officer will give a ticket or a warning for speeding is 0.52, what is the probability that a police officer will give a ticket for speeding?

$0.49$0.49​ Because it is impossible for a police officer to give a ticket and a warning for speeding, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find thatP(A)=P(A OR B)−P(B)So plugging in the given values, we findP(A)=0.52−0.03=0.49 The probability that a police officer will give a ticket for speeding is 0.49.

Randy wants to either ride share to work or drive his own car to work, but it is impossible for Randy to ride share and drive his own car in one trip. If the probability that Randy ride shares is 0.22, and the probability that Randy drives his own car is 0.42, what is the probability that Randy ride shares or drives his own car to work?

$0.64$0.64​ Because it is impossible for Randy to ride share and drive his own car in one trip, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) So we find thatP(A OR B)=P(A)+P(B)=0.22+0.42=0.64 The probability that Randy ride shares or drives his own car to work is 0.64.

Given that the probability of a student taking a statistics class is 0.83, and the probability of a student taking a calculus class and a statistics class is 0.66, what is the probability of a student taking a calculus class given that the student takes a statistics class? Round your answer to three decimal places.

$0.795$0.795​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging this, we findP(A|B)=P(A AND B)P(B)So if we think of A as being the event a student taking a calculus class and B as being the event a student taking a statistics class, then we can plug in the known information to findP(A|B)=0.660.83≈0.795

There are two known issues with a certain model of new car. The first issue, A, occurs with a probability of P(A)=0.1. B is another known issue with the car. If it is known that either event occurs with a probability of P(A OR B)=0.93, and that both events occur with a probability of P(A AND B)=0.07, calculate P(B).

$0.9$0.9​ Begin with the Addition Rule: P(A OR B)=P(A)+P(B)−P(A AND B) Rearranging to solve for P(B), we find that P(B)=P(A OR B)−P(A)+P(A AND B)=0.93−0.1+0.07=0.9 So P(B)=0.9.

Given that the probability of a student spending time reading is 0.59, and the probability of a student doing well on an exam and spending time reading is 0.58, what is the probability of a student doing well on an exam given that the student spends time reading? Round your answer to three decimal places.

$0.983$0.983​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging, we see thatP(A|B)=P(A AND B)P(B) So if we think of A as being the event of a student doing well on an exam and B as being the event of a student spending time reading, then we can plug in the known information to find P(A|B)=0.580.59≈0.983

Fill in the following contingency table and find the number of students who both have a cat AND have a dog. StudentshaveacatdonothaveacatTotalhaveadog46donothaveadog17Total4998

$14$14​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the row of totals in the table, we know that the unknown number plus 46 is 98, so the missing number must be 52. Continuing in this way, we can fill in the entire table: StudentshaveacatdonothaveacatTotalhaveadog143246donothaveadog351752Total494998 From this, we can see that the number of students who both have a cat and have a dog is 14.

Two companies offer their employees retirement plans. Let A be the event that a randomly chosen employee from the first company chooses to invest in the retirement plan. Let B be the event that a randomly chosen employee from the second company chooses to invest in the retirement plan. The probabilities of them investing in these plans are as follows: P(A)=0.30 and P(B)=0.48 You choose one employee randomly from each company. If A and B are independent events, what is P(A AND B), the probability that they both invest in the retirement plan? Give your answer as a percent, rounded to two decimal places if necessary.

$14.4\%$14.4%​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=0.30⋅0.48=0.144 To express our answer as a percent, we multiply by 100: 0.144×100=14.4%

Complete the following contingency table and find the number of students who both do not play sports AND do not play an instrument. StudentsPlay an instrumentDo not play an instrumentTotalPlay sports19 47Do not play sports 29Total 46

$18\text{ students}$18 students​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the first row in the table, we know that 19 added to the unknown number in the middle is 47, so that unknown number is 28. Continuing in this way, we can fill in the entire table: StudentsPlay an instrumentDo not play an instrumentTotalPlay sports192847Do not play sports111829Total304676 From this, we can see that the number of students who both do not play sports and do not play an instrument is 18.

Fill in the following contingency table and find the number of students who both do not have a cat AND do not have a dog. StudentshaveacatdonothaveacatTotalhaveadog20donothaveadog50Total4946

$26$26​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the second row in the table, we know that 20 added to the unknown number in the middle is 46, so that unknown number is 26. Continuing in this way, we can fill in the entire table: StudentshaveacatdonothaveacatTotalhaveadog252045donothaveadog242650Total494695 From this, we can see that the number of students who both do not have a cat and do not have a dog is 26.

Fill in the following contingency table and find the number of students who both do not go to the beach AND do not go to the mountains. StudentsgotothebeachdonotgotothebeachTotalgotothemountains13donotgotothemountains45Total4941

$28$28​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the second row in the table, we know that 13 added to the unknown number in the middle is 41, so that unknown number is 28. Continuing in this way, we can fill in the entire table: StudentsgotothebeachdonotgotothebeachTotalgotothemountains321345donotgotothemountains172845Total494190 From this, we can see that the number of students who both do not go to the beach and do not go to the mountains is 28.

Fill in the following contingency table and find the number of students who both play sports AND play an instrument.

$30$30​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the second row in the table, we know that 23 added to the unknown number in the middle is 48, so that unknown number is 25. Continuing in this way, we can fill in the entire table: From this, we can see that the number of students who both play sports and play an instrument is 30.

A survey of students in a class reports who plays video games and who plays board games. The following Venn diagram displays the results. A Venn diagram with a universal set contains two intersecting circles labeled Video Games and Board Games that form four regions. There are values in each region as follows, where the location of the region is given first and the content is given second: Video Games only, 15; Video Games and Board Games only, 9; Board Games only, 19; outside the circles only, 7. Given that a random student does not play board games, what is the probability that the person also does not play video games? Give your answer as a percent rounded to two decimal places.

$31.82\%$31.82%​ We are given that the student does not play board games, so we must look outside the red circle. This leaves 22 students: the 15 who play video games and the 7 who do not play video games and do not play board games. Therefore, the probability that a student does not play video games given that the student does not play board games is 722=31.82%.

Fill in the following contingency table and find the number of students who both read mysteries AND read comics. StudentsreadmysteriesdonotreadmysteriesTotalreadcomics32donotreadcomics61Total7057

$34$34​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the second row in the table, we know that 32 added to the unknown number in the middle is 57, so that unknown number is 25. Continuing in this way, we can fill in the entire table: StudentsreadmysteriesdonotreadmysteriesTotalreadcomics343266donotreadcomics362561Total7057127 From this, we can see that the number of students who both read mysteries and read comics is 34.

Fill in the following contingency table and find the number of students who both do not have a cat AND do not have a dog. StudentshaveacatdonothaveacatTotalhaveadog30donothaveadog65Total5655

$39$39​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the first row in the table, we know that 30 added to the unknown number in the middle is 56, so that unknown number is 26. Continuing in this way, we can fill in the entire table: StudentshaveacatdonothaveacatTotalhaveadog301646donothaveadog263965Total5655111 From this, we can see that the number of students who both do not have a cat and do not have a dog is 39.

Evaluate 12!5!.

$3991680$3991680​ Write out each factorial as a product of natural numbers, then simplify the fraction by cancelling like factors and finding the product of the remaining expression. 12!5!====12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅15⋅4⋅3⋅2⋅112⋅11⋅10⋅9⋅8⋅7⋅6⋅5!5!12⋅11⋅10⋅9⋅8⋅7⋅63,991,680 Notice how we canceled the 5! which appears in both the numerator and denominator of the fraction shown above. Answer: 3,991,680

The probability that a child watches cartoons after school is 0.56. The probability that the child requests a certain toy of a cartoon character given that the child watches cartoons after school is 0.89. What is the probability that a child will request a toy and watch cartoons after school? Give your answer as a percent, rounded to two decimal places if necessary.

$49.84\%$49.84%​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) So if we think of A as being the event of a child requesting a toy and B as being the event of a watching cartoons after school, then we can plug in the known information to findP(A AND B)=(0.89)(0.56)≈0.4984 To express our answer as a percent, we multiply by 100: 0.4984×100=49.84%

The results of an experiment are shown below. What is E? Note that for this particular distribution there is a simple way to determine E. Outcome246810Probability0.150.20.30.20.15

$6$6​ Note that the distribution of results is entirely symmetrical about the outcome =6, so E=6. However we can also use the formula: E=2⋅0.15+4⋅0.2+6⋅0.3+8⋅0.2+10⋅0.15=6.

The results of an experiment are shown below. Calculate E, the expected value of the experiment. Write your answer as a decimal. Outcome246810Probability110310210110310

$6.4$6.4​ E=2⋅110+4⋅310+6⋅210+8⋅110+10⋅310=6.4.

If A and B are independent events with P(A)=0.20 and P(A AND B)=0.12, find P(B). Give your answer as a percentage, rounded to two decimal places if necessary.

$60\%$60%​ Remember that for independent events, P(A AND B)=P(A)P(B) Solving for P(B), we find thatP(B)=P(A AND B)P(A)Now, plugging in the values we are given, we find thatP(B)=0.120.20=0.6 To change the decimal to a percent, we multiply by 100: 0.6×100=60%

A poker game requires players to draw a card and roll a die. The combination of card and die will determine if a player wins. There are: 13 different types of cards: Ace, 2,3,4,5,6,7,8,9,10, Jack, Queen, King 6 different numbers, one on each face of a die: 1,2,3,4,5,6 How many different combinations are there of one card and one die roll?

$78$78​ The Fundamental Counting Principle tells us that we multiple the number of cards by the number of faces on the die to get the total number of possibilities: Possibilities=13×6=78

A survey of a sample of recent hires at major tech companies aims to investigate which applicants are most likely to be hired for positions in data science. The findings of the survey are presented in the Venn diagram below. A Venn diagram with an unlabeled universal set contains two intersecting circles labeled Graduate degree and 10+ years experience that divides the universal set into four regions labeled as follows, where the label is given first and the content is given second: Graduate degree only, 16; Graduate degree and 10+ years experience only, 4; 10+ years experience only 20; outside the circles only, 7. Given that a random data scientist from the sample has less than ten years of experience in the industry, what is the probability that they have a graduate degree in a relevant discipline? Provide the final answer as a simplified fraction.

$\frac{16}{23}$16/23​​ From the Venn diagram, a total of 16+7=23 data scientists have less than ten years of experience. Of these, we want to know the number of them who have graduate degrees, which is 16. Given that a random data scientist from the sample has less than ten years of experience in the industry, the probability that they have a graduate degree in a relevant field is 16/23.

The table below represents the outcomes and probabilities for those outcomes. Find the expected value of the experiment. Outcome349Probability131313 Write your answer as an exact fraction.

$\frac{16}{3}$163​​ The expected value of this experiment is given by E=3⋅13+4⋅13+9⋅13=3+4+93=163.

If you roll a fair die and then roll a second fair die, what is the probability that you roll a 1, 2, or 3 on the first die and roll a 5 on the second die? Write your answer as a fraction.

$\frac{1}{12}$112​​ Note that these are independent events because the outcome when you roll a fair die does not affect the outcome when you roll a second fair die. So by the multiplication rule for independent events, we can take the probability of each event and multiply them.The probability that you roll a 1, 2, or 3 on the first die is 36, and the probability that you roll a 5 on the second die is 16. The probability that you roll a 1, 2, or 3 on the first die and roll a 5 on the second die is 36⋅16=112

Write the first five terms of the sequence whose general term is an=n!(n+1)!. Note: Enter your answers as simplified fractions. To enter your answer as a fraction, click in the first answer box and then use the fraction template that appears in the menu.

$\frac{1}{2},\ \frac{1}{3},\ \frac{1}{4},\ \frac{1}{5},\ \frac{1}{6}$12​, 13​, 14​, 15​, 16​​ We substitute the values 1,2,3,4,5 in order into the formula an=n!(n+1)!. ana1a1a1a1=n!(n+1)!=1!(1+1)!=(1)!(2)!=12⋅1=12ana2a2a2a2=n!(n+1)!=2!(2+1)!=(2)!(3)!=2⋅13⋅2⋅1=13ana3a3a3a3=n!(n+1)!=3!(3+1)!=(3)!(4)!=3⋅2⋅14⋅3⋅2⋅1=14ana4a4a4a4=n!(n+1)!=4!(4+1)!=(4)!(5)!=4⋅3⋅2⋅15⋅4⋅3⋅2⋅1=15ana5a5a5a5=n!(n+1)!=5!(5+1)!=(5)!(6)!=5⋅4⋅3⋅2⋅16⋅5⋅4⋅3⋅2⋅1=16 The first five terms of the sequence are 1/2,1/3,1/4,1/5, and 1/6.

An airline tracks each of its airplanes' stops for the day. A particular airplane can travel to one of the following cities for each of its stops: First StopSecond StopThird StopNew York BostonMiamiCharlotteDallasNew YorkChicago These options are mapped out in the tree diagram below: A tree diagram is shown. The top row reads New York. There are four arrows from New York, one arrow to each of Boston, Miami, Charlotte, and Dallas in the middle row. From each of the items in the middle row there are two arrows, one each to New York and Chicago. What is the probability that the stops include Boston and Chicago?

$\frac{1}{8}$1/8​​ There is 1 possible outcome that visits Boston and Chicago. There are 8 possible combinations (Possibilities = 1 first stop × 4 for the second stop × 2 for the third stop = 8). So the probability of visiting both Boston and Chicago is: P(Boston and Chicago)=1/8

A group of high school students reports who has a job and who plays sports. The information is presented in the following Venn diagram. A Venn diagram with universal set contains two intersecting circles labeled Job and Sports that divide the universal set into four regions labeled as follows, where the region is given first and the content is given second: Job only, 20; Job and Sports only, 8; Sports only, 16; Outside the circles only, 24. Given that a random student has a job, what is the probability that the student does not play sports? Provide the final answer as a fraction.

$\frac{20}{28}$20/28​​ A total of 8+20 students have a job, and of these, we want to know the number of students who do not play sports, which is 20. So the probability is 20/28.

Use the completed contingency table to find the probability that a citizen exercises, given that they are female. Express your answer as a fraction.

$\frac{22}{93}$2293​​ Because our given information is that the citizen is female we want to focus on that row of the table. In that row, there are 93 citizens total, and 22 of them exercise. ExercisesDoes not ExerciseTotalFemale227193Male115667Total33127160 Therefore, P(Exercises|Female)=2293.

Adam loves to visit his local coffee shop. When there, his two favorite drinks are cafe latte and espresso. Let A represent the event that he drinks a cafe latte and B represent the event that he drinks an espresso. If A and B are independent events with P(A)=0.5 and P(B)=0.8, find P(A AND B). Give your answer as a fraction in simplest form.

$\frac{2}{5}$25​​ Remember that for independent events, P(A AND B)=P(A)P(B) So, plugging in the values we are given, we find thatP(A AND B)=(0.50)(0.80)=0.4 To express this answer as a fraction in simplest form, we have 0.4=410=25 The probability Adam goes to the coffee shop and drinks both a cafe latte and an espresso is 25.

The event that it is a sunny day and the event that you are given a speeding ticket are independent events. Let A be the event that it is a sunny day with P(A)=35. Let B be the event that you are given a speeding ticket with P(B)=120. What is the probability that it is a sunny day and you are given a speeding ticket? Give your answer as a fraction in simplest form.

$\frac{3}{100}$3100​​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=35⋅120=3100

A survey on the educational backgrounds of a sample of working computer scientists produces the findings presented in the Venn diagram below. A Venn diagram with an unlabeled universal set contains two intersecting circles labeled Computer science and Mathematics that divides the universal set into four regions labeled as follows, where the label is given first and the content is given second: Computer science only, 65; Computer science and Mathematics only, 10; Mathematics only 13; outside the circles only, 42. Given that a random computer scientist from the sample does not have a degree in computer science, what is the probability that they do not have a degree in mathematics? Provide the final answer as a simplified fraction.

$\frac{42}{55}$42/55​​ From the Venn diagram, a total of 13+42=55 computer scientists do not have a degree in computer science. Of these, we want to know the number who do not have a degree in mathematics, which is 42. Given that a random computer scientist from the sample does not have a degree in computer science, the probability that they also do not have a degree in mathematics is 42/55.

First year nursing students are randomly assigned to classes in a nursing school. Let A be the event that they are assigned to an Anatomy course and B be the event that they are assigned to a Psychology course. The probabilities of them being assigned to these courses are as follows: P(A)=23 and P(B)=25 If A and B are independent events, what is P(A AND B), the probability that a student will be assigned to both an Anatomy and a Psychology course? Give your answer as a fraction.

$\frac{4}{15}$415​​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=23⋅25=415

A group of 120 students at a high school were asked if they prefer streaming to live TV. The results are shown in the table above. Given that a randomly selected survey participant is a male, what is the probability that this student prefers live TV? Give your answer as a fraction in simplest form.

$\frac{5}{23}$523​​ Because our given information is that the chosen participant is male, we want to focus on that column of the table. The total number of those that are males is 23. Of these, 5 prefer live TV to streaming. Therefore, the probability that a randomly selected male student that was surveyed prefers live TV is 523.

A class of eighth graders reports who plays music and who plays sports. They present the information in the following Venn diagram. A normal curve is over a horizontal axis and is centered on 0.00. Two points are labeled on the horizontal axis, one at negative 0.76 and another at 0.76. The area under the curve to the left of negative 0.76 and right of 0.76 is shaded. Given that a random student does not play music, what is the probability that the student does not play sports? Provide the final answer as a fraction.

$\frac{6}{16}$616​​ A total of 6+10 students do not play music, and of these, we want to know the number of students who do not play sports, which is 6. So the probability is 616.

A group of patients who are trying to lose weight report who exercises and who diets. The information is presented in the following Venn diagram. A Venn diagram with a universal set contains two intersecting circles labeled Exercise and Diet that divide the set into four regions labeled as follows, where the region is given first and the content is given second: Exercise only, 25; Exercise and Diet only, 30; Diet only, 15; Outside the circles only, 5. If a patient is selected at random from the group, what is the probability that the patient exercises or diets (or both)? Provide the final answer as a fraction.

$\frac{70}{75}$7075​​ A total of 25+30+15=70 people exercise or diet or both, and the total group size is 25+30+15+5=75. Therefore, the probability is7075.

A group of 155 students at a private university were asked if they are full-time or part-time and if they own a car. The following table shows the results. Part-TimeFull-TimeTotalOwns a Car226587Does Not Own a Car76168Total29126155 According to the table, what is the probability that a randomly chosen student is does not own a car given that they are Part-Time? Give your answer as a fraction in simplest form.

$\frac{7}{29}$729​​ Because our given information is that the is Part-Time, we want to focus on that column of the table. Part-TimeFull-TimeTotalOwn a Car226587Do Not Own a Car76168Total29126155 In that row, there are 29 students total, and 7 of them did not own a car, therefore the probability is 729.

A survey on the spending habits of a sample of households finds that two of the most common monthly expenses for households in the sample are mortgages (home loans) and car loans. The findings from the survey are presented in the Venn diagram below. Given that a random household from the sample does not have a mortgage, what is the probability that the household has a car loan? Provide the final answer as a simplified fraction.

$\frac{8}{11}$8/11​​ From the Venn diagram, a total of 24+9=33 households do not have mortgages. Of these, we want to know the number of households who have car loans, which is 24. Given that a random household from the sample does not have a mortgage, the probability that the household has a car loan is 24/33=8/11.

A survey on the spending habits of a sample of households finds that two of the most common monthly expenses for households in the sample are mortgages (home loans) and car loans. The findings from the survey are presented in the Venn diagram below. A Venn diagram with an unlabeled universal set contains two intersecting circles labeled Mortgages and Car loans that divides the universal set into four regions labeled as follows, where the label is given first and the content is given second: Mortgages only, 11; Mortgages and Car loans only, 85; Car loans only 24; outside the circles only, 9. Given that a random household from the sample does not have a mortgage, what is the probability that the household has a car loan? Provide the final answer as a simplified fraction.

$\frac{8}{11}$8/11​​ From the Venn diagram, a total of 24+9=33 households do not have mortgages. Of these, we want to know the number of households who have car loans, which is 24. Given that a random household from the sample does not have a mortgage, the probability that the household has a car loan is 24/33=8/11.

Write the first three terms of the sequence whose general term is an=1/n!(n−1)!.

$a_1=1,\ a_2=\frac{1}{2},\ a_3=\frac{1}{3}$a1​=1, a2​=12​, a3​=13​​ We substitute the values 1, 2, and 3 into the formula an=1n!(n−1)! to find ana1ana2ana3=1n!(n−1)!=(1−1)!1!=(0)!(1)!=11=1=1n!(n−1)!=(2−1)!2!=(1)!(2)!=12⋅1=12=1n!(n−1)!=(3−1)!3!=(2)!(3)!=2⋅13⋅2⋅1=13 The first three terms of the sequence are a1=-1, a2=1/2, and a3=1/3.

Write the first five terms of the sequence whose general term is an=6(n!).

$a_1=6,\ a_2=12,\ a_3=36,\ a_4=144,\ a_5=720$a1​=6, a2​=12, a3​=36, a4​=144, a5​=720​ We substitute the values 1, 2, 3, 4, and 5 into the formula an=6n! to find ana1ana2ana3ana4ana5=6n!=6(1)!=6⋅1=6=6n!=6(2)!=6⋅2⋅1=12=6n!=6(3)!=6⋅3⋅2⋅1=36=6n!=6(4)!=6⋅4⋅3⋅2⋅1=144=6n!=6(5)!=6⋅5⋅4⋅3⋅2⋅1=720 The first five terms of the sequence are a1=6, a2=12, a3=36, a4=144, and a5=720.

Write the first three terms of the sequence whose general term is an=(n+1)!4n.

$a_1=\frac{1}{2},\ a_2=\frac{3}{4},\ a_3=2$a1​=12​, a2​=34​, a3​=2​ We substitute the values 1, 2, and 3 into the formula an=(n+1)!4n to find ana1ana2ana3=(n+1)!4n=((1)+1)!4(1)=2⋅14=12=(n+1)!4n=((2)+1)!4(2)=3⋅2⋅18=34=(n+1)!4n=((3)+1)!4(3)=4⋅3⋅2⋅112=2 The first three terms of the sequence are a1=12, a2=34, and a3=2.

Write the first three terms of the sequence whose general term is an=(n−1)!(n+2)!.

$a_1=\frac{1}{6},\ a_2=\frac{1}{24},\ a_3=\frac{1}{60}$a1​=16​, a2​=124​, a3​=160​​ We substitute the values 1, 2, and 3 into the formula an=(n−1)!(n+2)! to find ana1ana2ana3=(n−1)!(n+2)!=(1−1)!(1+2)!=(0)!(3)!=13⋅2⋅1=16=(n−1)!(n+2)!=(2−1)!(2+2)!=(1)!(4)!=14⋅3⋅2⋅1=124=(n−1)!(n+2)!=(3−1)!(3+2)!=(2)!(5)!=2⋅15⋅4⋅3⋅2⋅1=160 The first three terms of the sequence are a1=16, a2=124, and a3=160.

Write the first three terms of the sequence whose general term is an=(n+1)!3n.

$a_1=\frac{2}{3},\ a_2=1,\ a_3=\frac{8}{3}$a1​=23​, a2​=1, a3​=83​​ We substitute the values 1, 2, and 3 into the formula an=(n+1)!3n to find ana1ana2ana3=(n+1)!3n=((1)+1)!3(1)=2⋅13=23=(n+1)!3n=((2)+1)!3(2)=3⋅2⋅16=1=(n+1)!3n=((3)+1)!3(3)=4⋅3⋅2⋅19=83 The first three terms of the sequence are a1=2/3, a2=1, and a3=8/3.

The probability that a debt holder has student loan debt, given that they also have credit card debt is 2242. If we know that 33 people in a sample of debt holders have student loan debt and 42 people have credit card debt, fill in the Venn diagram below with the number of debt holders to reflect this probability. Let Event A represent the people with student loans, and Event B represent the people with credit card debt.

1$$ 2$$ 3$$ 1$11$11​ 2$22$22​ 3$20$20​ We are given that P(A|B)=2242, so we know that there are 22 debt holders in A AND B, so we fill in 22 for Response 2. We are given that there are 33 people total who have student loans, so Response 1 is 33−22=11. We are given that there are 42 people total who have credit card debt, so Response 3 is 42−22=20.

The probability that a college graduate is offered a job at a particular company, given that they also completed an internship is 611. If we know that 21 college graduates and 11 previous interns (who did not attend college) were offered jobs with the company, fill in the Venn diagram below with the number of hired workers to reflect this probability. Let Event A represent the college graduates, and Event B represent the previous interns offered jobs.

1$$ 2$$ 3$$ 1$15$15​ 2$6$6​ 3$5$5​ We are given that P(A|B)=611, so we know that there are 6 hired employees in A AND B, so we fill in 6 for Response 2. We are given that there are 21 employees total who are college graduates, so Response 1 is 21−6=15. We are given that there are 11 employees total who are previous interns, so Response 3 is 11−6=5.

The probability that a debt holder has student loan debt, given that they also have credit card debt is 1130. If we know that 13 people in a sample of debt holders have student loan debt and 30 people have credit card debt, fill in the Venn diagram below with the number of debt holders to reflect this probability. Let Event A represent the people with student loans, and Event B represent the people with credit card debt.

1$$ 2$$ 3$$ 1$2$2​ 2$11$11​ 3$19$19​ We are given that P(A|B)=1130, so we know that there are 11 debt holders in A AND B, so we fill in 11 for Response 2. We are given that there are 13 people total who have student loans, so Response 1 is 13−11=2. We are given that there are 30 people total who have credit card debt, so Response 3 is 30−11=19.

The probability that a person uses public transit given that they also own a car is 814. If we know that 32 people in a sample use public transit and 14 people own cars, fill in the Venn diagram below with the number of people to reflect this probability. Let Event A represent the people who use public transit, and Event B represent the people who own cars.

1$$ 2$$ 3$$ 1$24$24​ 2$8$8​ 3$6$6​ We are given that P(A|B)=814, so we know that there are 8 people in A AND B, so we fill in 8 for Response 2. We are given that there are 32 people total who use public transit, so Response 1 is 32−8=24. We are given that there are 14 people total who own cars, so Response 3 is 14−8=6.

Alyssa likes to play roulette, but she doesn't like the low probability of betting on a single number. Therefore, she bets on a block of 4 numbers, increasing her probability of winning to 438. She generally places a $5chip on her block of 4. If one her 4 numbers come up, she wins $40 (and gets to keep her bet!). What is the expected value for Alyssa playing roulette? Round to the nearest cent. Do not round until your final calculation.

1$-0.26$−0.26​ Recall, the expected value is found by multiplying the value by the probability of the event. Here, Alyssa either wins $40 or loses $5. E(x)=40⋅438−5⋅3438 E(x)=4.21−4.47 E(x)=−0.26 Alyssa can expect to lose $0.26 each time she plays roulette.

Darryl plays a card game with his sister. He uses a 52 card deck and she chooses 1 card. If she chooses the 4 of hearts, she will win $10 and if she chooses any spade, she loses $2. If any other card is drawn, they break even. What is the expected value for Darryl's sister? Round to the nearest cent. Do not round until your final calculation. Note: within a deck of cards, there are 4 equal suits: hearts, clubs, diamonds, and spades.

1$-0.31$−0.31​ Recall, the expected value is found by multiplying the value and the probability for that event. Here, Darryl's sister can win $10, lose $2, or break even. E(x)=10⋅152−2⋅1352+0⋅3852 E(x)=1052−2652+0 E(x)=−1652=−0.30769...=0.31 She can expect to lose $0.31 each time she chooses a card.

Jim loves to play roulette! He bets $10 on a single number, usually the number 13. He has a 138 chance of winning. If the ball lands on his number, he wins $350 (and gets to keep his $10 bet!). If it lands on any other number, a 3738 probability, he loses his bet. What is the expected value of a game of roulette? Round to the nearest cent. Do not round until your final answer.

1$-0.53$−0.53​ Recall, expected value is found my multiplying the value by the probability of that event. Here, you are either winning $350 or losing $10. E(x)=350⋅138−10⋅3738 E(x)=9.21−9.74 E(x)=−0.53 You can expect to lose $0.53 each time you play roulette.

A scratch off lottery ticket can be purchased for $2. There are multiple prize options, shown along with the probability, in the table below. What is the expected win (or loss) for this lottery ticket? Round your answer to the nearest cent. Do not round until your final answer. WinProbability$5,0000.000125$2500.000375$750.002$100.0025$20.0200.975

1$-1.07$−1.07​ Recall, to find expected value, we multiply the value by the probability of that event. Also keep in mind, even though you may win a particular amount, it does cost you $2 to play. Therefore, we need to subtract 2 from the winning amount to accurately find the expected value. E(x)=4998⋅0.000125+248⋅0.000375+73⋅0.002+8⋅0.0025+0⋅0.02−2⋅0.975 E(x)=0.62475+0.093+0.146+0.02+0−1.95 E(x)=−1.06625=−1.07 You can expect to lose $1.07 each time you purchase one of these scratch off lottery tickets.

"Marble Madness" is a local carnival game, costing $2. There are 100 total marbles in a bag: 2 red, 8 orange, 10 yellow, 30 green, 30 blue, and 20 black. If a red marble is pulled, you win $6, an orange marble wins $4, and a yellow marble wins $2. A green, blue, and black marble result in a loss of the $2 cost to play. What is the expected value of a marble pull?

1$-1.36$−1.36​ Recall, expected value is found by multiplying the value times the probability of the event. E(x)=4⋅2100+2⋅8100+0⋅10100−2⋅80100 E(x)=0.08+0.16+0−1.6 E(x)=−1.36 You can expect to lose $1.36 each time you pull a marble.

A lottery ticket can be purchased for $5. There are multiple prize options, shown along with the number of prizes, in the table below. What is the expected win (or loss) for this lottery ticket? Round your answer to the nearest cent. Do not round until your final answer. PrizeNumber of Prizes$50,0005$5,000155010055,000Lose94,880

1$-1.45$−1.45​ Recall that expected value is found by multiplying the value times the probability of that event. Remember, it costs $5 to play, so subtract 5 from the win to determine the actual amount won by the lottery ticket winner. E(x)=49,995⋅5100,000+4,995⋅15100,000+45⋅100100,000+0⋅5,000100,000−5⋅94,880100,000 E(x)=2.49975+0.74925+0.045+0−4.744 E(x)=−$1.4 You can expect to lose $1.4 for each lottery ticket you purchase.

A community festival sells 20,000 raffle tickets for $10 each. There are 4 possible prize levels. The grand prize is a golf cart valued at $6,000. There are two winners of $1,000, three winners of $500, and ten winners of $100. What is the expected value of a raffle ticket? Round to the nearest cent. Do not round until the final calculation

1$-9.48$−9.48​ Recall, expected value is calculated by multiplying the value by the probability of the event. Remember, you pay $10 for the raffle ticket, so you must subtract 10 from the 'win' in order to find the expected value. E(x)=5990⋅120,000+990⋅220,000+490⋅320,000+90⋅1020,000−10⋅19,98420,000 E(x)=0.2995+0.099+0.0735+0.045−9.992 E(x)=−9.475=−9.48

A community festival sells 20,000 raffle tickets for $10 each. There are 4 possible prize levels. The grand prize is a golf cart valued at $6,000. There are two winners of $1,000, three winners of $500, and ten winners of $100. What is the expected value of a raffle ticket? Round to the nearest cent. Do not round until the final calculation.

1$-9.48$−9.48​ Recall, expected value is calculated by multiplying the value by the probability of the event. Remember, you pay $10 for the raffle ticket, so you must subtract 10 from the 'win' in order to find the expected value. E(x)=5990⋅120,000+990⋅220,000+490⋅320,000+90⋅1020,000−10⋅19,98420,000 E(x)=0.2995+0.099+0.0735+0.045−9.992 E(x)=−9.475=−9.48

A high school club is selling 15,000 raffle tickets at $10 each for a fundraiser. The table below shows the items given away, along with the value of the item and how many are available. Based on this information, what is the expected value of this raffle? ItemValueNumber AvailableTrip for 2$100014 tickets to a football game$50024 tickets to a baseball game$3002Full year sports pass to the high school$15010 Round to the nearest cent. Do not round until the final calculation.

1$-9.73$−9.73​ Recall, expected value can be found by multiplying the value by the probability for the event. Here, remember that you paid $10 for a raffle ticket. Therefore, you must subtract 10 from the value before finding the expected value. E(x)=990⋅115,000+490⋅215,000+290⋅215,000+140⋅1015,000−10⋅14,98515,000 E(x)=0.066+0.0653+0.0387+0.0933−9.99 E(x)=−9.7267=−9.73 You can expect to lose $9.73 for each raffle ticket.

Janelle is playing craps and places a $5 bet on the number 8 with a probability of winning at 536. However, the probability of losing is 16 and happens if a 7 is rolled. Any other number rolled means she doesn't win or lose and the game continues. If an 8 is rolled, she will win $6 (and keep her initial bet) but if a 7 is rolled, she loses all $5. What is the expected value for Janelle on a roll in this game of craps? Round to the nearest cent. Do not round until the final calculation.

1$0$0​ Recall, the expected value can be calculated by multiplying the value by the probability of that event. Here, Janelle can win $6 or lose $5. E(x)=6⋅5/36−5⋅6/36 E(x)=30/36−30/36 E(x)=0 Therefore, Janelle can expect to break even on each roll during a game of craps, if she only bets on the number 8.

Jessie is playing craps and bets $5 on the come out roll. If a 7 or 11 is rolled, she wins $5. This happens with a probability of 29. If a 2,3,or12 is rolled, she loses her $5. This has a probability of 19. If any other number is rolled, she does not win or lose, and the game continues. Find the expected value for Jessie on the come out roll. Round to the nearest cent. Do not round until the final calculation.

1$0.56$0.56​ Recall, expected value can be calculated by multiplying the value by the probability of that event. Here, Jessie can win $5, lose $5, or break even. E(x)=5⋅29−5⋅19+0⋅23 E(x)=109−59+0 E(x)=59=0.55555555...=0.56 Jessie can expect to win $0.56 on each come out roll.

The following incomplete contingency table shows a random sample of 120 employees and the routes they prefer. Complete the table. Correct! You nailed it. WalkBusTotalNew Employee$$18$$67$$85Seasoned Employee$$53035Total2397120

1$18$18​ 2$67$67​ 3$85$85​ 4$5$5​ Note that from the Bus column, we have that A+30=97⟹A=67 Note that from the Total column, we have that B+35=120⟹B=85 Plugging this in the New Employee row, we have that 67+C=85⟹C=18 Plugging this in the Walk column, we have that 18+D=23⟹D=5 If you complete the rest of the table, you should get WalkBusTotalNew Employee186785Seasoned Employee53035Total2397120

Fill in the following contingency table and find the number of students who both read mysteries AND read comics.

By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the row of totals in the table, we know that the unknown number plus 44 is 101, so the missing number must be 57. Continuing in this way, we can fill in the entire table: StudentsreadmysteriesdonotreadmysteriesTotalreadcomics133144donotreadcomics193857Total3269101 From this, we can see that the number of students who both read mysteries and read comics is 13.

Fill in the following contingency table and find the number of students who both play sports AND play an instrument. StudentsplaysportsdonotplaysportsTotalplayaninstrument67donotplayaninstrument10Total5799

By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the row of totals in the table, we know that the unknown number plus 67 is 99, so the missing number must be 32. Continuing in this way, we can fill in the entire table: StudentsplaysportsdonotplaysportsTotalplayaninstrument353267donotplayaninstrument221032Total574299 From this, we can see that the number of students who both play sports and play an instrument is 35.

A university is randomly assigning its 150 first-year students to writing classes or study skills seminars. 50 of the students will be randomly selected to take the writing seminar the first semester, and 15 of the students will be randomly assigned to the study skills seminar for the second semester. Only 5 students will be allowed to take both seminars. What is the probability that a student will be assigned to the study skills seminar given that they were chosen to participate in the writing seminar?

Let A be the event that the student is randomly selected to take the writing seminar, and note that P(A)=50150=13. Let B be the event that the student is selected to take the study skills seminar, and note that P(B)=15150=110. P(A AND B), the probability that a student takes both the writing and the study skills seminars, is 5150=130. Now to calculate P(B|A), the probability that a student will be assigned to the study skills seminar given that they were chosen to participate in the writing seminar: P(B|A)=P(B AND A)P(A)=13013=330=110 Conditional probabilities may also deal with changes to the sample space. For example, if we are interested in the probability of drawing two diamonds in a row from a standard deck of playing cards, we have to consider that drawing the first card reduces the number of cards in the pile from 52 to 51. Problems will designate "with replacement" or "without replacement" to help you identify what is happening to the sample space as cards are being drawn. If events (cards being drawn from a deck, marbles being chosen from a bag) are happening with replacement, an item from the population is chosen and then replaced before a second item is chosen. Here, the sample size does not change. If events are happening without replacement, the first item from the population that is chosen is not replaced before a second item is chosen. Here, the sample size changes.

The probability that a randomly chosen college student smokes is 0.2. The probability that a randomly chosen college student is an athlete is 0.15. The probability that a randomly chosen college student smokes given that the student is an athlete is 0.1. What is the probability that a randomly chosen college student is a smoker and an athlete?

Let S be the event that a randomly chosen student smokes. Let A be the event that a randomly chosen student is an athlete. We are told P(S)=0.2, P(A)=0.15, and P(S|A)=0.1. We are asked to find P(S AND A). Using what we are told and the formula for the compound probability above, we find that P(S AND A)=P(S|A)P(A)=(0.1)⋅(0.15)=0.015 So the probability that a randomly chosen student is both an athlete and a smoker is0.015. Note that we did not use the information about P(S). Sometimes you may be given extraneous information in problems, so you need to be careful to apply the formula correctly.

A nursing student is planning his schedule for next quarter. He needs to take three courses, and one must be Statistics. For his other courses, he can choose one of two science classes and one of three social sciences classes. MathScience Social Science Statistics Anatomy Biology Psychology Economics Sociology All of the possibilities for his schedule are shown in the tree diagram below: A tree diagram is shown. The top row reads "Statistics" and has two arrows pointing from it to the two items in the middle row, Anatomy, and Biology. From each of the two items in the middle row there are three arrows pointing to Psychology, Economics, and Sociology in the bottom row. What is the probability that a randomly chosen schedule has statistics, anatomy, and economics classes?

$$1/6​ Answer Explanation Correct answers:$\frac{1}{6}$16​​ Let the desired event, a statistics, anatomy, and economics schedule, be SAE. There is 1 possible outcome that matches this event. There are 6 possible outcomes (1 math × 2 science × 3 social sciences =6). So the probability of the schedule having statistics, anatomy, and economics is: P(SAE)=1/6

A couple has two children. Let A be the event that their first child is a boy, and note that P(A)=51.2%. Let B be the event that their second child is a girl, with P(B)=48.8%. A and B are independent events. What is the probability that the couple has a first child that is a boy and a second child that is a girl? Give your answer as a decimal rounded to two decimal places.

$0.25$0.25​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=51.2%⋅48.8%=0.512⋅0.488≈0.25

Write the first three terms of the sequence whose general term is an=n!n3.

$a_1=1,\ a_2=\frac{1}{4},\ a_3=\frac{2}{9}$a1​=1, a2​=14​, a3​=29​​ We substitute the values 1, 2, and 3 into the formula an=n!n3 to find ana1ana2ana3=n!n3=1!13=11=1=n!n3=2!23=2⋅18=14=n!n3=3!33=3⋅2⋅127=29 The first three terms of the sequence are a1=1, a2=14, and a3=29.

Write the first three terms of the sequence whose general term is an=(2n)!.

$a_1=2,\ a_2=24,\ a_3=720$a1​=2, a2​=24, a3​=720​ We substitute the values 1, 2, and 3 into the formula an=(2n)! to find ana1ana2ana3=(2n)!=(2(1))!=2!=2⋅1=2=(2n)!=(2(2))!=4!=4⋅3⋅2⋅1=24=(2n)!=(2(3))!=6!=6⋅5⋅4⋅3⋅2⋅1=720 The first three terms of the sequence are a1=2, a2=24, and a3=720.

Write the first five terms of the sequence whose general term is an=4n!.

$a_1=4,\ a_2=2,\ a_3=\frac{2}{3},\ a_4=\frac{1}{6},\ a_5=\frac{1}{30}$a1​=4, a2​=2, a3​=23​, a4​=16​, a5​=130​​ We substitute the values 1, 2, 3, 4, and 5 into the formula an=4n! to find ana1ana2ana3ana4ana5=4n!=41!=41=4=4n!=42!=42⋅1=2=4n!=43!=43⋅2⋅1=23=4n!=44!=44⋅3⋅2⋅1=16=4n!=45!=45⋅4⋅3⋅2⋅1=130 The first five terms of the sequence are a1=4, a2=2, a3=23, a4=16, and a5=130.

Find the expected value of the experiment shown in the table below. OutcomeProbability−$513200110$5320$15120$50120

1$0.75$0.75​ Recall, expected value can be found by multiplying the outcome by the probability. E(x)=−5⋅1320+0⋅110+5⋅320+15⋅120+50⋅120 E(x)=−3.25+0+0.75+0.75+2.5=0.75 You can expect to win $0.75 within this experiment.

Let R be the event that a randomly chosen person has visited Rome, Italy. Let G be the event that a randomly chosen person has visited Greece. Place the correct event in each response box below to show: Given that the person has visited Rome, Italy, the probability that a randomly chosen person has visited Greece.

1$G$G​ 2$R$R​ Remember that in general, P(A|B) is read as "The probability of A given B". Here we want to know the probability that a person has visited Greece given that the person has visited Rome, Italy, so the correct answer is P(G|R).

Tree diagram: a special type of graph, consisting of "branches" that are labeled with either frequencies or probabilities, used to determine the outcomes of an experiment and to make some probability problems easier to visualize and solve. Relative Frequency Method: To calculate the probability of a specific outcome, we count the number of times the event occurs and compare it to the total number of events . P(A)=number of times event A occurred/total number of events The Fundamental Counting Principle, also known as the Counting Principle, a method for determining the total number of possible outcomes by multiplying together the numbers of each possible event.

Tree diagram: a special type of graph, consisting of "branches" that are labeled with either frequencies or probabilities, used to determine the outcomes of an experiment and to make some probability problems easier to visualize and solve. Relative Frequency Method: To calculate the probability of a specific outcome, we count the number of times the event occurs and compare it to the total number of events . P(A)=number of times event A occurred/total number of events The Fundamental Counting Principle, also known as the Counting Principle, a method for determining the total number of possible outcomes by multiplying together the numbers of each possible event.

The Fundamental Counting Principle Suppose a first process has a possible outcomes and a second process has b possible outcomes. Then the total number of different outcomes is a×b. This extends to any number of outcomes. Adding a third process with c possible outcomes brings the total number to a×b×c.

The Fundamental Counting Principle Suppose a first process has a possible outcomes and a second process has b possible outcomes. Then the total number of different outcomes is a×b. This extends to any number of outcomes. Adding a third process with c possible outcomes brings the total number to a×b×c.

A local pizza parlor claims that they have over 100 pizza options, and advertise, "A pizza for everyone!" A group of friends visit the pizza parlor to check out their menu. They find: SizeSauceCheeseToppingsSmallMediumLargeGarlicMarinaraMozzarellaProvoloneParmesanPepperoniSausageHamPineappleOnionsPeppers To make a pizza, each customer must select one option from each menu column. Are there over 100 pizza possibilities? What is the probability that a randomly selected pizza by the group of friends will have pepperoni on it?

Yes. We use the Fundamental Counting Principle here, rather than draw a tree diagram with over 100 branches. Size×Sauce×Cheese×Toppings=3×2×3×6=108 Notice that pepperoni is 1 of 6 available toppings. If we were to draw a tree diagram, one out of every six of our final row of branches would be pepperoni. Therefore, the probability of pepperoni is 16.

A fair six-sided die is rolled, and a random variable Y represents the score. What is the name of the distribution of Y?

Discrete uniform distribution. It is discrete uniform distribution, characterized by its symmetry and the uniformity by which its probabilities are distributed.A finite number of values are likely to be observed, in the case of a die, {1,2,3,4,5,6}.Every one of n values has probability 1/n. In the case of a die, 1/6.

The probability that a product passes quality-control inspections is 0.88. The probability that it passes quality-control inspections and is packaged for shipment immediately is 0.03. What is the probability that it is packaged for shipment immediately given that it passed quality-control inspections? Round your answer to three decimal places.

$0.034$0.034​ The probability that a product passes quality-control inspections is 0.88. The probability that it is packaged for shipment immediately and passes quality-control inspections is 0.03. What is the probability that it is packaged for shipment immediately given that it passed quality-control inspections? Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) Rearranging, we see thatP(B|A)=P(B AND A)/P(A) So if we think of A as the product passing quality-control inspections and B as the event that the product is packaged to ship immediately, then we can plug in the known information to find P(B|A)=0.030.88≈0.034

Suppose A and B are mutually exclusive events, and that P(A)=0.09 and P(B)=0.03. Find P(A OR B).

$0.12$0.12​ Remember that when A and B are mutually exclusive events, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) So we find thatP(A OR B)=P(A)+P(B)=0.09+0.03=0.12

Suppose A and B are mutually exclusive events, and that P(B)=0.48 and P(A OR B)=0.98. Find P(A).

$0.5$0.5​ Remember that when A and B are mutually exclusive events, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find thatP(A)=P(A OR B)−P(B)So plugging in the given values, we findP(A)=0.98−0.48=0.5

At a certain school, intro to economics and intro to calculus meet at the same time, so it is impossible for a student take both classes. If the probability that a student takes intro to economics is 0.57, and the probability that a student takes intro to calculus 0.17, what is the probability that a student takes intro to economics or into to calculus?

$0.74$0.74​ Because it is impossible for a student to take both intro to economics and intro to calculus, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) So we find thatP(A OR B)=P(A)+P(B)=0.57+0.17=0.74 The probability that a student takes into to economics or intro to calculus is the sum of the individual probabilities, which is 0.74.

Sam is able to choose from two job positions at one company. He has the option to choose the day shift or the night shift, so it is impossible for Sam to work the day and night shift. If the probability that Sam chooses the day shift is 0.9, and the probability that Sam chooses the night shift is 0.03, what is the probability that Sam chooses the day or night shift?

$0.93$0.93​ Because it is impossible for Sam to work the day and night shift, we see that they are mutually exclusive events. Therefore, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) So we find thatP(A OR B)=P(A)+P(B)=0.9+0.03=0.93 The probability that Sam choose the day or night shift is 0.93.

Fill in the following contingency table and find the number of students who both watch comedies AND watch dramas. StudentswatchcomediesdonotwatchcomediesTotalwatchdramas34donotwatchdramas10Total3967

$16$16​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the row of totals in the table, we know that the unknown number plus 34 is 67, so the missing number must be 33. Continuing in this way, we can fill in the entire table: StudentswatchcomediesdonotwatchcomediesTotalwatchdramas161834donotwatchdramas231033Total392867 From this, we can see that the number of students who both watch comedies and watch dramas is 16.

A company is planning its spring advertising campaign. Its marketing team is planning to mail out different types of advertisements, at random, to their town's residents each week. They have the following types of advertisements ready: Week 1Week 2Week 3Small PosterBrochureLeafletPostcardBrochureBusiness CardPostcardSmall Poster How many different possible ways can a resident receive an advertisement?

$18$18​ Week 1Week 2Week 3Small PosterBrochureLeafletPostcardBrochureBusiness CardPostcardSmall Poster We could draw a tree diagram here to see all the possible outcomes, or we can use the Fundamental Counting Principle and quickly calculate the number of outcomes. The Fundamental Counting Principle tells us that we multiple the number of choices each week by one another to get the total number of possibilities. Possibilities=Week 1×Week 2×Week 3 Possibilities=2×3×3=18

A random variable X has the following probability distribution: Outcome123Probability0.20.40.4 Calculate E(X), the expected value of X.

$2.2$2.2​ Applying the formula, we have E(X)=1⋅0.2+2⋅0.4+3⋅0.4=2.2.

Fill in the following contingency table and find the number of students who both do not read mysteries AND do not read comics.

$24$24​ By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the first row in the table, we know that 13 added to the unknown number in the middle is 42, so that unknown number is 29. Continuing in this way, we can fill in the entire table: StudentsreadmysteriesdonotreadmysteriesTotalreadcomics131831donotreadcomics292453Total424284 From this, we can see that the number of students who both do not read mysteries and do not read comics is 24.

Given that P(B AND A)=0.03 and P(A)=0.11, what is P(B|A)? Give your answer as a percent. Round your answer to two decimal places.

$27.27\ \%$27.27 %​ Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) Rearranging, we see thatP(B|A)=P(B AND A)P(A)Now, plugging in the values we were given, we find thatP(B|A)=0.03/0.11≈0.2727 To rewrite this decimal as a percent, we multiply by 100: 0.2727×100=27.27%.

The results of an experiment are shown below. Calculate E, the expected value of the experiment. Write your answer as a decimal. Outcome2468Probability0.30.30.250.15

$4.5$4.5​ Using the definition for E we have E=2⋅0.3+4⋅0.3+6⋅0.25+8⋅0.15=4.5.

A college professor assigns a class a research project and gives guidelines on which sources are acceptable to use for the research. Students are expected to pick one resource from each of the following categories: News Scholarly Journal Magazine University Research Times Daily News Review Science Quarterly Psychology Report Sociology Digest Economics Journal News for the week Times and Errors Professor Smith's Research Professor Rubenstein's Research How many different possible groups of resources are there?

$48$48​ The Fundamental Counting Principle tells us that we multiple the number of choices in each category by one another to get the total number of possibilities. Possibilities=3 News×4 Scholarly Journals×2 Magazines×2 University Research=48

Let S={1,2,3,4,5,6,7,8} be a sample space with P(x)=k2x where x is a member of S, and k is a positive constant. Compute E(S). Round your answer to the nearest hundredths.

$5.67$5.67​ First we must compute k. By summing the probabilities and equating to one we have k2+2k2+3k2+...+8k2=1⟹36k2=1⟹k=16. So to compute E(S) we haveE(S)=1⋅136+2⋅236+3⋅336+4⋅436+5⋅536+6⋅636+7⋅736+8⋅836=5.666...which is 5.67 to 2d.p.

The events of a patient at a doctor's office having the flu and a patient at a doctor's office being a woman are independent. Let A be the event that a patient at a doctor's office has the flu with P(A)=0.09. Let B be the event that a patient at a doctor's office is a woman with P(B)=0.65. What is the probability that a patient at a doctor's office has the flu and is a woman? Give your answer as a percent rounded to two decimal places if necessary.

$5.85\%$5.85%​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=0.09⋅0.65=0.0585 Then, to express this answer as a percent, we multiply by 100: 0.0585×100=5.85%

Let S={1,4,8,16,32,64} be a sample space. If P(1)=132 and P(2k)=21−k for 2≤k≤6, find the expected value of the event E={1,8,32,64}. Give your answer as a fraction in its simplest form.

$\frac{193}{32}$193/32​​ We know that P(1)=132,P(8)=14,P(32)=116,P(64)=132, soE(E)=1⋅1/32+8⋅1/4+32⋅1/16+64⋅1/32=193/32.

A probability distribution for a random variable Y is given by P(Y=y)=cy, for y=1,2,4,5 and c is a constant. Find E(Y), the expected value of Y. Give your answer as a fraction in its simplest form.

$\frac{23}{6}$236​​ The first thing we must do is calculate c. This can be done by noting that all of the probabilities must add up to 1. Therefore c+2c+4c+5c=1, which solves to givec=112.Using this, we can now say thatP(Y=1)P(Y=2)P(Y=4)P(Y=5)=112,=212,=412,=512.We can now compute E(Y), thereby givingE(Y)=1⋅112+2⋅212+4⋅412+5⋅512=236.

The real estate manager of a commercial building needs to fill three empty spaces. The options for each store are: All of the possible options for the three stores are shown in the tree diagram below. What is the probability that the group of stores will have a restaurant or a hardware store, or both? Enter your answer as a fraction.

$\frac{2}{3}$2/3​​ There are 4 possible outcomes that include a restaurant, a hardware store, or both. There are 6 possible store combinations (Possibilities = 1 grocery × 3 for the second store × 2 for the third store = 6). So the probability of having a group of stores with a restaurant or a hardware store is: P(restaurant or a hardware store)=4/6=2/3

Use the completed contingency table to find the probability that an employee prefers the bus given that they are a seasoned employee. Express your answer as a fraction.

$\frac{30}{35}$30/35​​ Because our given information is that the employee is a seasoned employee we want to focus on that row of the table. In that row, there are 35 employees total, and 30 of them prefer riding the bus. WalkBusTotalNew Employee186785Seasoned Employee53035Total2397120 Therefore, P(Bus|Seasoned Employee)=30/35.

If you roll a fair die and then flip a fair coin, what is the probability that you roll greater than 1 and get heads with the coin?

$\frac{5}{12}$512​​ Note that these are independent events because the outcome when you roll a fair die does not affect the outcome when you flip a fair coin. So by the multiplication rule for independent events, we can take the probability of each event and multiply them.The probability that you roll greater than 1 is 56, and the probability that you get heads with the coin is 12 so the probability that you roll greater than 1 and get heads with the coin is 5/6⋅1/2=5/12

The probability that a car has a certain factory defect is 825. The probability that a car has a certain factory defect and needs an oil change is 750. What is the probability that a car needs an oil change given that it has a certain factory defect? Give your answer as a fraction in simplest form.

$\frac{7}{16}$7/16​​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging, we see thatP(A|B)=P(A AND B)P(B) So if we think of A as being the event that a car has a factory defect and B as being the event that the car needs an oil change, then we can plug in the known information to find P(A|B)=750825=716

A group of 155 seniors at a local university were asked if they took an economics class and if they had an internship. The following table shows the results. Economics ClassNo Economics ClassTotalInternship306898No Internship164157Total46109155 According to the table, what is the probability that a randomly chosen senior does not have an internship given that they took an economics class? Give your answer as a fraction in simplest form

$\frac{8}{23}$823​​ Because our given information is that the student took an economics class, we want to focus on that column of the table. Economics ClassNo Economics ClassTotalInternship306898No Internship164157Total46109155 In that row, there are 46 participants total, and 16 of them did not have an internship, therefore the probability is 1646=823

Write the first three terms of the sequence whose general term is an=n!n2.

$a_1=1,\ a_2=\frac{1}{2},\ a_3=\frac{2}{3}$a1​=1, a2​=12​, a3​=23​​ We substitute the values 1, 2, and 3 into the formula an=n!n2 to find ana1ana2ana3=n!n2=1!12=11=1=n!n2=2!22=2⋅14=12=n!n2=3!32=3⋅2⋅19=23 The first three terms of the sequence are a1=1, a2=1/2, and a3=2/3.

Write the first three terms of the sequence whose general term is an=(2n)!.

$a_1=2,\ a_2=24,\ a_3=720$a1​=2, a2​=24, a3​=720​ We substitute the values 1, 2, and 3 into the formula an=(2n)! to find ana1=(2n)!=(2(1))!=2!=2⋅1=2 ana2=(2n)!=(2(2))!=4!=4⋅3⋅2⋅1=24 ana3=(2n)!=(2(3))!=6!=6⋅5⋅4⋅3⋅2⋅1=720 The first three terms of the sequence are a1=2, a2=24, a3=720.

Your brother has a bag of marbles containing 1 red marble, 2 blue marbles, and 47 green marbles. He claims he'll give you $10 if you pull the red marble but just $5 if you pull a blue marble. However, if you pull a green marble, you will owe him $3. What is the expected value each time you pull a marble? Round to the nearest cent. Do not round until your final answer.

1$-2.42$−2.42​ Recall, expected value is found by multiplying the value times the probability of that event. Note, there are 50 total marbles. Pulling a red marble has a probability of 150 . Pulling a blue marble has a probability of 250. Pulling a green marble has a probability of 4750. Knowing the probabilities, we can find the expected value. E(x)=10⋅150+5⋅250−3⋅4750 E(x)=0.2+0.2−2.82=−2.42 You can expect to lose $2.42 each time you pull a marble.

A bicycle shop sells 10,000 raffle tickets at $5 each, raising money for a community park. The table below describes the prizes, value of each prize, and how many are being given away. PrizeValueNumber to be Given AwayBicycle and Helmet$18020Bicycle$15015Helmet$3010 What is the expected value of the raffle? Round to the nearest cent. Do not round until the final calculation.

1$-4.39$−4.39​ Recall, expected value is found by multiplying the value by the probability of the event. Remember, in this example, you paid $5 for the raffle ticket, so you must subtract 5 from the 'value' in order to find the expected value. E(x)=175⋅2010,000+145⋅1510,000+25⋅1010,000−5⋅9,95510,000 E(x)=0.35+0.2175+0.025−4.9775 E(x)=−4.385−−4.39 You can expect to lose $4.39 for each raffle ticket purchased.

Find the expected value of the experiment shown in the table below. OutcomeProbability−20.4300.0620.1840.2460.09

1$1$1​ Recall, expected value can be found by multiplying the outcome by the probability. E(x)=−2⋅0.43+0⋅0.06+2⋅0.18+4⋅0.24+6⋅0.09 E(x)=−0.86+0+0.36+0.96+0.54 E(x)=1 The expected value is 1 for this experiment.

How To Addition Rule for Probabilities If A and B are events defined on a sample space, then P(A OR B)=P(A)+P(B)−P(A AND B)

How To Addition Rule for Probabilities If A and B are events defined on a sample space, then P(A OR B)=P(A)+P(B)−P(A AND B)

A company is planning its spring advertising campaign. Its marketing team is planning to mail out different types of advertisements, at random, to their town's residents each week. They have the following types of advertisements ready: How many different possible ways can a resident receive an advertisement?

We could draw a tree diagram here to see all the possible outcomes, or we can use the Fundamental Counting Principle and quickly calculate the number of outcomes. The Fundamental Counting Principle tells us that we multiple the number of choices each week by one another to get the total number of possibilities. Possibilities=Week 1×Week 2×Week 3 Possibilities=2×3×3=18

We can extend this thinking to include another event. If we toss a second coin, how many outcomes do you think we will have?

We have a total of 6×2×2=24 possible outcomes. 1HH, 1HT, 1TH, 1TT, 2HH, 2HT, 2TH, 2TT, 3HH, 3HT, 3TH 3TT, 4HH, 4HT, 4TH, 4TT, 5HH, 5HT, 5TH, 5TT, 6HH, 6HT, 6TH, and 6TT. This is the basis of the Counting Rule detailed below. Sometimes, there may be too many possible outcomes to create a tree diagram. Say, for example, there were 20 different toppings for the brothers to choose from at the ice cream parlor. Then, there would be 120 different possibilities for sundaes. Drawing a tree diagram for that many possibilities would be very tedious. To avoid this, we can utilize something called the Fundamental Counting Principle.

Two friends are both pregnant, and find out they are each expecting twins! Let A be the event that one friend is pregnant with identical twins, and note that P(A)=0.0045. Let B be the event that the other friend is pregnant with fraternal twins, and note that P(B)=0.01. A and B are independent events. What is the probability that one friend is pregnant with identical twins, and one friend is pregnant with fraternal twins? Give your answer as a percent, rounded to four decimal places if necessary.

$0.0045\%$0.0045%​ Since we know that A and B are independent events, we can use the multiplication rule to find: P(A AND B)=P(A)P(B)=0.0045⋅0.01=0.000045 To express our answer as a percent, we multiply by 100: 0.000045×100=0.0045%

Given that P(B|A)=0.84 and P(A)=0.43, what is P(B AND A)?

$0.3612$0.3612​ Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) So plugging in the values that we know, we findP(B AND A)=(0.84)(0.43)=0.3612

If A and B are events with P(A)=0.4, P(A OR B)=0.89, P(A AND B)=0.01, find P(B).

$0.5$0.5​ First, it is helpful to write down the addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, rearranging this, we find thatP(B)=P(A OR B)+P(A AND B)−P(A)Plugging in the known values, we findP(B)=0.89+0.01−0.4=0.5

Given that a student takes algebra, what is the probability that the student does not take chemistry? Give your answer as a fraction. You may reduce it if you want, but it is not necessary.

$\frac{40}{50}$

A track team contains sprinters numbered 1,2,3,4,5 and distance runners numbered 1,2. Let S be the event of selecting a sprinter, D the event of selecting a distance runner, E the event of selecting an even numbered runner, and O the event of selecting an odd runner.Selecting the sprinter numbered 1 is one outcome of which of the following events? Select all correct answers.

D OR O E′ Because the runner is a sprinter and the number is odd, the runner is an example of S and O. Therefore, it is also an example of E′ (not even) and D OR O.

A group of friends take classes at a university in biology, chemistry, and physics. If you selected a friend at random, the probabilities that they take certain classes is given below: The probability that they take biology class is P(B)=13; The probability that the friend takes chemistry is P(C)=12; A student takes physics with the probability P(P)=16; A student takes both biology and physics with a probability of 112. What's the probability that a randomly chosen friend takes biology or physics, P(B OR P)?

Without a diagram, or information about how many friends there are, we have to use the Addition Rule for Probabilities to P(B OR P): P(B OR P)=P(B)+P(P)−P(B AND P)=13+16−112=512 So, the probability that a randomly chosen friend takes biology or physics is 512.

A student is mapping out their options for their post-secondary education. They list out all of their considerations. CollegeMaster's Degree Further Education Public University Private University Master's of Science Master's of Business Administration Financial Exams Ph. D post-baccalaureate studies How many possible combinations of college, master's degree, and further education are there?

$12$12​ The Fundamental Counting Principle tells us that we multiple the number of choices in each section by one another to get the total number of possibilities. Possibilities=2×2×3=12

A random number X has the following probability distribution: x01510yP(X=x)0.20.20.10.30.2 If E(X)=10, calculate the value of y.

$31.5$31.5​ The formula for E(X) gives us 10=0⋅0.2+1⋅0.2+5⋅0.1+10⋅0.3+y⋅0.2, which on simplification gives us10=3.7+0.2y,and solving this for y gives y=31.5.

Let S={1,2,3,4} be a sample space with P(x)=kx where x is a member of {1,3}, and P(x)=k(x−1) where x is a member of {2,4}, where k is a constant. Find E(S).

$3$3​ First we find k by summing the probabilities and equating to one, i.e. k+(2−1)k+3k+(4−1)k=1, which gives k=18. HenceE(S)=1⋅18+2⋅18+3⋅38+4⋅38=3.

A probability distribution for a random variable X is given by P(X=x)=cx, for x=1,2,3,4 and c is a constant. Find E(X), the expected value of X.

$3$3​ The first thing we must do is calculate c. This can be done by noting that all of the probabilities must add up to 1. Therefore c+2c+3c+4c=1, which solves to give c=0.1. Using this, we can now say thatP(X=1)=0.1,P(X=2)=0.2,P(X=3)=0.3,P(X=4)=0.4.We can now compute E(X), thereby givingE(X)=1⋅0.1+2⋅0.2+3⋅0.3+4⋅0.4=3.

Consider a set S={1,2,3,4,5} and P(x)=x−110, where x is a member of S. Calculate E(S).

$4$4​ Multiply each event by the probability of then event and then add up the products to find E(S). E(S)E(S)=1P(1)+2P(2)+3P(3)+4P(4)+5P(5)=1⋅0+2⋅110+3⋅210+4⋅310+5⋅410=4.

If A and B are independent events with P(A)=0.60 and P(A AND B)=0.30, find P(B). Give your answer as a percent, rounded to two decimal places if necessary.

$50\%$50%​ Remember that for independent events, P(A AND B)=P(A)P(B) Solving for P(B), we find thatP(B)=P(A AND B)P(A)Now, plugging in the values we are given, we find thatP(B)=0.300.60=0.50 To express this answer as a percent, we multiply by 100: 0.50×100=50%

The real estate manager of a commercial building needs to fill three empty spaces. The options for each store are: Store 1Store 2Store 3GroceryPet StoreRestaurantGymHardware StorePizza Shop All of the possible options for the three stores are shown in the tree diagram below. A tree diagram is shown. The top row reads "Grocery" and has three arrows pointing from it, each pointing to one item between Pet Store, Restaurant, and Gym in the middle row. From each item in the middle row there are two arrows pointing to on item, Hardware Store or Pizza Shop, in the bottom row. What is the probability that a randomly chosen group of stores would not include a pet store?

$\frac{2}{3}$23​​ There are 4 possible outcomes that do not include a pet store. There are 6 possible store combinations (Possibilities = 1 grocery × 3 for the second store × 2 for the third store = 6). So the probability of having a group of stores without a pet store is: P(no pet store)=4/6=2/3

135 physical therapists were asked if they use their complimentary gym membership and if they bring their own lunch to work. The following table shows the results. Brings Lunch to WorkDoes Not Bring Lunch to WorkTotalUses gym Membership247296Does Not Use Gym Membership182139Total4293135 According to the table, what is the probability that a randomly chosen physical therapist does not use the gym membership given that they bring their lunch to work? Give your answer as a fraction in simplest form.

$\frac{3}{7}$37​​ Because our given information is that the physical therapy brings lunch to work, we want to focus on that column of the table. Brings Lunch to WorkDoes Not Bring Lunch to WorkTotalUses gym Membership247296Does Not Use Gym Membership182139Total4293135 In that row, there are 42 participants total, and 18 of them do not use the complimentary gym membership, therefore the probability is 18/42=3/7.

The probability that a car has a certain factory defect is 825. The probability that a car has a certain factory defect and needs an oil change is 750. What is the probability that a car needs an oil change given that it has a certain factory defect? Give your answer as a fraction in simplest form.

$\frac{7}{16}$716​​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging, we see thatP(A|B)=P(A AND B)P(B) So if we think of A as being the event that a car has a factory defect and B as being the event that the car needs an oil change, then we can plug in the known information to find P(A|B)=750825=716

A survey of 130 college graduates was conducted. Students were asked whether they were employed and if they were looking for a job. The following table shows the results. Job SearchingNo Job SearchingTotalEmployed286189Not Employed182341Total4684130 According to the table, what is the probability that a randomly chosen student is not employed given that they are job searching? Give your answer as a fraction in simplest form.

$\frac{9}{23}$9/23​​ Because our given information is that the individual is job searching, we want to focus on that column of the table. Job SearchingNo Job SearchingTotalEmployed286189Not Employed182341Total4684130 In that row, there are 46 individuals total, and 18 of them are not employed, therefore the probability is 1846=9/23.

The brothers in the first example decide to try the "Lucky Sundae-" where the employees of the ice cream parlor choose the sundae for them. (a) What is the probability that the brothers will end up with a two-scoop, chocolate ice cream, fudge sauce sundae? (b) What is the probability that they will have a sundae with strawberry ice cream?

(a) Let the desired event, the brothers getting a two-scoop, chocolate ice cream, fudge sauce sundae, be S. There is 1 possible outcome that matches this event. There are 12 possible events. So the probability of the brothers getting this specific sundae is: P(S)=1/12 (b) Let the desired event, the brothers ending up with a strawberry ice cream sundae, be R. There are 4 possible outcomes that match this event. There are 12 possible events. So the probability of the brothers getting a sundae with strawberry ice cream is: P(R)=4/12=1/3

Suppose we roll a fair die and toss a coin and want to count the total number of outcomes. The die roll has six possible outcomes (1, 2, 3, 4, 5, 6), and the coin toss has two outcomes, heads (H) and tails (T). The outcomes of the die roll can occur with either outcome of the coin toss.

The total number of outcomes for the die roll and coin toss is 6×2=12, these being 1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, and 6T. These outcomes are shown in the tree diagram below.

Given that P(A AND B)=0.29 and P(A|B)=0.67, what is P(B)? Give your answer as a percent. Round to two decimal places.

$43.28\ \%$43.28 %​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging, we find thatP(B)=P(A AND B)P(A|B)Plugging in the values we were given, we find thatP(B)=0.290.67≈0.4328 To convert the decimal to a percent, we can multiply by 100: 0.4328×100=43.28%

If A and B are events with P(A)=0.5, P(A OR B)=0.65, P(A AND B)=0.15, find P(B).

$0.3$0.3​ First, it is helpful to write down the addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, rearranging this, we find thatP(B)=P(A OR B)+P(A AND B)−P(A)Plugging in the known values, we findP(B)=0.65+0.15−0.5=0.3

The results of an experiment are shown below. Calculate E, the expected value of the experiment. Write your answer as an exact fraction in its simplest form. Outcome248Probability162316

$\frac{13}{3}$133​​ Using the definition for E we have E=2⋅16+4⋅23+8⋅16=133.

The probability that a randomly chosen student plays in the band is 150. The probability that a randomly chosen student sings in the choir and plays in the band is 1200. What is the probability that a randomly chosen student sings in the choir, given that the student plays in the band?

If B is the event that a random student plays in the band, and C is the event that a random student sings in the choir. Then we are told P(B)=150 and P(C AND B)=1200. We are asked for P(C|B). Using the formula, we find that P(C|B)=P(C AND B)P(B)=1/2001/50=50200=14 So the probability that a randomly chosen student sings in the choir given that he or she plays in the band is 14.

Suppose A and B are mutually exclusive events, and that P(B)=0.24 and P(A OR B)=0.27. Find P(A).

$0.03$0.03​ Remember that when A and B are mutually exclusive events, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find thatP(A)=P(A OR B)−P(B)So plugging in the given values, we findP(A)=0.27−0.24=0.03

If A and B are events with P(A)=0.5, P(B)=0.3, P(A OR B)=0.67, find P(A AND B).

$0.13$0.13​ First, it is helpful to write down the addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, rearranging this, we find thatP(A AND B)=P(A)+P(B)−P(A OR B)Plugging in the known values, we findP(A AND B)=0.5+0.3−0.67=0.13

A group of 150 students in a high school were asked if they prefer texting or calling. The following table shows approximate numbers. MaleFemaleTotalTexting225880Calling86270Total30120150 According to the table, what is the probability that a randomly chosen student prefers calling, given that they are male? Give your answer as a fraction in simplest form.

$\frac{4}{15}$415​​ Because our given information is that the participant was male, we want to focus on that column of the table. MaleFemaleTotalTexting225880Calling86270Total30120150 In that row, there are 30 participants total, and 8 of them prefer calling, therefore the probability is 830=415.

Write the first three terms of the sequence whose general term is an=(3n)!.

$a_1=6,\ a_2=720,\ a_3=362880$a1​=6, a2​=720, a3​=362880​ We substitute the values 1, 2, and 3 into the formula an=(3n)! to find ana1ana2ana3=(3n)!=(3(1))!=3!=3⋅2⋅1=6=(3n)!=(3(2))!=6!=6⋅5⋅4⋅3⋅2⋅1=720=(3n)!=(3(3))!=9!=9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1=362,880 The first three terms of the sequence are a1=6, a2=720, and a3=362,880.

The probability that a person has life insurance, given that they also have health insurance is 1626. If we know that 18 people have in a sample have life insurance and 26 people have health insurance, fill in the Venn diagram below with the number of insurance holders to reflect this probability. Let Event A represent the people who have life insurance, and Event B represent the people who have health insurance.

1$$ 2$$ 3$$ 1$2$2​ 2$16$16​ 3$10$10​ We are given that P(A|B)=1626, so we know that there are 16 insurance holders in A AND B, so we fill in 16 for Response 2. We are given that there are 18 people total who have life insurance, so Response 1 is 18−16=2. We are given that there are 26 people total who have health insurance, so Response 3 is 26−16=10.

The probability that a woman works full-time, given that she has children is 527. If we know that 30 women in a sample work full-time and 27 women have children, fill in the Venn diagram below with the number of women to reflect this probability. Let Event A represent working full-time, and Event B represent having children.

1$$ 2$$ 3$$ 1$25$25​ 2$5$5​ 3$22$22​ We are given that that P(A|B)=527, so we know that there are 5 women in A AND B, so we fill in 5 for Response 2. We are given that there are 30 women total who work full-time, so Response 1 is 30−5=25. We are given that there are 27 women total who have children, so Response 3 is 27−5=22.

A community festival organizes a fundraiser for their local park. They sold 25,000 tickets at $20 each. The grand prize is $5,000. There are two prizes valued at $2,000, three prizes valued at $1,000, five prizes of $500, and ten prizes of $100. What is the expected value of this raffle? Round to the nearest cent. Do not round until the final calculation.

1$-19.38$−19.38​ Recall, expected value can be found by multiplying the value by the probability of the event. Here, remember that you paid $20 for the ticket, so you must subtract 20 from the value before finding the expected value. E(x)=4,980⋅125,000+1,980⋅225,000+980⋅325,000+480⋅525,000+80⋅1025,000−20⋅24,97925,000 E(x)=0.1992+0.1584+0.1176+0.096+0.032−19.9832 E(x)=−19.38 You can expect to lose $19.38 for each raffle ticket.

A NPR/Robert Wood Johnson Foundation/Harvard T.H. Chan School of Public Health poll asked adults whether they participate in a sport. Given that the probability that a randomly chosen adult is a woman is 0.51, and the probability that a randomly chosen adult is a woman and plays a sport is 0.08, what is the probability of an adult playing a sport given that the adult is a woman? Round your answer to two decimal places.

1$0.16$0.16​ Let M be the event that a randomly chosen adult plays a sport, and let W be the event that a randomly chosen adult is a woman. Then according to the formula for conditional probability, P(M|W)=P(M AND W)P(W) So plugging in what we know, we find that P(M|W)=0.080.51=0.16 So the probability that a random adult plays a sport given that the adult is a woman is 0.16.

Janet plays a dice game. If she rolls a 1, she wins $3 while if she rolls a 2, she wins $1. If she rolls a 3,4,or5, she loses $1 and if a 6 is rolled, she doesn't win or lose. Find the expected value of this dice game. Round to the nearest cent. Do not round until the final calculation.

1$0.17$0.17​ Recall, expected value is found by multiplying the value by the probability of the event. Here Janet can win 3, win 1, lose 1, or break even. E(x)=3⋅16+1⋅16−1⋅12+0⋅16 E(x)=36+16−36+0=16 E(x)=16=0.166666...=0.17 Janet can expect to win $0.17 each time she rolls the dice.

The probability that a high school athlete is offered admissions to a college, given that they were also involved in music, is 715. If we know that 25 athletes and 15 musicians were offered admission, fill in the Venn diagram below with the number of students to reflect this probability. Let Event A represent the athletes offered admission, and Event B represent the musicians offered admission.

1$18$18​ 2$7$7​ 3$8$8​ We are given that that P(A|B)=715, so we know that there are 7 students in A AND B, so we fill in 7 for Response 2. We are given that there are 25 students total that are athletes, so Response 1 is 25−7=18. We are given that there are 15 students total that are musicians, so Response 3 is 15−7=8.

In a game show, contestants are given the opportunity to win a new car if they correctly choose the winning door three times in a row. In the first round, they must choose between 3 doors, one of which is labeled "win." In the second round and third rounds, they will choose from 2 doors, with one labeled "win." Their options are shown in the tree diagram below. They are blind folded, and are not given an opportunity to see the doors beforehand. What is the probability that a contestant will NOT win a new car?

11/12 To win the new car, a contestant must choose a "win" door three times in a row. We can see from the tree diagram that there is only one possibility of doing this, but there are 12 possible outcomes. So, there are 12−1=11 ways to NOT win. The probability of a contestant not winning is then 11/12.

Conditional Probability: The probability of one event based on the probability of another event already happening. The Multiplication Rule for Conditional Probability: The probability of A given B is the probability of A AND B divided by the probability of B. P(A|B)=P(A AND B)P(B) With Replacement: When a population element can be selected more than one time, so after choosing an element it is put back in before the next one is taken. Without Replacement: When a population element can be selected only one time, so after choosing an element it is set aside and is no longer available to be chosen again.

Conditional Probability: The probability of one event based on the probability of another event already happening. The Multiplication Rule for Conditional Probability: The probability of A given B is the probability of A AND B divided by the probability of B. P(A|B)=P(A AND B)P(B) With Replacement: When a population element can be selected more than one time, so after choosing an element it is put back in before the next one is taken. Without Replacement: When a population element can be selected only one time, so after choosing an element it is set aside and is no longer available to be chosen again.

Employees at a company are randomly assigned certain shifts during the busy holiday season. Let A represent the shift between the hours of 8 a.m. and 12 p.m., and B represent the shift between the hours of 12 and 4 p.m. If the shifts are assigned with the following probabilities: P(A)=0.28; P(B)=0.83; P(A OR B)=0.93 What is P(A AND B), the probability that an employee will randomly be assigned both shifts?

Here, we are given slightly different information than in the rule above, but note that we can rearrange the rule to solve for P(A AND B): P(A AND B)=P(A)+P(B)−P(A OR B)=0.28+0.83−0.93=0.18 So we find that P(A AND B)=0.18. This means that the probability that an employee will randomly be assigned both shifts is 0.18, or 18%.

If we were to select one friend at random, what is the probability that they would be in either statistics or writing, but not both?

To solve this problem, it is pretty simple for us to count the number of friends that are not in both classes. There are 5 out of the 6 friends who are only in one of the courses, so P(NOT both classes)=56 However, there will be many other instances in probability where it isn't as easy to calculate combinations of events. If we don't have a Venn diagram, or information about how many friends there are, we have to use something else to help us find the probability. The following rule will help us to calculate probabilities for OR :

A bag of marbles contains 2 gold marbles, 10 silver marbles, and 30 black marbles. You decide to play the following game: Randomly select one marble from the bag. If it is gold, you win $5. If it is silver, you win $2. If it is black, you lose $2. What is the expected value if you play the game? Express losses as negative values and do not include the dollar sign in your answer. Round to the nearest cent.

$-0.71$

Given that P(A AND B)=0.70 and P(A|B)=0.94, what is P(B)? Round your answer to three decimal places.

$0.745$0.745​ Remember the multiplication rule for conditional probability: P(A AND B)=P(A|B)P(B) Rearranging, we find thatP(B)=P(A AND B)P(A|B)Plugging in the values we were given, we find thatP(B)=0.700.94≈0.745

Given that P(B AND A)=0.07 and P(B|A)=0.20, what is P(A)? Express your answer as a percent, rounded to two decimal places if necessary.

$35\%$35%​ Remember the multiplication rule for conditional probability: P(B AND A)=P(B|A)P(A) Rearranging, we find thatP(A)=P(B AND A)P(B|A)Plugging in the values we were given, we find thatP(A)=0.070.20=0.35 To express our answer as a percent, we multiply by 100: 0.35×100=35%

Suppose A and B are mutually exclusive events, and that P(B)=0.03 and P(A OR B)=0.52. Find P(A).

$0.49$0.49​ Remember that when A and B are mutually exclusive events, we know that P(A AND B)=0 (this is the definition of mutually exclusive events). So for mutually exclusive events, the probability addition rule becomes P(A OR B)=P(A)+P(B)−P(A AND B)=P(A)+P(B) Rearranging, we find thatP(A)=P(A OR B)−P(B)So plugging in the given values, we findP(A)=0.52−0.03=0.49

The probability that a person catches the flu given that they've had a flu shot is 945. If we know that 40 people caught the flu and 45 people received flu shots, fill in the Venn diagram below with the number of people to reflect this probability. Let Event A those who received a flu shot, and Event B represent those who caught the flu.

1$$ 2$$ 3$$ 1$36$36​ 2$9$9​ 3$31$31​ We are given that the probability that a person catches the flu (event B) given that they've had a flu shot (event A) is P(B|A)=945, so we know that there are 9 people in A AND B, so we fill in 9 for Response 2. We are given that there are 45 people total received flu shots (event A), so Response 1 is 45−9=36. We are given that there are 40 people total who caught the flu (event B), so Response 3 is 40−9=31.

A university surveys 50 undergraduates to learn more about students' class preferences. They find that 25 of the students like to take online classes and 35 of the students are freshman. Let Event A be students that like to take online classes; Let Event B represent first year students. Given the following conditional probabilities: P(A|B)=1535; P(B|A)=1525; Fill in the Venn diagram with the number of students for events A, B, A AND B, and not A or B:

1$$ 2$$ 3$$ 4$$ 1$10$10​ 2$15$15​ 3$20$20​ 4$5$5​ We are given that there are 50 undergraduate students, P(A|B)=1535, and P(B|A)=1525. Since P(A|B) is a comparison of how many students are in event A AND B total to how many students are in just B, we know that A AND B=15 and 35=15+B. So, there are 20 students in just B. Next, we know that P(B|A)=1525, so again, this is a comparison of students in A AND B to how many students in A total. We already know that A AND B=15, and we have 25=15+A. So there are 10 students in just A. We now have 15+10+20=45 students that either like online classes or are first year students. This means that 50−45=5 students that do not fit in any category.

The following incomplete contingency table shows a random sample of 160 citizens were asked if they exercise. Complete the table.

1$22$22​ 2$11$11​ 3$56$56​ 4$33$33​ Note that from the Female row, we have that A+71=93⟹A=22 Note that from the Total row, we have that B+127=160⟹B=33 Plugging this in the Exercises column, we have that 22+C=33⟹C=11 Plugging this in the Male row, we have that 11+D=67⟹D=56 If you complete the rest of the table, you should get ExercisesDoes not ExerciseTotalFemale227193Male115667Total33127160

A lawyer has numbered the cases that he is working on. He has criminal cases numbered 1,2,3,4 and civil numbered 1,2,3,4,5. Let R be the event of selecting a criminal case, C the event of selecting a civil case, E the event of selecting an even numbered case, and O the event of selecting an odd case.Selecting the criminal case number 4 is one of the outcomes in which of the following events? Select all correct answers.

C′ R AND E Because the case is criminal and the number is even, the case is an outcome of R and E. Therefore, it is also an outcome of R AND E and C′.

A standard six-sided die shows a number, 1, 2, 3, 4, 5, or 6, on each of its sides. You roll the die once. Let E be the event of rolling the die and it showing an even number on top and L be the event of rolling a number less than 4. Rolling a 3 is an outcome of which of the following events? Select all correct answers.

E OR L E′ AND L E′ OR L′ To roll a 3: It is an outcome of E′, that is NOT even. It is an outcome of L, less than 4. So, correct the only correct "AND" answer is: E′ AND L. There are many more correct "OR" answers: E′E′E∪L∪L′∪L

Two fair dice are rolled, one blue, (abbreviated B) and one red, (abbreviated R). Each die has one of the numbers {1,2,3,4,5,6} on each of its faces. The dots in the Venn diagram below show the number and the color of the dice. Let A be the event of rolling an even number on either of the dice. Let B be the event of rolling a number greater than 4 on either of the dice. Move the dots on the Venn diagram to place the dots in the correct event, A, B,or A AND B. Note that you might not use all of the dots.

Event A is the event of rolling an even number on either of the dice, so A should contain elements {R2,B2,R4,B4,R6,B6}. Event B is the event of rolling a number greater than 4 on either of the dice, so B should contain outcomes {R5,B5,R6,B6}. Event A AND B should therefore contain all outcomes that are greater than 4 AND even. Therefore, A AND B should contain the outcomes {R6,B6}. Notice that the outcomes R1, B1, R3, and B3 do not fall into either of these events. They should therefore be outside of the Venn diagram.

A bag of candy contains 3 blue candies, 4 green candies, and 6 red candies. If you draw two candies, one at a time and without replacement, what is the probability that you will draw a green candy and a blue candy?

Let G be the event of drawing a green candy on the first draw and B be the event of drawing a blue candy on the second draw. Then P(B|G) is the probability of drawing a blue candy given that the first candy was green, and P(B AND G) is the probability of drawing a green and a blue candy. P(G)=413; P(B|G)=312=14. Notice that here there are 3 possible blue candies and 12 candies left after drawing the first green candy (since we drew without replacement). We want to know P(B AND G), the probability of drawing a green and a blue candy. The Multiplication Rule for Conditional Probabilities gives that: P(B|G)=P(B AND G)P(G) Rearranging using algebra, we have: P(B AND G)=P(B|G)⋅P(G) Plugging in the probabilities from above, we have P(B AND G)=14⋅413=113

A NPR/Robert Wood Johnson Foundation/Harvard T.H. Chan School of Public Health poll asked adults whether they participate in a sport. Given that the probability that a randomly chosen adult is a woman is 0.51, and the probability that a randomly chosen adult is a woman and plays a sport is 0.08, what is the probability of an adult playing a sport given that the adult is a woman? Round your answer to two decimal places.

Let M be the event that a randomly chosen adult plays a sport, and let W be the event that a randomly chosen adult is a woman. Then according to the formula for conditional probability, P(M|W)=P(M AND W)P(W) So plugging in what we know, we find that P(M|W)=0.080.51=0.16 So the probability that a random adult plays a sport given that the adult is a woman is0.16.

Let D be the event that a randomly chosen person has seen a dermatologist. Let S be the event that a randomly chosen person has had surgery for skin cancer. Identify the answer which expresses the following with correct notation: The probability that a randomly chosen person has had surgery for skin cancer, given that the person has seen a dermatologist.

P(S|D) Remember that in general, P(A|B) is read as "The probability of A given B". Here we are given that the person has seen a dermatologist, so the correct answer is P(S|D).

Write the first five terms of the sequence whose general term is an=2(n!).

$a_1=2,\ a_2=4,\ a_3=12,\ a_4=48,\ a_5=240$a1​=2, a2​=4, a3​=12, a4​=48, a5​=240​ We substitute the values 1, 2, 3, 4, and 5 into the formula an=2n! to find ana1ana2ana3ana4ana5=2n!=2(1)!=2⋅1=2=2n!=2(2)!=2⋅2⋅1=4=2n!=2(3)!=2⋅3⋅2⋅1=12=2n!=2(4)!=2⋅4⋅3⋅2⋅1=48=2n!=2(5)!=2⋅5⋅4⋅3⋅2⋅1=240 The first five terms of the sequence are a1=2, a2=4, a3=12, a4=48, and a5=240.

If A and B are independent events with P(A)=0.90 and P(A AND B)=0.54, find P(B). Give your answer as a decimal rounded to two decimal places.

$0.60$0.60​ Remember that for independent events, P(A AND B)=P(A)P(B) Solving for P(B), we find thatP(B)=P(A AND B)P(A)Now, plugging in the values we are given, we find thatP(B)=0.540.90=0.60