PSYC 245 Exam 3

This event is impossible. It can never occur.

0

Probability works not by compensating for imbalances but by overwhelming them. Suppose that the first 10 tosses of a coin give 10 tails and that tosses after that are exactly half heads and half tails. (Exact balance is unlikely, but the example illustrates how the first 10 outcomes are swamped by later outcomes.) (Use decimal notation. Give your answers as exact numbers.) What is the proportion of heads after the first 10 tosses?

0 0/10=0

You flip a coin for which the probability of heads is 0.5 and the probability of tails is 0.5. If the coin comes up heads, you win $1. Otherwise, you lose $1. Determine the expected value of your winnings. (Use decimal notation. Give your answer as an exact number.)

0 The probability of earning $1 is 0.5, the probability of losing $1 is 0.5. Multiply the outcomes by their probabilities and add the products to obtain the expected value. 1⋅0.5−1⋅0.5=0 Thus, the average winnings in the long run are $0.

This event is very unlikely, but it will occur once in a while in a long sequence of trials.

0.01

A roulette wheel has 38 slots, numbered 0, 00, and 1 to 36. The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and, at the same time, rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. What is the probability of any one of the 38 possible outcomes to three d

0.026 1/38=0.026

A roulette wheel has 38 slots, numbered 0, 00, and 1 to 36. The slots 0 and 00 are colored green, 1818 of the others are red, and 1818 are black. The dealer spins the wheel and, at the same time, rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. What is the probability of any one of the 38 possible outcomes to thr

0.026 1/38=0.026

The table of random digits (Table A) was produced by a random mechanism that gives each digit probability 0.10.1 of being a 0.0. What proportion of the 400 digits in lines 125 to 134 in the table are 0s? This proportion is an estimate, based on 400 repetitions, of the true probability, which in this case is known to be 0.1. (Use decimal notation. Give your answer to two decimal places.)

0.08 First, calculate the number of 0s in lines 125 to 134 in the table. The number of 0s in lines 125 to 134 is 33. Then divide the number of 0s in lines 125 to 134 by the number of digits in these lines. 33/400=0.08 Therefore, the proportion of 0s in lines 125 to 134 equals 0.08. This is an estimate of the true probability based on 400 repetitions.

Suppose that 80% of American Airline flights are on time (this is approximately the percentage of American Airline flights on time in November 2018). You check 10 American Airline flights chosen at random. What is the probability that all 10 are on time? Simulate 25 repetitions, starting at line 129 of Table A. Use the assignment from the previous question: 0 to 7 will represent flights that are on time, and 8 and 9 will represent flights that are not on time. What is your estimate 𝑝 of the p

0.12 Starting at Line 129 of the table, the first 10 digits are 36759 5898436759 58984 Since either 8 or 9 represents a flight that is not on time, there are 4 such digits in this set. Therefore, there are 6 flights simulated to be on time in this sample. Repeat this with the next 24 sets of 10 digits, wrapping around to the next line as needed. The counts obtained from the 25 simulated samples are shown in the table. Determine the number of times all 10 flights were on time out of the 25. This gives 3 out of the 25 simulations, thus the probability all 1010 flights are on time is 3/25=0.12

A couple plan to have three children. There are eight possible arrangements of girls and boys. For example, GGB means the first two children are girls and the third child is a boy. All eight arrangements are (approximately) equally likely. Write down all eight arrangements of the sexes of three children. What is the probability of any one of these arrangements? Enter your answer to three decimal places.

0.125 1/8=0.125

Sue Bird of the WNBA Seattle Storm makes 39% of her three‑point shots. In an important game, she shoots four three‑point shots late in the game and misses all of them. The fans think she was nervous, but the misses may simply be chance. Let's shed some light by estimating a probability. Simulate 50 repetitions of the four three‑point shots and record the number missed on each repetition. Use Table A, starting at Line 135. Use the probability assignment where 00 to 38 represents a successfu

0.18 To simulate 50 repetitions of four three‑point shots, go to Line 135 of Table A. To simuluate four shots, you need 4 pairs of digits. Thus, 8 digits are needed for the first repetition. The first eight digits of Line 135 are 6692555666925556 Separate into 4 pairs. Any pair that is between 0000 and 38, that counts as a successful shot. Any pair from 39 to 99, that counts as a miss. In this simulation of four shots, there are zero pairs between 00 and 38 (inclusive). The number of pairs corresponds to the number of successful free‑throw shots. Continue with the next pair. The next eight digits of Line 135 are 58391007 Separate into 4 pairs. Any pair that is between 00 and 38, that counts as a successful shot. Any pair from 39to 99, that counts as a miss. In this simulation of four shots, there are two pairs between 00 and 38 (inclusive). The number of pairs corresponds to the number of successful free‑throw shots. If none of the pairs of digits fall in the range 00 to 38,

Select a first‑year college student at random and ask how many hours during a typical week they spent studying or doing homework during their last year in high school. Probabilities for the outcomes are Less than one hour - 0.10 1 to 5 hours - 0.50 6 to 10 hours - 0.20 More than 10 hours - ? What must be the probability that a randomly chosen first‑year college student says they spent more than 10 hours per week studying or doing homework during their last year in high school? (Use decimal n

0.2 𝑃(Less than 10 hours)=0.1+0.5+0.2=0.8 1-0.8=0.2

Select a first‑year college student at random and ask how they would characterize their political views. Here are the probabilities, based on proportions from a large sample survey in 2016 of first‑year students: Far left: 0.04 Liberal: 0.31 Middle of the road: 0.43 Conservative: 0.20 Far right: 0.02 What is the probability that a first‑year student had political views that are either conservative or far right? (Use decimal notation. Give your answer as an exact number.)

0.22 𝑃(conservative or far right)=𝑃(conservative)+𝑃(far right)=0.20+0.02=0.22

The game of craps is played with two dice. The player rolls both dice and wins immediately if the outcome (the sum of the faces) is 7 or 11. If the outcome is 2, 3, or 12, the player loses immediately. If he rolls any other outcome, he continues to throw the dice until he either wins by repeating the first outcome or loses by rolling a 7. Use Table A, beginning at line 115, to simulate 10 plays and estimate the probability that the player wins. (Use decimal notation. Give your answer as an exact

0.3 For a single fair die, random digits 1-6 represent the result of the roll. If you get 0 or 7-9, ignore and move to the next digit in the table of random numbers. To simulate a roll of two fair dice, repeat the single die process twice and add to find the total. Start at line 115 of Table A. The player's results in 10 plays would be Win, Loss, Loss, Loss, Loss, Win, Loss, Loss, Win, Loss. Therefore, the player would win 3 times. To determine the estimated probability that the player wins, divide the number of wins by the total number of plays. 3/10=0.3 Thus, the estimated probability that the player wins equals 0.3. Note that more repetitions would be likely to converge around the true probability of winning.

A roulette wheel has 38 slots, numbered 0, 00, and 1 to 36. The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and, at the same time, rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. The slot numbers are laid out on a board on which gamblers place their be

0.316 The column with multiples of 3 consists of a total of 12 slots because there are 12 multiples of 3 between 1 and 36. The probability of winning the column bet is 12/38=0.316.

A couple plan to have three children. There are eight possible arrangements of girls and boys. For example, GGB means the first two children are girls and the third child is a boy. All eight arrangements are (approximately) equally likely. What is the probability that the couple's children are two girls and one boy? Enter your answer to three decimal places.

0.375 3/8=0.375

Sociologists studying social mobility in the U.S. find that the probability that someone who began their career in the bottom 10% of earnings remains in the bottom 10% fifteen years later is 0.59. What is the probability that such a person moves to one of the higher income classes fifteen years later? (Use decimal notation. Give your answer as an exact number.)

0.41 1-0.59=0.41 The probability that an event occurs and the probability that it does not occur always add to 1. Subtract the probability that a person remains in the bottom 10% of earnings from 1 to obtain the required probability.

Choose an acre of land in Canada at random. The probability is 0.38 that it is forest and 0.07 that it is agricultural. What is the probability that it is either forest or agricultural? Enter your answer to two decimal places.

0.45 0.38+0.07=0.45

Probability works not by compensating for imbalances but by overwhelming them. Suppose that the first 10 tosses of a coin give 10 tails and that tosses after that are exactly half heads and half tails. (Exact balance is unlikely, but the example illustrates how the first 10 outcomes are swamped by later outcomes.) (Use decimal notation. Give your answers as exact numbers.) What is the proportion of heads after 100 tosses if half of the last 90 produce heads (45 heads)?

0.45 45/100=0.45

Government data assign a single cause for each death that occurs in the United States. Data from 2016 show that the probability is 0.23 that a randomly chosen death was due to heart disease and 0.22 that it was due to cancer. What is the probability that a death was due either to heart disease or to cancer? Enter your answer to two decimal places.

0.45 The probability that a death was due either to heart disease or to cancer equals the sum of the probability of death due to heart disease and the probability of death due to cancer. 0.23+0.22=0.45

A roulette wheel has 38 slots, numbered 0, 00, and 11 to 36. The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and, at the same time, rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. If you bet on "red," you win if the ball lands in a red slot. What is th

0.474 18/38=0.474

A roulette wheel has 38 slots, numbered 0, 00, and 1 to 36. The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and, at the same time, rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. A friend tells you that the odds that a bet on "red" will win are 10 to 9

0.474 Your friend is right because the probability of an event with 10:9 odds equals the probability of winning on "red". The odds of 10 to 9 means that out of 19 spins, you would predict 10 loses and 9 wins. The probability associated with these odds would be 9/19 or 0.474. The probability of winning on "red" is 18/38 or 0.474. Therefore, your friend is right because the probability of an event with 10 : 9 odds equals the probability of winning on "red".

Probability works not by compensating for imbalances but by overwhelming them. Suppose that the first 10 tosses of a coin give 10 tails and that tosses after that are exactly half heads and half tails. (Exact balance is unlikely, but the example illustrates how the first 10 outcomes are swamped by later outcomes.) (Use decimal notation. Give your answers as exact numbers.) What is the proportion of heads after 1000 tosses if half of the last 990 produce heads?

0.495 495/1000=0.495

Probability works not by compensating for imbalances but by overwhelming them. Suppose that the first 10 tosses of a coin give 10 tails and that tosses after that are exactly half heads and half tails. (Exact balance is unlikely, but the example illustrates how the first 10 outcomes are swamped by later outcomes.) (Use decimal notation. Give your answers as exact numbers.) What is the proportion of heads after 10,000 tosses if half of the last 9990 produce heads?

0.4995 4995/10000=0.4995

Winning numbers for the "Straight" DC‑4 lottery game are reported on television and in local newspapers. You pay $1.00 and pick a four‑digit number. The state chooses a four‑digit number at random and pays you $5000 if your number is chosen. What are the expected winnings from a $1.00 Straight DC‑4 wager? Please round your final answer to the nearest cent.

0.50 The expected winnings are the expected value from the probability model. First, determine the number of possible outcomes for a four‑digit number. Because each digit is equally likely in a lottery game, there will be 10×10×10×10=10000 possible four‑digit outcomes. Because the lottery selects only one four‑digit number, the probability of winning is 1/10000, or 0.0001. The probability of not winning is 1−.0001=0.9999. The table shows the probability model for the "Straight" DC‑4 lottery game. To calculate the expected winnings, multiply each possible outcome by its corresponding probability. Expected winnings =($0×0.9999)+($5000×0.0001)=$0.50=($0×0.9999)+($5000×0.0001)=$0.50 In the long run, the average winnings with a $1.00 wager is $0.50.

Government data assign a single cause for each death that occurs in the United States. Data from 2016 show that the probability is 0.23 that a randomly chosen death was due to heart disease and 0.22 that it was due to cancer. What is the probability that the death was due to some other cause? Enter your answer to two decimal places.

0.55 The probability that the death was due to some other cause equals one minus the probability that the death was due to cancer or heart disease. From the previous question, the probability that a death was due to cancer or heart disease is 0.55.is 0.55. 1−0.45=0.55

This event will occur more often than not

0.6

Choose an acre of land in Canada at random. The probability is 0.38 that it is forest and 0.07 that it is agricultural. What is the probability that the acre chosen is not forested? Enter your answer to two decimal places.

0.62 1-0.38=0.62

The Census Bureau gives this distribution for the number of a family household's related children under the age of 18 in American households in 2016: In this table, 4 actually represents four or more (the Census Bureau does not provide separate information on households with 5,6, or more children under the age of 18). But for purposes of this exercise, assume that it means only households with exactly four children under the age of 18.18. This is also the probability distribution for the number

0.76 expected value=0⋅0.58+1⋅0.18+2⋅0.16+3⋅0.06+4⋅0.02=0.76 So, the expected value of children under 1818 is 0.760.76 per household.

Here is another basic rule of probability: if several events are independent, the probability that all of the events happen is the product of their individual probabilities. We know, for example, that a child has probability 0.49 of being a girl and probability 0.51 of being a boy and that the sexes of successive children are independent. So the probability that a couple's two children are two girls is (0.49)(0.49)=0.2401. A couple plan to have three children. You can use this multiplication rul

0.8762 0.1274+0.1274+0.1225+0.1274+0.1225+0.1225+0.1176=0.8672

An American roulette wheel has 38 slots, of which 18 are black, 18 are red, and 2 are green. When the wheel is spun, the ball is equally likely to come to rest in any of the slots. Gamblers bet on roulette by placing chips on a table that lays out the numbers and colors of the 38 slots in the roulette wheel. The red and black slots are arranged on the table in three columns of 12 slots each. A $1 column bet wins $3 if the ball lands in one of the 12 slots in that column. What is the expected amo

0.95 The expected winnings are the expected value from the probability model. Because the roulette wheel has 38 slots, the probability of winning with one of the columns is 12/38.. The probability of not winning is 1−12/38=26/38. Expected winnings =($0×26/38)+($3×12/38)=$0.95

Select a first‑year college student at random and ask how they would characterize their political views. Here are the probabilities, based on proportions from a large sample survey in 2016 of first‑year students: Far left: 0.04 Liberal: 0.31 Middle of the road: 0.43 Conservative: 0.20 Far right: 0.02 What is the probability that a randomly chosen first‑year college student had political views that are not far left? (Use decimal notation. Give your answer as an exact number.)

0.96 𝑃(not far left)=1−𝑃(far left)=1−0.04=0.96

An opinion poll asks an SRS of 501 teenagers, "Generally speaking, do you approve or disapprove of legal gambling or betting?" Suppose that, in fact, exactly 50% of all teenagers would say Approve if asked. (This is close to what polls show to be true.) The poll's statisticians tell us that the sample proportion who say Approve will vary in repeated samples according to a Normal distribution with mean 0.50. and standard deviation about 0.022. Using the 68-95-99.7 rule, what is the probability th

0.975 Two standard deviations, 0.022, above the mean of 0.5 is 0.544 or 54.4%. According to the 68-95-99.7 rule, 95% of teenagers will lie within two standard deviations from the mean. Also, 2.5% of teenagers will lie more than two standard deviations below the mean. Thus, the probability that fewer than 54% of teenagers say Yes is 97.5%=95%+2.5% or 0.975.

This event is certain. It will occur on every trial.

1

Select a first‑year college student at random and ask how they would characterize their political views. Here are the probabilities, based on proportions from a large sample survey in 2016 of first‑year students: Far left: 0.04 Liberal: 0.31 Middle of the road: 0.43 Conservative: 0.20 Far right: 0.02

1 0.04+0.31+0.43+0.20+0.02=1 Since the probabilities in the table are numbers between 0 and 1 and all possible outcomes have probability 1, the table provide a legitimate assignment of probabilities to outcomes.

The probability of a head in tossing a coin is 1‑in‑2. This means that as we make more tosses, the proportion of heads will eventually get close to 0.5. It does not mean that the count of heads will get close to one‑half the number of tosses. (Use decimal notation. Give your answers as whole positive numbers.) The proportion of heads is 0.49 in 100 tosses. What is the difference between the number of heads and half the number of tosses?

1 0.49x100=49 50-49

When there are few data, we often fall back on personal probability. There had been just 24 space shuttle launches, all successful, before the Challenger disaster in January 1986. The shuttle program management thought the chances of such a failure were only 11 in 100,000. Suppose 1 in 100,000 is a correct estimate of the chance of such a failure. If a shuttle was launched every day, about how many failures would one expect in 300 years? Round your answer to the nearest whole number.

1 300(365)(1/100,000)=1.095

Matt has lots of experience taking multiple‑choice exams without doing much studying. He is about to take a quiz that has 10 multiple‑choice questions, each with four possible answers. Here is Matt's personal probability model. He thinks that in 75% of questions, he can eliminate one answer as obviously wrong; then he guesses from the remaining three. He then has probability 1‑in‑3 of guessing the right answer. For the other 25% of questions, he must guess from all four answers, with pro

1 The tree diagram for one question has two branches on the first stage; one for the 75% of questions Matt can eliminate one answer as obviously wrong and one branch for the 25% of questions Matt must guess from all four answers. From each of these branches, there will be two more branches in stage 2: one for a correct answer and one for an incorrect answer. A simulation of 10 questions using two-digit numbers gives a score of 11 when starting on line 103. Reading across line 103 and using the labeling given in the question and tree diagram, the following results are obtained. Question 1: The stage 1 digits are 45 and the stage 22 digits are 46. The 45 tells you whether Matt can eliminate one answer choice as obviously wrong. The 46 tells you, based on the previous digits, whether Matt answered correctly or not. The two pairs of digits translate to incorrectly answering the question. Question 2: The stage 1 digits are 77 and the stage 2 digits are 17, which translate to correctly answe

The Census Bureau gives this distribution for the number of people in American households in 2016. Note: In this table, 7 actually represents households of size 7 or greater. But for purposes of this exercise, assume that it means only households of size exactly 7. This is also the probability distribution for the size of a randomly chosen household. The expected value of this distribution is the average number of people in a household. What is this expected value?

2.44 The expected value is found by multiplying each outcome by its probability and then adding all the products. expected value=1⋅0.28+2⋅0.35+3⋅0.15+4⋅0.13+5⋅0.06+6⋅0.02+7⋅0.01=2.44 The expected value is 2.44 people in a household.

The probability of a head in tossing a coin is 1‑in‑2. This means that as we make more tosses, the proportion of heads will eventually get close to 0.5. It does not mean that the count of heads will get close to one‑half the number of tosses. (Use decimal notation. Give your answers as whole positive numbers.) The proportion of heads is 0.4969 in 10,000 tosses. What is the difference between the number of heads and half the number of tosses?

31 0.4969x10000=4969 5000-4969=31

Suppose that 44% of all adults think that airline travel is safer than driving. An opinion poll plans to ask a simple random sample (SRS) of 1021 adults about airplane safety. The proportion of the sample who think that airline travel is safer than driving will vary if we take many samples from the same population. The sampling distribution of the sample proportion is approximately Normal with mean of 0.44 and standard deviations about 0.016. Sketch this Normal curve and use it to answer the fol

5% Begin by determining the number of standard deviations that is represented by the area within 3.2% of the mean. That is, the area between 40.8% (=44%−3.2%) and 47.2% (=44%+3.2%). Since the standard deviation is equal to 0.016 or 1.6% and given that 2×1.6% equals 3.2%, these percentages are within 2 standard deviations of the mean. Based on the 68‑95‑99.7 rule, 95% of the population is within two standard deviations of the truth 44%. That is, the probability is 95% that between 40.8% and 47.2% of the people think that airline travel is safer than driving. Therefore, the probability of getting a sample that misses the truth (44%) by more than 3.2% is equal to the area outside the area within two standard deviations of the mean. This can be obtained by taking the difference between the total area of the normal curve (which is equal to 1) and the 0.95: 1−.95=.05 or 5%.

The probability of a head in tossing a coin is 1‑in‑2. This means that as we make more tosses, the proportion of heads will eventually get close to 0.5. It does not mean that the count of heads will get close to one‑half the number of tosses. (Use decimal notation. Give your answers as whole positive numbers.) The proportion of heads is 0.493 in 1000 tosses. What is the difference between the number of heads and half the number of tosses?

7 0.493x1000=493 500-493=7

Suppose that 44% of all adults think that airline travel is safer than driving. An opinion poll plans to ask a simple random sample (SRS) of 1021 adults about airplane safety. The proportion of the sample who think that airline travel is safer than driving will vary if we take many samples from the same population. The sampling distribution of the sample proportion is approximately Normal with mean of 0.44 and standard deviations about 0.016. Sketch this Normal curve and use it to answer the fol

97.5 (0.408−0.44)/0.016=-2 This means 40.8% is two standard deviations below the mean of 44%. Using the 68‑95‑99.7 rule, 95% of the population is between 40.8% (two standard deviations below the mean) and 47.2% (two standard deviations above the mean). This means that 5% is outside that range with half below 40.8% and half above 47.2%. Therefore, the percent above 40.8% includes the 95% in between 2 standard deviations, as well as the 5%/2=2.5% above the upper end point of 47.2%. 95%+2.5%=97.5% Thus, there is a 97.5% chance of being above 40.8%.

Which of the following is true of probability?

A probability of 0 means the outcome never occurs. A probability of 1 means the outcome always occurs.

Recent surveys of college students suggest that 40% are liberal, 40% are middle‑of‑the‑road (MOTR), and 20% are conservative. How would you assign digits for a simulation to determine the political affiliation (liberal, middle‑of‑the‑road, or conservative) of a college student chosen at random from the population of all college students? Select the correct assignments.

Assign 0, 3, 5, 7 to liberal, 1, 4, 8, 9 to MOTR, and 2, 6 to conservative. Assign 1, 2, 3, 4 to liberal, 5, 6, 7, 8 to MOTR, and 9, 0 to conservative.

Select a first‑year college student at random and ask how many hours during a typical week they spent studying or doing homework during their last year in high school. Probabilities for the outcomes are Less than one hour - 0.10 1 to 5 hours - 0.50 6 to 10 hours - 0.20 More than 10 hours - ? To simulate the responses of randomly chosen first‑year college students, how would you assign digits to represent the four possible outcomes listed? Select the correct assignment.

Assign 3 to "less than 1 hour," 2, 7, 9, 6,8 to "1 to 55 hours," 1, 4 to "6 to 10 hours," and 5, 0 to "more than 10 hours."

Sue Bird of the WNBA Seattle Storm makes 39% of her three‑point shots. In an important game, she shoots four three‑point shots late in the game and misses all of them. The fans think she was nervous, but the misses may simply be chance. Let's shed some light by estimating a probability. Describe how to simulate four independent three‑point shots with the same probability 0.39 for a successful shot.

Assign two‑digit pairs 00 to 38 to be a successful shot, and 39 to 99 to be a miss. Select a Line of Table A, such as Line 101, and select the first two‑digit pair to assign a successful shot or a miss. Move on to the next pair by skipping over both used digits, and so on for four total pairs.

Sue Bird of the WNBA Seattle Storm makes 39% of her three‑point shots. In an important game, she shoots four three‑point shots late in the game and misses all of them. The fans think she was nervous, but the misses may simply be chance. Let's shed some light by estimating a probability. Describe how to simulate a single three‑point shot if the probability of making each shot is 0.39.

Assign two‑digit pairs 00 to 38 to be a successful shot, and 39 to 99 to be a miss. Select a Line of Table A, such as Line 125, and select the first two digits. If this number is between 00 and 38, it is a successful free‑throw. If this number is between 39 and 99, it is a missed free‑throw. The assignment of 00 to 38 for a successful shot constitutes 39 of 100 values (01 to 38 is 38 values plus 0000 is one more to make 39), and thus satisfies the required probability. The remaining values 39 to 99 contain all other possible two‑digit values. This makes the sum of all the probabilities (success and failure) add up to 1. An assignment of 00 to 39 is actually 40 values and so does not represent the required probability. An assignment of single digits 0 to 4 is 5 values of out 10 and so does not represent the required probabilty. An assignment of 000 to 390 is 391 values (001 to 390 is 390 values plus 000 is one more making 391) of out 1000 values. While this is close, it is not e

Suppose that 80% of American Airline flights are on time (this is approximately the percentage of American Airline flights on time in November 2018). You check 10 American Airline flights chosen at random. What is the probability that all 10 are on time? Give a probability model for checking 10 flights chosen independently of each other. Select the best description of the model.

Each flight has a probability of 0.80 of being on time, and 0.20 chance of being delayed. Each flight is independent of each other. 0.80+0.20=1 The problem states that the outcomes are independent, so that the outcome of one flight does not affect the outcome of another. This is reasonable in a randomly selected sample even though, in the real world, cancelled or delayed flights due to weather might occur in bunches. This is because the random sample is not choosing consecutive flights.

Suppose that 80% of American Airline flights are on time (this is approximately the percentage of American Airline flights on time in November 2018). You check 10 American Airline flights chosen at random. What is the probability that all 10 are on time? Select the assignment of random digits that can validly represent the probability distribution for this question.

Let 0 to 7 represent on time, and 8 and 9 represent delayed. For a single digit assignment, there are 1010 possible single digits. {0,1,2,3,4,5,6,7,8,9} If 8 and 9 are for the delayed situation, that represents the probability of 2/10 or 20%. Counting up the numbers 0 to 7 represent 8 digits, and therefore, they represent the probability 8/10 or 80%. Thus, this assignment correctly reflects the probability distribution needed.

The game of craps is played with two dice. The player rolls both dice and wins immediately if the outcome (the sum of the faces) is 7 or 11. If the outcome is 2, 3, or 12, the player loses immediately. If he rolls any other outcome, he continues to throw the dice until he either wins by repeating the first outcome or loses by rolling a 7. Explain how to simulate the roll of a single fair die. (Hint: Just use digits 1 to 6 and ignore the others.) Then explain how to simulate a roll of two fair di

Let 1-6 represent the result of a single die. If 0 or 7-9, ignore and move to the next digit. To simulate a roll of two fair dice, repeat the single die process twice and sum to find the total.

The figure displays several assignments of probabilities to the six faces of a die. We can learn which assignment is actually correct for a particular die only by rolling the die many times. However, some of the assignments are not legitimate assignments of probability. That is, they do not obey the rules. Which model is legitimate?

Model 2 Model 2 is the only legitimate probability model. Each individual probability is between 0 and 1. The sum of all the probabilities is 1. Both of these are requirements for a legitimate probability model. Model 1 is not legitimate because the sum of all the probabilities is less than 1. Model 3 is not legitimate because the sum of all the probabilities is greater than 1. Model 4 is not legitimate because the individual probabilities are not between 0 and 1.

The baseball player Jose Altuve gets a hit about 31.6% of the time over an entire season. After he has failed to hit safely in nine straight at‑bats, the TV commentator says, "Jose is due for a hit by the law of averages."

No. Knowing the outcome of previous trials does not change the probabilities of future trials. The probability of a hit does not change from one trial to the next. Therefore, knowing the outcome of previous trials does not change the probabilities of future trials. It is incorrect to conclude that future outcomes must make up for the imbalance of nine failed hits. Such a conclusion represents a misinterpretation of the law of averages. Since each trial is independent, knowing the outcome of previous trials does not change the probabilities of future trials. Therefore, although Jose failed the first nine hits, he is not more likely to fail future hits. Probabilities describe only what happens in the long run.

During the Michigan versus Ohio State football game in 2018, news media reported that Jim Harbaugh and Urban Meyer, the head coaches of Michigan and Ohio State, respectively, were born in the same hospital in Toledo, Ohio. That a pair of coaches from two arch rivals were also born in the same hospital was reported as an extraordinarily improbable event. Should this fact (that they are the head coaches of two famous arch rivals and both were born in the same hospital) surprise you?

No. The event is actually chance behavior. The event is actually chance behavior. Random phenomena are unpredictable in the short run, but exhibit predictable behavior in the long run. Given the necessary information, we could determine the probability that at least one pair of head coaches from two arch rivals were born in the same hospital. There is not enough information to determine the probability that the event will occur.

I toss a coin 1000 times and observe the outcome "heads" 481 times. Which of the following can be concluded from this result?

Our best estimate of the probability of heads is 0.481, but a longer sequence of flips should yield a better estimate of this probability.

A six‑sided die has two green and four red faces and is balanced so that each face is equally likely to come up. You must choose one of the following three sequences of colors: RGRRR RGRRRG GRRRRR Now, start rolling the die. You will win $25 if the first rolls give the sequence you chose. Because the $25 payoff is fixed, the most probable sequence has the highest expected value. Which sequence has the highest probability? Why?

RGRRR has the highest probability because the sequence is the shortest.

A six‑sided die has two green and four red faces and is balanced so that each face is equally likely to come up. You must choose one of the following three sequences of colors: RGRRR RGRRRG GRRRRR Now, start rolling the die. You will win $25 if the first rolls give the sequence you chose. In a psychological experiment, 63% of 260 students who had not studied probability chose the second sequence. Based on the discussion of "myths about chance behavior" in Chapter 17, explain why most students

Students assume that after a long sequence of one outcome, another outcome is "due". Since the second sequence has three Rs in a row, followed by a G, that outcome fits the assumption.

Elaine is enrolled in a self-paced course that allows three attempts to pass an examination on the material. She skims the online reading and then takes the exam. On the first try she has probability 0.4 of passing. If she fails on the first try, her probability on the second try increases to 0.5 because she learned something from her first attempt. If she fails on two attempts, the probability of passing on a third attempt is 0.6. She will stop as soon as she passes. The course rules force her

The first step is to identify the probability model and assign digits to match the model for the simulation. Starting with the first attempt, use Table A (selecting a random row) and the assigned digits to determine if Elaine passed on the first attempt. If she did not pass on the first attempt, move to the next value and see if she passed on the second attempt with the new model and digit assignments.

(b) After listening to you explain why red and black are still equally likely after five reds on the roulette wheel, the gambler moves to a poker game. He is dealt five straight red cards. He remembers what you said and assumes that the next card dealt in the same hand is equally likely to be red or black. Is the gambler right or wrong, and why?

The gambler is wrong because cards being dealt are not independent events. The gambler is wrong with the reasoning in the poker game because the cards being dealt are not independent events. Unlike the roulette wheel, each card dealt is not an independent event. A deck of cards has a finite number of trials (52 deals from one deck). After each card is dealt, the probability of getting either red or black changes. If 55 red cards are dealt, there are 21 red cards left and 26 black cards left (47 cards total). The probability of the next card being red would be 21/47, while the probability of getting a black card would be 26/47. There is a higher probability that the next card will be black than red. Five red cards could be dealt from a shuffled deck just by chance. It is true that the law of averages does not come into play until many cards have been drawn. However, the law of averages is only relevant with independent events, not dependent events like the cards being dealt.

The Census Bureau gives this distribution for the number of people in American households in 2016. Note: In this table, 7 actually represents households of size 7 or greater. But for purposes of this exercise, assume that it means only households of size exactly 7. Calculate the probability distribution for the household size lived in by individual people. Describe the shape of this distribution. What does this shape tell you about household structure?

The graph is skewed right, which means most families have 1 to 2 individuals.

Which of the following is true of any legitimate probability model?

The probabilities of the individual outcomes must be numbers between 0 and 1, and they must sum to exactly 1.

(a) A gambler knows that red and black are equally likely to occur on each spin of a roulette wheel. He observes five consecutive reds occur and bets heavily on black at the next spin. Asked why, he explains that black is "due by the law of averages." What is wrong with the gambler's reasoning?

The probability of getting a red or black does not change based on what was previously spun. The probability of getting a red or black does not change based on what was previously spun. The roulette wheel does not contain memory of what was previously spun. After each spin, the odds of getting a red or black remain the same, unaffected by whatever the previous spins were. The law of averages states that after a very large number of trials, the proportions become more stable. In this example, after a large number of roulette spins, the proportions of black and red spins should be about equal. The law of averages cannot be applied to a small number of samples, such as the five the gambler is thinking about. The law of averages can be in reference to proportions or averages, so either categorical or numeric data.

You read in a book on poker that the probability of being dealt four-of-a-kind (a hand containing 4 cards of the same value, such as four kings) in a five‑card poker hand is about 0.00024. Explain in simple language what this means.

The proportion of times the outcome of four‑of‑a‑kind will occur in many repetitions of dealing a five‑card poker hand is 0.00024.

Which of the following is true of density curves?

The total area under a density curve is 1. Areas under a density curve determine probabilities of outcomes. The Normal curve is a density curve.

Elaine is enrolled in a self-paced course that allows three attempts to pass an examination on the material. She skims the online reading and then takes the exam. On the first try she has probability 0.4 of passing. If she fails on the first try, her probability on the second try increases to 0.5 because she learned something from her first attempt. If she fails on two attempts, the probability of passing on a third attempt is 0.6. She will stop as soon as she passes. The course rules force her

Use the table that defines the digits that are successes and failures for each attempt and Table A to calculate the probability. The following is an example of how to use the table to calculate the probabilities for four repetitions. Starting at line 110, the first 9 digits are 384484878. The first 3 digits represent the first repetition. The first digit (3) counts as a success, the second digit (8) counts as a fail, but the third digit (4) counts as a success. Thus, the first repetition results in passing the exam. Next is the second repetition. The fourth digit (4) counts as a failure for the first attempt, the fifth digit (8) counts as a failure for the second attempt, but the sixth digit (4) counts as a success for the third attempt. Next is the third repetition. Finally, the seventh, eighth, and ninth (878) count as fails for the first, second, and third attempts respectively. This is the fourth repetition. Out of these four repetitions, there were three successes (or passes) and

To find the probability of any event,

add up the probabilities of the outcomes that make up the event.

The law of large numbers says that the mean outcome in many repetitions of a random phenomenon having numerical outcomes

gets close to the expected value as the number of repetitions increases.

Two random phenomena are independent if

knowing the outcomes of one does not change the probabilities for outcomes of the other.

I simulate a random phenomenon that has numerical outcomes many, many times. If I average together all the outcomes I observe, this average

should be close to the expected value of the random phenomenon.

A tree diagram

specifies a probability model in graphical form.

The expected value of a random phenomenon that has numerical outcomes is

the average of all possible outcomes, each weighted by its probability.

The probability of a specific outcome of a random phenomenon is

the proportion of times it occurs in very many repetitions of the phenomenon.

Here is another basic rule of probability: if several events are independent, the probability that all of the events happen is the product of their individual probabilities. We know, for example, that a child has probability 0.49 of being a girl and probability 0.51 of being a boy and that the sexes of successive children are independent. So the probability that a couple's two children are two girls is (0.49)(0.49)=0.2401. A couple plan to have three children. You can use this multiplication rul

𝑃(BBB)=0.51⋅0.51⋅0.51=0.132651=0.1327 𝑃(BBG)=0.51⋅0.51⋅0.49=0.127449=0.1274 𝑃(BGB)=0.51⋅0.49⋅0.51=0.127449=0.1274 𝑃(BGG)=0.51⋅0.49⋅0.49=0.122451=0.1225 𝑃(GBB)=0.49⋅0.51⋅0.51=0.127449=0.1274 𝑃(GBG)=0.49⋅0.51⋅0.49=0.122451=0.1225 𝑃(GGB)=0.49⋅0.49⋅0.51=0.122451=0.1225 𝑃(GGG)=0.49⋅0.49⋅0.49=0.117649=0.1176