Basic Probability
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?
0.015 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
If A and B are events with P(A)=0.5 P(B)=0.3 P(A OR B) =0.67 What is P(A AND B).?
0.13 Begin with the Addition Rule: P(A AND B) = P(A) + P(B) − P(A OR B) P(A AND B) = 0.5 + 0.3 - 0.67 0.13 = 0.5 + 0.3 − 0.67
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?
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)= (0.08/0.51) = 0.16
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 shifts 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?
0.18, or 18%. 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%.
Kelsey, a basketball player, hits 3-point shots on 37.2% of her attempts. If she takes 12 attempts at 3-point shots in a game, what is the probability that she hits exactly 5 of them?
0.217 binomial probability. In this case, we want to find the probability of 5 successes, where a success is hitting the 3-point shot. BINOMPDF ( 12, 0.372, 5)= 0.217
If A and B are events with: P(A)=0.5 P(A OR B)=0.65, P(A AND B) =0.15 What is P(B).?
0.3 Begin with the Addition Rule: P(A AND B) = P(A) + P(B) − P(A OR B) 0.15 = 0.5 + P(B) - 0.65 rearrange P(B)=P(A OR B)+P(A AND B)−P(A) P(B) = 0.65 + 0.15 − 0.5 = 0.3 P(B) = 0.3
In a recent baseball season, Ron was hit by pitches 19 times in 596 plate appearances during the regular season. Assume that the probability that Ron gets hit by a pitch is the same in the playoffs as it is during the regular season. In the first playoff series, Ron has 22 plate appearances. What is the probability that Ron will get hit by a pitch exactly once?
0.355 binomial probability. In this case, we want to find the probability of 1 success, where a success is getting hit by a pitch. BINOMPDF ( 22, 19/596, 1) = 0.355
Given that: P(B/A)=0.84 P(A) = 0.43 what is P(B AND A)?
0.3612 The multiplication rule for conditional probability: P(B AND A)= P(B/A)P(A) plug in the values, we find: P(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 What is P(B).?
0.5 Begin with the Addition Rule: P(A AND B) = P(A) + P(B) − P(A OR B) 0.01 = 0.4 + P(B?) - 0.89 rearrange P(B)=P(A OR B)+P(A AND B)−P(A) P(B)= 0.89 +0.01 - 0.4 P(B) = 0.5
A roulette wheel has 38 slots, numbered 1 to 36, with two additional green slots labeled 0 and 00. Jim, a dealer at a roulette table, tests the roulette wheel by spinning a ball around the wheel repeatedly and seeing where the ball lands. The ball has an equally likely chance of landing in each slot. If Jim spins the ball around the wheel 20 times, what is the probability that the ball lands in a green slot at most once?
0.716 BINOMCDF(20, 1/19, 1)
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 aterm-15 statistics class is 0.66, what is the probability of a student taking a calculus class given that the student takes a statistics class?
0.795 p(b)=0.83 and P(C and B)= 0.66 equation P(C/B)= p( C and B)/ P(B) Plug it in= 0.83/0.66 = 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, What is P(B).?
0.9. Begin with the Addition Rule: P(A AND B) = P(A) + P(B) − P(A OR B) 0.07 = 0.1 + p(B) - 0.93 Rearranging to solve for P(B), we find that P(B)=P(A OR B)−P(A)+P(A AND B) = 0.1 + 0.07- 0.93 = 0.9 P(B)=0.9.
Bryan is performing data entry for an insurance company by taking handwritten records and entering them into the database. Thus far, 15.6% of the handwritten records have had missing data that prevented them from being entered into the database. If Bryan takes 10 handwritten records, what is the probability that at most three records will be incomplete?0.
0.943 we want to find the probability of 0 to 3 successes, inclusive, where a success is a handwritten record being incomplete. - BINOMCDF ( 10, 0.156, 3) = 0.943
A database system assigns a 32-character ID to each record, where each character is either a number from 0 to 9 or a letter from A to F. Assume that each number or letter is equally likely. Find the probability that at least 16 characters in the ID are numbers
0.948 choosing a number out of the possible character choices, which is either a number from 0 to 9 or a letter from A to F. so the probability of success is 10/16= 0.625. SECOND: BINOMCDF (32, 0.625, 15) = 0.0520221042 round three decimal places= 0.052 THIRD SUBTRACT: Subtract the probability from 1 gives the probability 1−0.052=0.948. So, the probability of at least 16 characters being numbers is 0.948.
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?
1/10 Let A be the event that the student is randomly selected to take the writing seminar, and note that: P(A)=50/150 = 1/3. Let B be the event that the student is selected to take the study skills seminar, and note that P(B)=15/150 = 1/10. The probability that a student takes both the writing and the study skills seminars: P(A AND B)= 5/150 = 1/30. 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) =(1/30)/(1/3) =3/30 =1/10
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?
1/13 Let G be the event of drawing a green candy on the first draw. Let 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)=4/13; P(B/G)=3/12 =1/4. 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)=1/4⋅4/13 = 1/13
You roll a standard die (with the numbers {1,2,3,4,5,6} on its faces). Let A be the event that you roll an odd number. What is the probability of the event A?
1/2 The event A consists of the outcomes {1,3,5} and there are a total of six possible outcomes S={1,2,3,4,5,6}. Therefore, P(A)=3/6 = 1/2
An urn contains 6 red beads, 4 blue beads, and 2 green beads. If a single bead is picked at random, what is the probability that the bead is blue?
1/3 4 balls are blue/ 12 total amount of balls 4/12 = 1/3
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 only is P(B) = 13 The probability that the friend only takes chemistry is P(C) = 12 A student only takes physics with the probability P(P) = 16 A student takes both biology and physics with a probability = 14. What's the probability that a randomly chosen friend takes biology or physics, P(B OR P)?
1/4 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(P) - P(B AND P) = 1/3 + 1/6 − 1/4 = 1/4
Martin is a food critic reviewing a restaurant that has 6 main courses, 3 desserts, and 5 appetizers he has yet to try. Martin asks the server to surprise him with one of the remaining items. What is the probability that Martin is served a main course?
3/7 There are 6 main courses / total of 6+3+5=14 the probability is 6/14 = 3/7
Given that: P(A AND B) = 0.29 P(A/B) = 0.67 what is P(B)? Give your answer as a percent. Round to two decimal places.
43.28% To find P(B); mulitply 0.29 x 0.67 the multiplication rule for conditional probability: P(A AND B) = P(A/B)P(B) Rearranging, we find that P(B)= P(A AND B)P(A/B) Plugging in the values we were given, we find that P(B)=0.290.67≈0.4328 To convert the decimal to a percent, we can multiply by 100: 0.4328×100= 43.28%
Alex wants to test the reliability of "lie detector tests," or polygraph tests. He performs a polygraph test on a random sample of 10 individuals. If there is more than a 95% chance that the tests result in no false positives (that is, the test does not result in a true statement being recorded as a lie), Alex will conclude that the tests are reliable. If the probability of a lie detector test resulting in a false positive is 9.1%, what will Alex conclude?
Alex will conclude that the test is not reliable since the probability of no false positives is less than 0.95. we want to find the probability of 0 successes, where a success is a false positive. BINOMPDF ( 10, 0.091, 0) =0.3851579196= 0.385 rounded to three decimal places. SO = This probability is less than 0.95, so Alex will conclude that the test is not reliable.
According to a poll, 37 out of every 100 people support a new bill about to be signed into law. Identify what method of probability was used to reach this statement. theoretical relative frequency
Relative frequency Out of the 100 people observed in this poll, 37 reported being in favor of the bill. The estimated probability from this poll is the number of people who support the bill divided by the total number of people polled= 37/100
During last year, the probability of any day being rainy in one city was 105365. Identify the method of probability that was used to reach this statement. theoretical relative frequency
Relative frequency Since it was observed over the past year that it rained on 105 of 365 days, the probability can be directly measured. The probability is calculated as number of rainy days/total number of days in the year = 105/365
In the year 2013, the state of Oklahoma reported that the probability of a randomly selected adult being obese was 32.5%. Identify the method of probability that was used to reach this conclusion. theoretical relative frequency
Relative frequency The state of Oklahoma gathered data on adult weights. Researchers determined from the data that in the year 2013, 32.5% of adults were obese.
On a standard roulette wheel, the probability of rolling an odd number is 1838, or 919. Identify the method of probability used that resulted in this statement. theoretical relative frequency
Theoretical There are a total number of 38 possible equally likely outcomes on a roulette wheel. Out of the 38 outcomes, there are 18 odd-numbered outcomes. The probability is determined from number of odd outcomes/total number of outcomes = 18/38 = 9/19 Since the probability is describing the chance that the event will occur, the theoretical method was used.
The Census Bureau reports that if you were to select any random person from the United States, there is a 77.5% chance that the person is 18 or older. Identify the method of probability that was used in order to reach this conclusion. theoretical relative frequency
relative frequency The census is a survey of the population that gathers various details about individuals, such as age. This survey observed that 77.5% of the population was 18 years old or older. Since the probability can be directly measured, the relative frequency method was used.
A dairy farmer milks his two cows every day. He determined the chance that he gets anywhere between 12 and 14 gallons of milk in one day is around 32%. Identify the method of probability the farmer used to reach this conclusion.
relative frequency There is no way the farmer could know before milking the cows how much milk he is going to get on any day. He would need to record the amount of milk gathered on previous days and estimate the probability of obtaining 12 to 14 gallons of milk. Thus, he used the relative frequency method.
A soda bottling company's manufacturing process is calibrated so that 2% of bottles are not filled within specification. Every hour, 12 random bottles are taken from the assembly line and tested. If 3 or more bottles in the sample are not within specification, the assembly line is shut down for re-calibration. What is the probability that the assembly line will be shut down for re-calibration?
the resulting probability is .001 BINOMCDF (12, .02,3)
Kevin works for a company that manufactures solar panels. In a large batch of solar panels, about 1 in 300 is defective. Suppose that Kevin selects a random sample of five solar panels from this batch. What is the probability that none of the solar panels are defective?
the resulting probability is 0.983 BINOMPDF (5, 1/300,0)
When you roll a pair of six-sided die, the probability of rolling an 8 is 536. Identify the method of probability that was used to reach this statement.
theoretical Note that there are 36 equally likely outcomes when rolling two six-sided dice, and 5 of the possible 36 outcomes are 8. The probability is determined by number of ways to roll 8 / total number of outcomes = 53/6. Thus, the theoretical method was used.
When a coin is tossed three times, the chance that all three tosses are heads is 18. Identify the method of probability used to reach this statement. theoretical relative frequency
theoretical Three heads is 1 of the possible 8 equally likely outcomes of tossing a coin three times. The probability is calculated by number of ways to get three heads / total number of outcomes =18 Since the probability is describing the chance that the event will occur, the theoretical method was used.
