Binomial distribution

In a coin game, you toss a coin three times. If all three coin tosses are heads or all three tosses are tails, you win $15. Otherwise, you lose $3. What is the expected profit for one round of the coin game? Round your answer to the nearest cent. Enter an expected loss as a negative number.

$1.50

The questions on a test consist of 10 multiple choices, 3 essays, and 9 free responses. If the questions are ordered randomly, what is the probability that the first question is a free response?

9/22

Let S be the event that a randomly chosen voter supports the president. Let W be the event that a randomly chosen voter is a woman. Identify the answer which expresses the following with correct notation: The probability that a randomly chosen voter is a woman, given that the voter supports the president. P(W|S) P(S AND W) P(W) AND P(S) P(S|W)

P(W|S) Remember that in general, P(A|B) is read as "The probability of A given B". Here we are given that the voter supports the president, so the correct answer is P(W|S).

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?

0.002 BINOMCDF(12,.02,3)= 0.001 Subtract 1 gives the probability = 1−0.99846297 = 0.00153703, or 0.002 at least 3 bottles in the sample are not within specification.

find P(B) If A and B are events: P(A)=0.3 P(A OR B)=0.63 P(A AND B)=0.17

0.5 addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) 0.63= 0.3 + P(B)- 0.17 rearrange to solve: P(B)=0.65+0.17−0.3 =0.3

Consider how the following scenario could be modeled with a binomial distribution, and answer the question that follows. 54.4% of tickets sold to a movie are sold with a popcorn coupon, and 45.6% are not. You want to calculate the probability of selling exactly 6 tickets with popcorn coupons out of 10 total tickets (or 6 successes in 10 trials). What value should you use for the parameter p?

0.544. The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a movie ticket with a popcorn coupon, so p=0.544.

Fill in the following contingency table and find the number of students who both do not read mysteries AND do not read comics. Students readcomics donotreadcomics Total readmysteries 41 donotreadmysteries 28 62 Total 50

34

two dice are rolled and you note the numbers on each dice. 1) list all possible outcomes 2) Let A be event of rolling the first dice as a two, what are the there outcomes of event A?

1) 1,2,3,4,5,6 2) (2,1), (2,2), (2,3) (1st dice 2, with the 2nd dice of the three outcomes).

Given that the probability of a student eating at the cafeteria and a student living off campus is 0.07, and the probability of a student eating at the cafeteria given that the student lives off campus is 0.20, what is the probability of a student living off campus?

0.35 Remember the multiplication rule for conditional probability: P(B AND A)=P(B/A)P(A) Rearranging, we find that P(A)=P(B AND A)P(B/A) So if we think of A= the event a student lives off campus and B = event a student eats at the cafeteria, then we can plug in the known information to find: P(A)=0.07 / 0.20 = 0.35

Trial best fits which of the following descriptions? Select the correct answer below: 1) a particular result of an experiment 2) a subset of the set of all outcomes of an experiment 3) one repetition or instance of an experiment 4) the set of all possible outcomes of an experiment

3) one repetition or instance of an experiment A trial is defined as one repetition or instance of an experiment.

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?

Solution: Let G = event of drawing a green candy on the first draw P(G)=4/13. Let B = event of drawing a blue candy on the second draw P(B/G)= 3/12 = 1/4. We want to know P(B AND G), 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 x 4/13 = 1/13

Identify the parameter n in the following binomial distribution scenario. A basketball player has a 0.429 probability of making a free throw and a 0.571 probability of missing. If the player shoots 20 free throws, we want to know the probability that he makes no more than 12 of them. (Consider made free throws as successes in the binomial distribution.)

n=20 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, the total number of trials, or free throws, is n=20.0.

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

There are 3 red balls, 5 green balls, and 1 yellow ball in a hat. If you draw the yellow ball, you gain $10. If you draw any of the red balls, you gain $4. If you draw a green ball, you lose $3. If you were to play this game 25 times, replacing the drawn ball each time, how much money can you expect to gain or lose? Round to the nearest cent. Do not round until the final answer. Enter an expected loss as a negative number.

19.5

Given that P(B|A)=0.76 and P(A)=0.41, what is P(B AND A)? Round to three decimal places.

.312

For the below problem, which values would you fill in the blanks of the function B(x,n,p)? The probability of saving a penalty kick from the opposing team is 0.617 for a soccer goalie. If 7 penalty kicks are shot at the goal, what is the probability that the goalie will save 5 of them? B(0.617;5,7) B(5;7,0.617) B(7;0.617,5) B(7;5,0.617)

B(5;7,0.617). The parameters of a binomial distribution are: n = the number of trials x = the number of successes experiment p = the probability of a success The parameters should be in the order of x, n, p in the binomial function B(x;n,p). So, in this case, you should input B(5;7,0.617).

What term is best described as the property of two events in which the knowledge that one of the events occurred does not affect the chance the other occurs? complement mutually exclusive dependent independent

Independent

Given that P(A|B)=0.22 and P(B)=0.46, what is P(A AND B)? Round your answer to three decimal places.

0.101 Remember the multiplication rule for conditional probability: P(A AND B)= P(A/B)P(B) So plugging in the values that we know, we find P(A AND B) =(0.22)(0.46) ≈0.101

Fill in the following contingency table and find the number of students who both do not watch comedies AND do not watch dramas. Students | WatchDrama | Don'tWatchD | Total WatchComedie | 36 75 Don'twatchC | 56 Total | 64

25

according to a Gallup poll, 60% of American adults prefer saving over spending. Let X= the number of American adults out of a random sample of 50 who prefer saving to spending. What is the mean (μ) and standard deviation (σ) of X?

μ=30 and σ≈3.46 mean is μ=np = μ=(50)(0.6)=30 find variance = σ=(50)(0.6)(0.4)=12 stand deviation= √=12 ≈3.46

Give the numerical value of the parameter p in the following binomial distribution scenario.The probability of winning an arcade game is 0.632 and the probability of losing is 0.368. If you play the arcade game 10 times, we want to know the probability of winning no more than 8 times.Consider winning as a success in the binomial distribution. Do not include p= in your answer.

0.632 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is winning a game, so p=0.632.

A softball pitcher has a 0.487 probability of throwing a strike for each pitch. If the softball pitcher throws 29 pitches, what is the probability that no more than 14 of them are strikes? Insert the correct symbol to represent this probability. P(X____14)

P(X≤14) This probability could be represented by P(X≤14) or P(X<15)

Jacqueline will spin a fair spinner with the numbers 0, 1, 2, 3, and 4 a total of 3 times. If Event A = spinner lands on numbers all greater than 2 and Event B = total sum of 9, which of the following best describes events A and B? Select the correct answer below: independent dependent mutually exclusive complement

dependent Events are independent if the knowledge of one event occurring does not affect the chance of the other event occurring. In this case, the chance of Event B occurring is greater if Event A occurs, so these events are dependent.

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? Give your answer as a fraction.

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? Give your answer as a fraction.

1/3 There are 4 possible outcomes (any one of the blue beads). There are 6 + 4 + 2 = 12 total possible outcomes (any of the beads). Therefore, the probability is 4/12 = 1/3.

Ronald finished his daily jog and wants a drink. In his refrigerator, there are 2 cans of energy drink, 9 cans of soda, and 11 bottles of water. If Ronald chooses a drink at random, what is the probability that he chooses a bottle of water? Give your answer as a fraction.

11/22 There are 11 bottles of water and a total of 2 + 9 + 11 = 22 drinks. So, the probability of choosing a bottle of water is 11/22

Male Female Total Orange 33 64 97 Green 15 23 38 Total 48 87 135 A group of 135 students at an elementary school were asked if they prefer the color orange to the color green. 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 the color green? (Enter your answer in fraction form.)

15/48

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 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

65% of the people in Missouri pass the driver's test on the first attempt. A group of 7 people took the test. What is the probability that at most 3 in the group pass their driver's tests in their first attempt? Round your answer to three decimal places.

0.200 The parameters of this binomial experiment are: n = 7 trials p = 0.65 x = at most 3 successes 2nd ..Vars.. BINOMCDF (7, .65, 3) 0.19984 = .rounded = 200

In a card game, there are 15 cards laid out on a table. 6 of the cards are blank, 2 of the cards are labeled $3, and the remaining cards are labeled $1. When you select a card at random, you earn what is labeled on the card. After each game, you return the card and the cards are shuffled. Suppose you play this game 10 times. How much money can you expect to gain? Round to the nearest cent. Do not round until the final answer.

$8.67. The probability of selecting: 6 cards are blank $0 = 6/16 2 cards labeled $3 = 2/15. 7 cards labeled $1 = 7/15. =E1×P1 + E2×P2 + E3×P3 =$0×25 + $3×215 + $1×715 ≈$0.00 + $0.40 + $0.4667 ≈ $0.8667 Play the game 10 times = $0.8667×10≈$8.67 From 10 games, you can expect to gain $8.67.

A softball pitcher has a 0.431 probability of throwing a strike for each pitch. If the softball pitcher throws 22 pitches, what is the probability that exactly 12 of them are strikes? Round your answer to 2 decimal places.

0.09 Binompdf (22, 0.431, 12)=0.0945 = 0.09

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.

0.16 Let M = event that a randomly chosen adult plays a sport, let W= event that a randomly chosen adult is a woman. = 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 So the probability that a random adult plays a sport given that the adult is a woman is 0.16.

Given that P(B|A)=0.76 and P(A)=0.49, what is P(B AND A)? Round to three decimal places.

0.372

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: 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 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.

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: biology class only: P(B)=1/3; chemistry class only: P(C)=1/2; physics class only: P(P)=1/6; biology & physics classes: 1/4. What's the probability that a randomly chosen friend takes biology or physics, P(B OR P)?

1/4 Addition Rule for Probabilities to P(B OR P): P(B OR P) = P(B) + P(P) − P(B AND P) P(B OR P) = 1/3 + 1/6 − 1/4 = 1/4 So, the probability that a randomly chosen friend takes biology or physics is 14.

Diana is packing a lunch that will include an orange. Diana has 3 navel oranges, 5 mandarin oranges, and 4 Valencia oranges. If Diana selects an orange at random, what is the probability that she selects a navel orange?

3/12

Let F be the event that a randomly employee is female. Let M be the event that a randomly chosen employee is a manager. Place the correct event in each response box below to show: Given that the employee is a manager, the probability that a randomly chosen employee is female. Provide your answer below: P( | )

P(F | M ) Remember that in general, P(A|B) is read as "The probability of A given B". Here we want to know the probability that an employee is female given that the employee is a manager, so the correct answer is P(F|M).

In a game, you toss a fair coin and a fair six-sided die. If you toss a heads on the coin and roll either a 3 or a 6 on the die, you win $30. Otherwise, you lose $6. What is the expected profit of one round of this game? Round your answer to the nearest cent. Enter an expected loss as a negative number.

$0.00 There are 12 possible outcomes in this game. Of those: there are 2 outcomes ({3H, 6H}) = win $30. This corresponds to a probability of 2/12, or 1/6. The remaining 10 outcomes = lose $6, so the probability is 10/12, or 5/6. the expected profit of one game. expected value= E1×P1 + E2×P2 =$30×16+(−$6)×56 =$5.00−$5.00=$0.00

Lindsey is a manager of the collections department for a store credit card. She randomly selects 245 accounts of customers who have had an account for 5 years or more. She records the total number of missed payments, x, and the probability of each value, P(x), as shown in the table provided. Find the mean and the standard deviation of the probability distribution using a TI-83 or TI-84 graphing calculator round to 3 decimal places: x P(X=x) 0 0.351 1 0.261 2 0.184 3 0.106 4 0.045 5 0.025 6 0.012 7 0.008 8 0.004 9 0.004

The mean: 1.448, standard deviation: 1.613. Go to STAT and then press ENTER. Clear the existing values in lists L1 and L2. Enter the values of X in list L1 and the corresponding probabilities in list L2. Press STAT, go to CALC, and press ENTER. Enter list L1 using 2nd and 1. Then enter a comma. Enter list L2 using 2nd and 2. Press ENTER. Identify the mean and standard deviation in the output. The mean, rounded to three decimal places, is 1.448, and the standard deviation, rounded to three decimal places, is 1.613.

A computer graphics card. If a sample of 100 graphics cards manufactured using the new process has a less than 10% chance of having 4 or more defective graphics cards, then the manufacturer will switch to the new process. Otherwise, the manufacturer will stay with its existing process. If the probability of a defective graphics card using the new process is 1.5%, will the manufacturer switch to the new production process? Select the correct answer below: Yes, because the probability of having 4 or more defective graphics cards is less than 0.10. Yes, because the probability of having 4 or more defective graphics cards is greater than 0.10. No, because the probability of having 4 or more defective graphics cards is less than 0.10. No, because the probability of having 4 or more defective graphics cards is greater than 0.10.

Yes, because the probability of having 4 or more defective graphics cards is less than 0.10. 1. First press 2ND and then VARS for the DISTR menu. Finf BINOMCDF 2. enter the values BINOMCDF (100, 0.015, 3) in that order, separated by commas, enter 3. The calculator should then display the probability. Here, the resulting probability is 0.9357840654, which is 0.936 rounded to three decimal places. To find the probability of having 4 or more defective graphics cards, subtract this probability from 1. The probability of having 4 or more defective graphics cards is 1−0.936=0.064, which is less than 0.10. So, the manufacturer will switch to the new process.

Identify the parameter p in the following binomial distribution scenario. The probability of buying a movie ticket with a popcorn coupon is 0.702, and the probability of buying a movie ticket without a popcorn coupon is 0.298. If you buy 23 movie tickets, we want to know the probability that more than 15 of the tickets have popcorn coupons.

p=0.702

A hat contains 6 red balls, 4 yellow balls, and 2 green balls. If you draw a red ball, you lose $5. If you draw a yellow ball, you win $1. If you draw a green ball, you win $7. What is the expected profit of one draw? Round your answer to the nearest cent. Enter an expected loss as a negative number.

$1.00 There are a total of 12 balls in the hat. 6 red balls = You lose $5 = 6/12 4 yellow balls = you win $1 = 4/12 2 green balls = you win $7 = 2/12 Multiply the events by the probabilities to determine the expected value of one draw. =E1×P1+E2×P2+E3×P3 = (−$5) × 1/2 + $1 × 1/3 + $7 × 1/6 ≈ −$2.50 + $0.33 + $1.17 = −$1.00

Let W be the event that a randomly chosen person works for the city government. Let V be the event that a randomly chosen person will vote in the election. Place the correct event in each response box below to show: Given that the person works for the city government, the probability that a randomly chosen person has will vote in the election.

p( V/W )

In a die game, you roll a fair six-sided die. If you roll a 2, you win $25. If you roll anything else, you lose $5. What is the expected profit of one roll in this game? Round your answer to the nearest cent. Enter an expected loss as a negative number.

$0.00 Rolling a 2 is 1/6 = winning $25 remaining is 5/6 = lose - $5. E1×P1+E2×P2 =$25×16+(−$5)×56 ≈$4.17−$4.17 =$0.00

In a die game, you roll a standard 6-sided die twice. If the second number rolled is the same as the first number rolled, you win $25. Otherwise, you lose $2. If you were to play the game 100 times, how much money can you expect to make? Do not round until the final answer. Enter an expected loss as a negative number.

$250. If you play the game 100 times, you can expect to make $250.00. Out of the 36 possibilities of two dice rolls, 6 of these consist of the same number twice. This corresponds to a probability of 6/36, or 1/6. The probability of earning $25 is 16. For the remaining 30 combinations, you lose $2. This corresponds to a probability of 30/36, or 5/6. Multiply the outcome values by the probabilities to get the expected profit from one game. expected value = E1 × P1 + E2 × P2 =$25×16+(−$2)×5/6 ≈$4.17 − $1.67 = $2.50 This is the expected profit of one game. To get the expected profit of 100 games, multiply the expected value by 100. $2.50×100=$250.00

Jamie is practicing free throws before her next basketball game. The probability that she makes each shot is 0.6. If she takes 10 shots, what is the probability that she makes exactly 7 of them?

0.215. Binompdf (10,0.6,7) 2nd vars .. list binompdf enetr details.

Give the numerical value of the parameter n in the following binomial distribution scenario. The probability of buying a movie ticket with a popcorn coupon is 0.597 and without a popcorn coupon is 0.403. If you buy 18 movie tickets, we want to know the probability that no more than 13 of the tickets have popcorn coupons. Consider tickets with popcorn coupons as successes in the binomial distribution. Do not include n= in your answer.

18 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, the total number of trials, or movie tickets, is n=18.

A group of 150 students in a high school were asked if they prefer texting or calling. Male Female Total Texting 22 58 80 Calling 8 62 70 Total 30 120 150 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.

4/15 In the male row, there are 30 participants total, and 8 of them prefer calling, therefore the probability is 8/30=4/15.

Is the statement below true or false?Independent is the property of two events in which the knowledge that one of the events occurred does not affect the chance the other occurs.

true

At a certain fast food restaurant, 81.3% of the customers order items from the value menu. If 15 customers are randomly selected, what is the probability that at least 10 customers ordered an item from the value menu?

0.954 this is a cumulative binomial probability. In this case. To determine the probability 1. First press 2ND and then VARS for the DISTR menu.select BINOMCDF( 15, 0.813 9), probability is 0.0457563736, = 0.046 rounded to three decimal places. Subtracting this probability from 1 gives the probability 1−0.046=0.954 that at least 10 customers ordered an item from the value menu.

The probability that a car has a certain factory defect is 8/25. The probability that a car has a certain factory defect and needs an oil change is 7/50. 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.

7/16 P(B|A) = P(B AND A)/P(A) plugging in the values P(B|A)= (7/50) / (8/25) ≈ 7/16 working out the fraction: reciprocal = 7/50 x 25/8 = (7/2) x (1/8) = 7/16

Let U be the event that a randomly chosen person lives in the United States. Let E be the event that a randomly chosen person lives in the Eastern Time Zone. Place the correct event in each response box below to show: Given that the person lives in the United States, the probability that a randomly chosen person lives in the Eastern Time Zone. P( | )

P( E | U ) 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 lives in the Eastern Time Zone given that the person lives in the United States, so the correct answer is P(E|U).

A softball pitcher has a 0.42 probability of throwing a strike for each pitch. If the softball pitcher throws 20 pitches, what is the probability that exactly 7 of them are strikes? Round your answer to three decimal places.

P(X=7)=0.150 In this case, we have that x=7, n=20, and p=0.42. Thus, the probability can be found by, Binompdf (20, .42, 7)= 0.150

Identify the parameter p in the following binomial distribution scenario. The probability of winning an arcade game is 0.403 and the probability of losing is 0.597. If you play the arcade game 24 times, we want to know the probability of winning more than 7 times.

P= 0.403 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is winning a game, so p=0.403.

Jackie is practicing free throws after basketball practice. She makes a free throw shot with probability 0.7. She takes 20 shots. We say that making a shot is a success. What are p, q, and n in this context?

p=0.7, q=0.3, n=20 Remember that p is the probability of success, which is the probability of making a shot, 0.7. The probability of failure is q=1−p=0.3. n is the number of repetitions, which is 20.

If the probability of a student selling brownies is 0.2, the probability of selling rootbeer floats is 0.4, and the probability of selling brownies and rootbeer floats is 0.01. what is the probability of a student selling brownies or rootbeer floats?

0.59 selling brownies = 0.2 rootbeer floats = 0.4, selling brownies & rootbeer floats = 0.01 P(A OR B)=P(A)+P(B)−P(A AND B) plug in the known values: P(A OR B)=0.2+0.4−0.01=0.59 So the probability of selling brownies or rootbeer floats is0.59.

If A and B are events with P(A)=0.2, P(A OR B)=0.62, and P(A AND B)=0.18, find P(B).

0.6

Give the numerical value of the parameter p in the following binomial distribution scenario.A softball pitcher has a 0.692 probability of throwing a strike for each pitch and a 0.308 probability of throwing a ball. If the softball pitcher throws 29 pitches, we want to know the probability that more than 17 of them are strikes.Consider strikes as successes in the binomial distribution. Do not include p= in your answer.

0.692 The parameters p and n represent the probability of success on any given trial and the total number of trials, respectively. In this case, success is a strike, so p=0.692.

In a coin and die game, you roll a fair six-sided die and toss a coin. If you roll a 6 and toss a tails, you gain $110. Otherwise, you lose $10. If you were to play the game 45 times, how much money can you expect to gain or lose? Round the expected value for one game to the nearest cent. Enter an expected loss as a negative number.

$0.00 Since there are 6 possible outcomes from rolling the dice and 2 possible outcomes from flipping the coin: 6×2=12 equally likely outcomes. rolling a 6 and tossing a tails= 1/12 to win $110 is other outcomes= 11/12 lose - $10 E 1× P 1 + E2 × P2 =$110×1/12 + (−$10)×11/12 ≈$9.17 − $9.17 =$0.00 You can expect to gain $0.00 in one game. This means the game is fair. For any number of games you play, you can expect to gain $0.00.

You toss a coin three times. If you toss heads exactly two times, you win $2. If you toss heads all three times, you win $8. Otherwise, you lose $3. What is the expected payout for one round of this game? Round your answer to the nearest cent. Enter an expected loss as a negative number.

$0.25 The sample space for tossing three coins= 8 chances {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}. possible outcomes of tossing a coin three times: three heads is 1/8 = winning $8 two heads 3/8 = winning $2 the rest = 4/8 = lose $3 Multiplying the events by the probabilities, we get the expected profit for each time the game is played. = E1×P1 +E2×P2 + E3×P3 =$8×1/8 + $2×3/8 + (−$3)×1/2 =$1.00+$0.75−$1.50 =$0.25

Suppose you select a random card from a standard deck of 52 cards and no jokers. If you draw a face card (king, queen, or jack), you win $5. If you draw an ace, you win $20. If you draw any other card, you lose $2. What is the expected profit from any one draw? Round your answer to the nearest cent. Enter an expected loss as a negative number. Note: There are 12 face cards and 4 aces in a standard deck of cards.

$1.31 out of all 52 playing cards; 12 are face cards 12/52 or 3/13 = win $5 4 aces in a 4/52 or 1/13 = win $20 36 remaining 36/52 or 9/13 = lose $2. Multiply the outcome values by the probabilities for the expected profit from one game. expected value = (E1×P1)+(E2×P2)+(E3×P3) = $5×313 + $20×113 + (−$2)×913 ≈$1.15 + $1.54 − $1.38 ≈$1.31

A flood insurance company sells policies for $700 per year. If a customer's house is flooded, they are given $250,000 for repairs. The insurance company has calculated the chances that a house is flooded to be 12,500 over the year. How much money can the insurance company expect to make with each policy sold? Round your answer to the nearest cent. Enter an expected loss as a negative number.

$600 The probability of a house not being flooded is 2,4992,500 with a net gain of $700. The probability of a house being flooded is 12,500. In this case, the insurance company is paid $700 and pays out $250,000 for a net loss of −$249,300. = E1×P1 + E2×P2 =$700 × 2,499/2,500 + (−$249,300) × 1/2,500 =$699.72 − $99.72 =$600

Suppose you play a game where you toss three fair coins. If you get three tails, you win $10. Otherwise, you lose $2. If you were to play this game 15 times, how much would you expect to gain or lose? Do not round until the final answer. Enter an expected loss as a negative number.

-$7.50. The sample space for tossing three coins is as follows, where H represents heads and T= 8 chances {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}. Three tails is 1/8 = winning $10 the rest is 7/8. = lose -$2 = E1×P1 + E2×P2 =$10 × 1/8 + (−$2) × 7/8 = $1.25 − $1.75 = −$0.50 multiply by playing the game 15 times: = −$0.50 × 15 = −$7.50 If you play the game 15 times, you'll lose $7.50.

In a large city's recent mayoral election, 131,506 out of 309,153 registered voters actually turned out to vote. If 20 registered voters are randomly selected, find the probability that 7 of them voted in the mayoral election. Use a TI-83, TI-83 plus, or TI-84 calculator to find the probability.

.146

The probability of winning on an arcade game is 0.612. If you play the arcade game 21 times, what is the probability of winning no more than 10 times?

.146 Binomcdf (21,.612,10)= .146

A roulette wheel has 38 slots, numbered 1 to 36, with two additional green slots labeled 0 and 00. 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?

0.176 green ball= 2 chances out of 38 balls in total. 2/38 = 1/19. calc: 2nd Vars - Distribu - BINOMCDF (20, 1/19,1) = 0.716 rounded to three decimal places

On a multiple-choice quiz, a correct answer is awarded 4 points, but an incorrect answer costs the student 1 point. Suppose each question on the quiz has 4 choices and no question has multiple correct answers. If a student were to guess on every question, what is the number of points the student should expect to get per question? Do not round the answer.

0.25 1 of the 4 choices on the quiz question is correct, so the probability of guessing a correct answer is 1/4 earning 4 points is 1/4. For the other 3 choices, f losing 1 point is 3/4. =E1 × P1 + E2 ×P2 = 4 × 1/4 + (−1) × 3/4 = 1 − 0.75 = 0.25

If A and B are events with P(A)=0.9, P(B)=0.3, P(A OR B)=0.95, find P(A AND B)

0.25 formula: P(A OR B)=P(A)+P(B)−P(A AND B) Now, rearranging this, we find that P(A AND B)=P(A)+P(B)−P(A OR B) Plugging in the known values, we find P(A AND B)=0.9+0.3−0.95=0.25

what is P(B)? If A and B are events: P(A)=0.5 P(A OR B)=0.65 P(A AND B)=0.15

0.3 addition rule for probabilities: P(A OR B)=P(A)+P(B)−P(A AND B) Now, rearranging this, we find that P(B)=P(A OR B)+P(A AND B)−P(A) Plugging in the known values, we find P(B)=0.65+0.15−0.5 =0.3

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

0.361 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 find P(B AND A)= (0.84)(0.43 )= 0.3612

find P(B): If A and B are events: P(A)=0.6 P(A OR B)=0.98 P(A AND B)=0.02

0.4 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 that P(B)=P(A OR B)+P(A AND B)−P(A) Plugging in the known values, we find P(B)=0.98+0.02−0.6=0.4

Employees at a company are randomly assigned certain shifts during the busy holiday season. Let A = the shift between the hours of 8 a.m. and 12 p.m., and B =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?

Solution: 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%.

Harrison is a sports statistician and is interested in the number of strikeouts by a starting pitcher over the past five years for a certain professional baseball league. He randomly selects 471 games and records the number of strikeouts by one of the starting pitchers in the game, where the pitcher for the home or visiting team is selected at random. Harrison records the total number of strikeouts by the starting pitcher in that game, x, and the probability of each value, P(x), as shown in the table provided. Find the mean and the standard deviation of the probability distribution Round the mean and standard deviation to three decimal places. Number of strikeouts x P(x) 0 0.016 1 0.045 2 0.059 3 0.083 4 0.098 5 0.117 6 0.130 7 0.115 8 0.100 9 0.074 10 0.047 11 0.038 12 0.034 13 0.023 14 0.013 15 0.006 16 0.002

The mean: 6.339, standard deviation : 3.246. Use the calculator to determine the mean and standard deviation of the probability distribution. Go to STAT and then press ENTER. Clear the existing values in lists L1 and L2. Enter the values of X in list L1 and the corresponding probabilities in list L2. Press STAT, go to CALC, and press ENTER. Enter list L1 using 2nd and 1. Then enter a comma. Enter list L2 using 2nd and 2. Press ENTER. Identify the mean and standard deviation in the output. The mean μ, identified by x¯: 6.339, standard deviation σ, identified by σx : 3.246.