Question 1

The set of all possible outcomes of an experiment is

Accepted Answer

The set of all possible outcomes of an experiment is called the sample space.

Question 2

Each customer entering a department store will either buy or not buy some merchandise. An experiment consists of following 3 customers and determining whether or not they purchase any merchandise. The number of sample points in this experiment is

Accepted Answer

Each customer has 2 possible outcomes: buy or not buy. For 3 customers, the total number of sample points is 2^3 = 8.

Question 3

Two events with nonzero probabilities

Accepted Answer

If two events have nonzero probabilities, they cannot be both mutually exclusive and independent. This is because if they were mutually exclusive, the occurrence of one event would make the occurrence of the other event impossible, meaning that they would be dependent. Conversely, if they were independent, the occurrence of one event would not affect the probability of the other event occurring, meaning that they could not be mutually exclusive. Therefore, it is not possible for two events with nonzero probabilities to be both mutually exclusive and independent.

Question 4

Two events, A and B, are mutually exclusive and each have a nonzero probability. If event A is known to occur, the probability of the occurrence of event B is

Accepted Answer

If events A and B are mutually exclusive, it means that they cannot occur at the same time. Therefore, if event A has occurred, it is impossible for event B to occur. Thus, the probability of event B occurring given that event A has occurred is zero.

Question 5

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

Accepted Answer

The addition law of probability states that the probability of the union of two events is the sum of their individual probabilities minus the probability of their intersection. Therefore, the addition law is very helpful when we are interested in computing the probability of the union of two events.

Question 6

The sum of the probabilities of two complementary events is

Accepted Answer

The sum of the probabilities of two complementary events is always equal to 1. This is because the probability of an event and its complement must add up to the total probability space of 1.

Question 7

If P(A) = 0.4, P(B | A) = 0.35, P(A $\cup$ B) = 0.69, then P(B) =

Accepted Answer

Using the law of total probability: P(A F1F1F1S1F1F1F10 B) = P(B | A) * P(A) + P(B | A^c) * P(A^c) Where A^c represents the complement of event A. We can rearrange and solve for P(B): P(B) = [P(A F1F1F1S1F1F1F10 B) - P(B | A^c) * (1 - P(A))] / P(A) Substituting the given values: P(B) = [0.69 - P(B | A^c) * 0.6] / 0.4 We don't have enough information to directly solve for P(B | A^c), but we can use the fact that P(B) must be between 0 and 1: 0 < P(B) < 1 0 < [0.69 - P(B | A^c) * 0.6] / 0.4 < 1 0 < 1.725 - 1.5P(B | A^c) < 2.5 -1.725 < -1.5P(B | A^c) < 0.775 1.15 > P(B | A^c) > -0.517 Since probabilities must be between 0 and 1, we can conclude that: 0.35 < P(B) < 1 Eliminating the impossible lower bound, we get: 0.43 < P(B) < 1 The only answer choice within this range is B, 0.43.

Question 8

Since the sun must rise tomorrow, then the probability of the sun rising tomorrow is

Accepted Answer

The probability of any event is always between 0 and 1, inclusive. Since the sun rising is a certain event, its probability is 1, which is not an option provided.

Question 9

If a dime is tossed four times and comes up tails all four times, the probability of heads on the fifth trial is

Accepted Answer

The probability of heads on the fifth trial is 1/2 regardless of the previous outcomes because each flip is independent and not affected by the previous outcomes. The probability of heads on any given coin flip is always 1/2, assuming the coin is fair.

Question 10

If A and B are independent events with P(A) = 0.65 and P(A $\cap$ B) = 0.26, then, P(B) =

Accepted Answer

P(A ∩ B) = P(A) × P(B)0.26 = 0.65 × P(B)P(B) = 0.26/0.65 = 0.4

Question 11

Three 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

Accepted Answer

Each applicant can be either male or female, giving 2 choices per applicant. With three applicants, the total number of combinations is 2^3 = 8.

Question 12

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

Accepted Answer

The number of ways to choose 2 letters out of 5 is given by the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of items to choose from, and k is the number of items to choose. Here, C(5, 2) = 5! / (2!(5-2)!) = 10.

Question 13

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

Accepted Answer

The probability of rolling a "3" on the third trial is independent of the previous two trials. This means that the probability is still 1/6, the same as any other outcome. The previous trials do not influence the probability of the third trial.

Question 14

Given that event E has a probability of 0.31, the probability of the complement of event E

Accepted Answer

The complement of event E is the probability of not E occurring, which is equal to 1 minus the probability of event E occurring. Therefore, the complement of event E has a probability of 1 - 0.31 = 0.69.

Question 15

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

Accepted Answer

If there are n possible outcomes on the first step, m possible outcomes on the second step, and k possible outcomes on the third step, then the total number of experimental outcomes is n x m x k. Using this formula, we get 4 x 3 x 2 = 24 possible outcomes for this experiment. Therefore, the best choice is C.

Question 16

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 ace?

Accepted Answer

There are 2 black aces in a deck of 52 cards (Ace of Spades and Ace of Clubs), so the probability is 2/52.

Question 17

Initial estimates of the probabilities of events are known as

Accepted Answer

Initial estimates of probabilities are known as prior probabilities, which serve as the basis for calculating posterior probabilities through the use of Bayes' theorem.

Question 18

Events A and B are mutually exclusive if their joint probability is

Accepted Answer

Mutually exclusive events cannot occur at the same time, hence their joint probability is zero.

Question 19

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

Accepted Answer

Each coin has 2 possible outcomes (head or tail). So, for 4 coins tossed successively, the total number of possible outcomes will be 2x2x2x2 = 16. Therefore, the correct choice is A.

Question 20

Assume your favorite soccwr team has 2 games left to finish the season. The outcome of each game can be win, lose or tie. The number of possible outcomes is

Accepted Answer

If the team can either win, lose, or tie each of the two remaining games, then there are three possibilities for each game, making a total of 3 x 3 = 9 possible outcomes. Therefore, the best choice is D, 9.

Quiz 4: Introduction to Probability

The set of all possible outcomes of an experiment is

Each customer entering a department store will either buy or not buy some merchandise. An experiment consists of following 3 customers and determining whether or not they purchase any merchandise. The number of sample points in this experiment is

Two events with nonzero probabilities

Two events, A and B, are mutually exclusive and each have a nonzero probability. If event A is known to occur, the probability of the occurrence of event B is

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

The sum of the probabilities of two complementary events is

If P(A) = 0.4, P(B | A) = 0.35, P(A $\cup$ B) = 0.69, then P(B) =

Since the sun must rise tomorrow, then the probability of the sun rising tomorrow is

If a dime is tossed four times and comes up tails all four times, the probability of heads on the fifth trial is

If A and B are independent events with P(A) = 0.65 and P(A $\cap$ B) = 0.26, then, P(B) =

Three 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

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

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

Given that event E has a probability of 0.31, the probability of the complement of event E

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

Initial estimates of the probabilities of events are known as

Events A and B are mutually exclusive if their joint probability is

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

Assume your favorite soccwr team has 2 games left to finish the season. The outcome of each game can be win, lose or tie. The number of possible outcomes is

Quiz 4: Introduction to Probability

The set of all possible outcomes of an experiment is

Each customer entering a department store will either buy or not buy some merchandise. An experiment consists of following 3 customers and determining whether or not they purchase any merchandise. The number of sample points in this experiment is

Two events with nonzero probabilities

Two events, A and B, are mutually exclusive and each have a nonzero probability. If event A is known to occur, the probability of the occurrence of event B is

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

The sum of the probabilities of two complementary events is

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

Since the sun must rise tomorrow, then the probability of the sun rising tomorrow is

If a dime is tossed four times and comes up tails all four times, the probability of heads on the fifth trial is

If A and B are independent events with P(A) = 0.65 and P(A ∩\cap∩ B) = 0.26, then, P(B) =

Three 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

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

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

Given that event E has a probability of 0.31, the probability of the complement of event E

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

Initial estimates of the probabilities of events are known as

Events A and B are mutually exclusive if their joint probability is

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

Assume your favorite soccwr team has 2 games left to finish the season. The outcome of each game can be win, lose or tie. The number of possible outcomes is

If P(A) = 0.4, P(B | A) = 0.35, P(A $\cup$ B) = 0.69, then P(B) =

If A and B are independent events with P(A) = 0.65 and P(A $\cap$ B) = 0.26, then, P(B) =