Question 1

Suppose you roll three fair six-sided dice. What is the probability of each die showing an even number (i.e., you roll no odd numbers)?
A) 1/3
B) 1/6
C) 1/8
D) 1/2
E) 0

Accepted Answer

The probability of rolling an even number on one die is 3/6 or 1/2. The probability of rolling an even number on all three dice is (1/2) * (1/2) * (1/2) = 1/8, which is option C.

Question 2

Bunches of bananas arriving from their supplier reach a grocer with probability 0.12 of being too ripe to sell. To simulate the event that a single bunch of bananas arriving at the grocery is too ripe to sell, the produce manager could use two digits from a random generator with the convention (choose the best answer):
A) 00, 01, 02, … , 09, 10, 11 $\rightarrow$ too ripe
12, 13, 14, … , 97, 98, 99 $\rightarrow$ acceptable
B) 01, 02, 03, … , 10, 11, 12 $\rightarrow$too ripe
13, 14, 15, … , 98, 99, 00 $\rightarrow$acceptable
C) 00, 01, 02, … , 85, 86, 87 $\rightarrow$ too ripe
88, 89, 90, … , 97, 98, 99 $\rightarrow$ defective
D) Any answer choice is correct.
E) None of the answer choices is correct.

Accepted Answer

D

Question 3

Suppose flights at a large metropolitan airport are on-time 68% of the time and late 32% of the time. We want to use a table of random digits to simulate flight status (on-time or late), so we'll assign 01, 02, … , 68 to represent on-time flights, and 69, 70, … , 99, 00 to represent late flights.
You want to estimate the probability that 15 or more flights out of 20 will be on-time at this airport. You simulate 20 flights 25 times and get the following numbers of on-time flights:

What is your estimate of the probability?
A) 0.16
B) 0.48
C) 0.52
D) 0.56
E) 13

Accepted Answer

Out of the 25 simulations, there are 2 where 15 or more flights were on-time:

- Simulation 15: 15 on-time flights 
- Simulation 17: 16 on-time flights

Therefore, the estimate of the probability that 15 or more flights out of 20 will be on-time is 2/25 = 0.08.

However, we need to account for the fact that there are two categories (on-time and late) and they are not equally likely (68% vs. 32%). To do this, we can use the binomial distribution with parameters n = 20 and p = 0.68:

P(X ≥ 15) = 1 - P(X < 15) = 1 - P(X ≤ 14)

Using a binomial table or calculator, we find that:

P(X ≤ 14) ≈ 0.51

Therefore,

P(X ≥ 15) ≈ 1 - 0.51 = 0.49

which is closest to choice C (0.52).

Question 4

China has 1.36 billion people. Soft drink manufacturers work constantly to try to carve out a portion of that huge market. A Euromonitor study in 2014 showed that Coca-Cola held 63% of the entire Chinese soft drink market. A researcher wants to simulate choosing 10 Chinese soft-drink consumers at random and asking each if he or she purchases Coca-Cola.
Use the correct assignment of digits from the previous question and the random digits below to simulate the answers of 10 Chinese citizens. Read across the row of random digits from left to right. How many of these 10 Chinese soft-drink consumers purchase Coca-Cola?

A) 4
B) 5
C) 6
D) 7
E) 8

Accepted Answer

The answer of China has 1.36 billion people. Soft drink...

Question 5

Computer voice recognition software is getting better. Some companies claim that their software correctly recognizes 98% of all words spoken by a trained user.
To simulate recognizing a single word when the probability of being correct is 0.98, you could use random digits as follows:
A) Two digits simulate one word; 00 to 97 mean "correct."
B) Two digits simulate one word; 00 to 98 mean "correct."
C) One digit simulates one word; 0 to 9 mean "correct."
D) Three digits simulate one word; 001 to 098 mean "correct."

Accepted Answer

To simulate a 98% probability, two digits can represent 100 possible outcomes (00 to 99). Using 00 to 97 for "correct" covers 98 outcomes, which corresponds to a 98% probability.

Question 6

In a small Colombian village, 20 percent of the adults own a car, and 75 percent of the adults attend church regularly. Suppose that car ownership is independent of church attendance.
What is the probability that a randomly selected adult from this village owns a car and attends church regularly?
A) 0.95
B) 0.55
C) 0.20
D) 0.15
E) 0.05

Accepted Answer

Since car ownership and church attendance are independent, their probabilities can be multiplied to get the probability of both events occurring.

P(owns a car and attends church regularly) = P(owns a car) × P(attends church regularly)
= 0.2 × 0.75 = 0.15

Therefore, the probability that a randomly selected adult from this village owns a car and attends church regularly is 0.15. The answer is choice D.

Question 7

A researcher estimates that 30 percent of students at his college would be willing to pay for Google Fiber, the high-speed, superfast Internet access. The researcher chooses a student at random and asks that student if she would pay. To simulate the outcome, the researcher could use one random digit as follows:
A) Simulation 1: 0, 1, 2, 3 meaning "Yes"
B) Simulation 2: 1, 2, 3 meaning "Yes"
C) Simulation 3: 0, 1, 2 meaning "Yes"
D) Both simulations 2 and 3 are correct.
E) None of these answer choices is correct.

Accepted Answer

The correct answer is D because both simulations 2 and 3 are correct. In simulation 2, the digits 1, 2, and 3 represent "Yes," which means that 3 out of 10 digits (30%) represent a "Yes" outcome. In simulation 3, the digits 0, 1, and 2 represent "Yes," which also means that 3 out of 10 digits (30%) represent a "Yes" outcome. Therefore, both simulations 2 and 3 correctly represent the researcher's estimate that 30% of students would be willing to pay for Google Fiber.

Question 8

Computer voice recognition software is getting better. Some companies claim that their software correctly recognizes 98% of all words spoken by a trained user.
The program recognizes words (or not) independently. To simulate the program's performance on 10 words, use two digits to simulate one word with 02 to 99 meaning "correct" and these random digits:

The number of words correct out of the 10 is
A)10.
B)9
C)8
D)2
E) 1.

Accepted Answer

To simulate the program's performance recognizing 10 words, we need to convert each word into a two-digit code. Since the software claims to have a correct recognition rate of 98%, we can assume that for every ten words spoken by a trained user, on average, only one word is incorrect. Therefore, out of the ten simulated words, we can expect only one to be incorrect, resulting in nine correctly recognized words. Looking at the options, choice B is the only one that matches this expectation.

Question 9

Suppose you toss a fair coin six times. What is the probability of all six tosses resulting in heads (i.e., you toss no tails)?
A) 1/64
B) 1/32
C) 1/6
D) 1/2
E) 0

Accepted Answer

Each individual toss has a probability of 1/2 of landing on heads. Since each toss is independent from the others, the probability of getting heads on all six tosses is (1/2)^6 = 1/64.

Question 10

I want to use simulation to estimate the probability of getting exactly one head and one tail in two tosses of a fair coin. I assign the digits 0, 1, 2, 3, 4 to heads and 5, 6, 7, 8, 9 to tails. Using the following random digits to simulate, what is the estimate of the probability?

A) 1/10
B) 1/20
C) 0.6
D) 2/3
E) 0.5

Accepted Answer

The answer of I want to use simulation to estimate...

Question 11

Suppose we need to flip a coin, but we don't have one. However, we do have a random number table that we could use.
To simulate the outcome of tossing a coin you could assign random digits as follows:
A) one digit simulates one toss; even digits are heads, and odd digits are tails.
B) one digit simulates one toss; 0, 1, 2, 3, 4 are heads, and 5, 6, 7, 8, 9 are tails.
C) one digit simulates one toss; 0, 1, 4, 6, 9 are heads, and 2, 3, 5, 7, 8 are tails.
D) two digits simulate one toss; 00 to 49 are heads, and 50 to 99 are tails.
E) Any of the above would work.

Accepted Answer

Each option provides a fair method to simulate a coin toss, ensuring an equal probability of landing heads or tails by evenly distributing outcomes between the two possibilities.

Question 12

The WNBA player Temeka Johnson of the Los Angeles Sparks (2015) is a career 82 percent free-throw shooter. Suppose Tameka is fouled while attempting a three-point shot so she is awarded three free throws. What is the probability that she will make all three of her free throws?
A) 0.6724
B) 0.5514
C) 0.82
D) 0.18
E) 0

Accepted Answer

The answer of The WNBA player Temeka Johnson of the...

Question 13

In a small Colombian village, 20 percent of the adults own a car, and 75 percent of the adults attend church regularly. Suppose that car ownership is independent of church attendance.
What is the probability that a randomly selected adult from this village owns a car but does not attend church regularly?
A) 0.95
B) 0.55
C) 0.20
D) 0.15
E) 0.05

Accepted Answer

Since car ownership and church attendance are independent, the probability that an adult owns a car and does not attend church regularly is the product of the probability of owning a car (20% or 0.20) and the probability of not attending church regularly (100% - 75% = 25% or 0.25). Thus, 0.20 * 0.25 = 0.05 or 5%.

Question 14

Suppose you toss a fair coin and roll a fair six-sided die. What is the probability your actions result in tails on the coin and 3 on the die?
A) 1/8
B) 2/3
C) 1/6
D) 1/12
E) 0

Accepted Answer

The probability of getting tails on the coin is 1/2.
The probability of getting 3 on the die is 1/6.
The probability of these two independent events occurring together is the product of their individual probabilities:
1/2 x 1/6 = 1/12
Therefore, the answer is D) 1/12.

Question 15

In a small Colombian village, 20 percent of the adults own a car, and 75 percent of the adults attend church regularly. Suppose that car ownership is independent of church attendance.
What is the probability that a randomly selected adult from this village does not own a car and does not attend church regularly?
A) 0.95
B) 0.55
C) 0.20
D) 0.15
E) 0.05

Accepted Answer

The answer of In a small Colombian village, 20 percent...

Question 16

A multiple choice exam offers four choices for each question. Jason just guesses the answers, so he has probability 1/4 of getting any one answer right. You want to simulate whether Jason's answers to 10 questions are right or wrong. One correct way to do this is
A) one digit from the random digit table simulates one answer, with 4 = right and all other digits = wrong. Ten digits from the table simulate 10 answers.
B) one digit from the random digit table simulates one answer, with 0 or 1 = right and all other digits = wrong. Ten digits from the table simulate 10 answers.
C) one digit from the random digit table simulates one answer, with odd = right and even = wrong. Ten digits from the table simulate 10 answers.
D) two digits from the random digit table simulate one answer, with 00 to 24 = right and 25 to 99 = wrong. Ten pairs of digits simulate 10 answers.
E) two digits from the random digit table simulate one answer, with 00 to 04 = right and 05 to 99 = wrong. Ten pairs of digits simulate 10 answers.

Accepted Answer

Using two digits allows for a more accurate representation of the 1/4 probability, with 00 to 24 representing the 25% chance of getting an answer right.

Question 17

Suppose flights at a large metropolitan airport are on-time 68% of the time and late 32% of the time. We want to use a table of random digits to simulate flight status (on-time or late), so we'll assign 01, 02, … , 68 to represent on-time flights, and 69, 70, … , 99, 00 to represent late flights.
Use the information above and these random digits to simulate 20 flights at this airport:

How many of the simulated flights were on-time?
A) 19
B) 15
C) 11
D) 10
E) 9

Accepted Answer

The answer of Suppose flights at a large metropolitan airport...

Question 18

Suppose you toss a fair coin twenty-six times and each time you observe heads (i.e., 26 heads in a row). What is the probability of heads on your next toss?
A) 1
B) 0.962
C) 0.5
D) 0.038
E) 0

Accepted Answer

Each toss of a fair coin has an independent 0.5 probability of landing heads. Thus, the previous tosses have no bearing on the probability of the next toss. The probability of heads on the next toss is still 0.5, regardless of the previous 26 tosses.

Question 19

A book about bridge says that the probability that each of the four players is dealt exactly one ace is about 0.11. To simulate an outcome with probability 0.11, a person could do which of the following simulations:
A) Simulation 1: look at two digits in the random number table; the outcome occurs if the digits are 11.
B) Simulation 2: look at two digits in the random number table; the outcome occurs if the digits are any of 00, 01, ... , 11.
C) Simulation 3: look at two digits in the random number table; the outcome occurs if the digits are any of 00, 01, ... , 10.
D) Simulation 4: look at two digits in the random number table; the outcome occurs if the digits are any of 01, 02, ... , 11.
E) Both Simulations 3 and 4 are correct simulations.

Accepted Answer

Simulation 3 and Simulation 4 both correctly simulate an event with a probability of about 0.11. Simulation 3 does this by defining the event to occur for any two-digit number from 00 to 10 out of 100 possible outcomes (11 out of 100, or 0.11). Simulation 4 does this by defining the event to occur for any two-digit number from 01 to 11 out of 100 possible outcomes (11 out of 100, or 0.11). Both methods provide an equal chance of about 11% for the event to occur, matching the given probability.

Question 20

China has 1.36 billion people. Soft drink manufacturers work constantly to try to carve out a portion of that huge market. A Euromonitor study in 2014 showed that Coca-Cola held 63% of the entire Chinese soft drink market. A researcher wants to simulate choosing 10 Chinese soft-drink consumers at random and asking each if he or she purchases Coca-Cola.
One correct way to assign random digits to simulate the answer is
A) one digit simulates one person's answer; odd means "Yes" and even means "No."
B) one digit simulates one person's answer; 0 to 6 mean "Yes" and 7 to 9 mean "No."
C) two digits simulate one person's answer; 00 to 61 mean "Yes" and 62 to 99 mean "No."
D) two digits simulate one person's answer; 00 to 62 mean "Yes" and 63 to 99 mean "No."
E) There are 1.2 billion possible answers, which is too many to simulate.

Accepted Answer

C) is correct because it accurately reflects the proportion of the market that Coca-Cola holds (63%). Using two digits (00 to 99) gives 100 possible outcomes, and assigning 00 to 61 for "Yes" responses simulates a 62% chance, which is the closest to the actual 63% market share within the constraints of using two-digit numbers.

Quiz 19: Simulation

Suppose you roll three fair six-sided dice. What is the probability of each die showing an even number (i.e., you roll no odd numbers) ?

Computer voice recognition software is getting better. Some companies claim that their software correctly recognizes 98% of all words spoken by a trained user. To simulate recognizing a single word when the probability of being correct is 0.98, you could use random digits as follows:

In a small Colombian village, 20 percent of the adults own a car, and 75 percent of the adults attend church regularly. Suppose that car ownership is independent of church attendance. What is the probability that a randomly selected adult from this village owns a car and attends church regularly?

Suppose you toss a fair coin six times. What is the probability of all six tosses resulting in heads (i.e., you toss no tails) ?

I want to use simulation to estimate the probability of getting exactly one head and one tail in two tosses of a fair coin. I assign the digits 0, 1, 2, 3, 4 to heads and 5, 6, 7, 8, 9 to tails. Using the following random digits to simulate, what is the estimate of the probability?

Suppose we need to flip a coin, but we don't have one. However, we do have a random number table that we could use. To simulate the outcome of tossing a coin you could assign random digits as follows:

The WNBA player Temeka Johnson of the Los Angeles Sparks (2015) is a career 82 percent free-throw shooter. Suppose Tameka is fouled while attempting a three-point shot so she is awarded three free throws. What is the probability that she will make all three of her free throws?

Suppose you toss a fair coin and roll a fair six-sided die. What is the probability your actions result in tails on the coin and 3 on the die?

A multiple choice exam offers four choices for each question. Jason just guesses the answers, so he has probability 1/4 of getting any one answer right. You want to simulate whether Jason's answers to 10 questions are right or wrong. One correct way to do this is

Suppose you toss a fair coin twenty-six times and each time you observe heads (i.e., 26 heads in a row) . What is the probability of heads on your next toss?

A book about bridge says that the probability that each of the four players is dealt exactly one ace is about 0.11. To simulate an outcome with probability 0.11, a person could do which of the following simulations: