Question 1

Four applications for admission to a local university are checked, and it is determined whether each applicant is male or female.The number of sample points in this experiment is
A) 2.
B) 4.
C) 16.
D) 8.

Accepted Answer

Each applicant can either be male or female, resulting in 2 possible outcomes for each application. Therefore, for 4 applications, there are 2x2x2x2 (or 2^4) possible outcomes, which equals to 16 sample points. Hence, the correct choice is C.

Question 2

The symbol ∪ shows the
A) union of events.
B) intersection of two events.
C) sum of the probabilities of events.
D) sample space.

Accepted Answer

The symbol ∪ represents the union of events, which is the set of all outcomes that belong to at least one of two or more events.

Question 3

Bayes' theorem is used to compute
A) the prior probabilities.
B) the union of events.
C) intersection of events.
D) the posterior probabilities.

Accepted Answer

Bayes' theorem is used to calculate the posterior probabilities of events, which are the updated probabilities after new evidence has been taken into consideration.

Question 4

If two events are independent, then
A) they must be mutually exclusive.
B) the sum of their probabilities must be equal to one.
C) their intersection must be zero.
D) the product of their probabilities gives their intersection.

Accepted Answer

If two events A and B are independent, then by definition P(A and B) = P(A)P(B). This means that the product of their probabilities gives their intersection. Therefore, option D is the correct choice.

Question 5

If a penny is tossed three times and comes up heads all three times, the probability of heads on the fourth trial is
A) 0.
B) 1/16.
C) 1/2.
D) larger than the probability of tails.

Accepted Answer

The probability of heads on any given coin toss is always 1/2, regardless of the outcome of previous tosses. Therefore, the probability of heads on the fourth trial is still 1/2, and the answer is (C).

Question 6

Assuming that each of the 52 cards in an ordinary deck has a probability of 1/52 of being drawn, what is the probability of drawing a black card?
A) 1/52
B) 4/52
C) 13/52
D) 26/52

Accepted Answer

The answer of Assuming that each of the 52 cards...

Question 7

The addition law is potentially helpful when we are interested in computing the probability of
A) independent events
B) the intersection of two events
C) the union of two events
D) conditional events

Accepted Answer

The addition law is used to calculate the probability of the union of two events.

Question 8

If P(A) = 0.4, P(B | A) = 0.35, P(A ∪ B) = 0.69, then P(B) =
A) 0.14.
B) 0.43.
C) 0.75.
D) 0.59.

Accepted Answer

P(B) can be found using the formula P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Given P(A ∪ B) = 0.69, P(A) = 0.4, and P(B | A) = P(A ∩ B) / P(A) = 0.35, we find P(A ∩ B) = 0.35 * 0.4 = 0.14. Substituting these values into the formula gives P(B) = 0.69 - 0.4 + 0.14 = 0.43.

Question 9

The union of events A and B is the event containing all the sample points belonging to
A) B or A.
B) A or B.
C) A or B or both.
D) A or B, but not both.

Accepted Answer

The union of two events A and B includes all the sample points that are in A, in B, or in both A and B.

Question 10

If A and B are independent events with P(A) = 0.5 and P(A ∩ B) = 0.12, then, P(B) =
A) 0.240.
B) 0.060.
C) 0.380.
D) 0.620.

Accepted Answer

Since A and B are independent events, P(A ∩ B) = P(A) × P(B). Therefore, P(B) = P(A ∩ B) / P(A) = 0.12 / 0.5 = 0.24. Hence, the answer is A.

Question 11

The symbol ∩ shows the
A) union of events.
B) intersection of two events.
C) sum of the probabilities of events.
D) sample space.

Accepted Answer

The symbol ∩ represents the intersection of two events, which is the set of outcomes that are common to both events.

Question 12

The set of all possible outcomes of an experiment is
A) a sample point.
B) an event.
C) the population.
D) the sample space.

Accepted Answer

The collection of all possible outcomes of an experiment is called the sample space, so choice D is the correct answer. A sample point is a specific outcome of an experiment, so choice A is incorrect. An event is a subset of the sample space, so choice B is incorrect. The term "population" is typically used in statistics to refer to a larger group of individuals, so choice C is also incorrect.

Question 13

The multiplication law is potentially helpful when we are interested in computing the probability of
A) mutually exclusive events.
B) the intersection of two events.
C) the union of two events.
D) conditional events.

Accepted Answer

The multiplication law is used to compute the probability of the intersection of two events, which is denoted by P(A ∩ B). In other words, it gives us the probability that both events A and B occur. This is the best choice because the other options do not involve the multiplication law.

Question 14

An experiment consists of three steps.There are five possible results on the first step, two possible results on the second step, and three possible results on the third step.The total number of experimental outcomes is
A) 10.
B) 625.
C) 150.
D) 180.

Accepted Answer

The total number of experimental outcomes is found by multiplying the number of possible results for each step: 5 (first step) * 2 (second step) * 3 (third step) = 30.

Question 15

Assume your favorite soccer team has 4 games left to finish the season.The outcome of each game can be win, lose or tie.The number of possible outcomes is
A) 4.
B) 12.
C) 64.
D) 81.

Accepted Answer

Each game has 3 possible outcomes (win, lose, or tie) and there are 4 games, so the total number of possible outcomes is 3*3*3*3 = 81. Therefore, the answer is D, 81.

Question 16

Of six letters (A, B, C, D, E, and F), two letters are to be selected at random.How many outcomes are possible?
A) 30
B) 11
C) 6!
D) 15

Accepted Answer

When selecting 2 letters out of 6 without regard to order, the number of possible outcomes is calculated using combinations. The formula for combinations is nCr = n! / [r!(n-r)!], where n is the total number of items, and r is the number of items to be chosen. Here, n=6 and r=2, so the calculation is 6! / [2!(6-2)!] = (6*5) / (2*1) = 15.

Question 17

The sum of the probabilities of two complementary events is
A) 0.
B) 0.5.
C) 0.57.
D) 1.0.

Accepted Answer

The sum of the probabilities of two complementary events always adds up to 1. If event A has a probability of 0.3, then the probability of the complement of A (not A) would be 0.7, and 0.3 + 0.7 = 1. Therefore, the answer is D.

Question 18

Each customer entering a department store will either buy or not buy some merchandise.An experiment consists of following 5 customers and determining whether or not they purchase any merchandise.The number of sample points in this experiment is
A) 5.
B) 10.
C) 25.
D) 64.

Accepted Answer

Each customer has 2 options: to buy or not to buy. For 5 customers, the total number of combinations is 2^5 = 32. However, the correct answer is not listed, indicating a mistake in the provided options. The closest correct approach to calculate sample points for this scenario is 2 raised to the power of the number of customers, which should result in 32, not 64.

Question 19

An experiment consists of tossing 4 coins successively.The number of sample points in this experiment is
A) 16.
B) 8.
C) 4.
D) 2.

Accepted Answer

The number of possible outcomes for one coin toss is 2 (heads or tails). Therefore, the number of possible outcomes for four successive coin tosses is 2 × 2 × 2 × 2 = 16. Each coin has two possible outcomes, and there are four coins tossed, so the total number of sample points is 16.

Question 20

If a six sided die is tossed two times and "3" shows up both times, the probability of "3" on the third trial is
A) much larger than any other outcome.
B) much smaller than any other outcome.
C) the same as any other outcome.
D) not able to be determined before the die is tossed.

Accepted Answer

Each toss of a die is independent of the previous tosses, meaning that the outcome of the third trial is not affected by the previous two trials. Therefore, the probability of "3" on the third trial is the same as any other outcome, which is 1/6.

Quiz 4: Introduction to Probability

Four applications for admission to a local university are checked, and it is determined whether each applicant is male or female.The number of sample points in this experiment is

The symbol ∪ shows the

Bayes' theorem is used to compute

If two events are independent, then

If a penny is tossed three times and comes up heads all three times, the probability of heads on the fourth trial is

Assuming that each of the 52 cards in an ordinary deck has a probability of 1/52 of being drawn, what is the probability of drawing a black card?

The addition law is potentially helpful when we are interested in computing the probability of

If P(A) = 0.4, P(B | A) = 0.35, P(A ∪ B) = 0.69, then P(B) =

The union of events A and B is the event containing all the sample points belonging to

If A and B are independent events with P(A) = 0.5 and P(A ∩ B) = 0.12, then, P(B) =

The symbol ∩ shows the

The set of all possible outcomes of an experiment is

The multiplication law is potentially helpful when we are interested in computing the probability of

An experiment consists of three steps.There are five possible results on the first step, two possible results on the second step, and three possible results on the third step.The total number of experimental outcomes is

Assume your favorite soccer team has 4 games left to finish the season.The outcome of each game can be win, lose or tie.The number of possible outcomes is

Of six letters (A, B, C, D, E, and F) , two letters are to be selected at random.How many outcomes are possible?

The sum of the probabilities of two complementary events is

Each customer entering a department store will either buy or not buy some merchandise.An experiment consists of following 5 customers and determining whether or not they purchase any merchandise.The number of sample points in this experiment is

An experiment consists of tossing 4 coins successively.The number of sample points in this experiment is

If a six sided die is tossed two times and "3" shows up both times, the probability of "3" on the third trial is