Question

Question 1 of 4

A family plans to have 3 children. For each birth, assume that the probability of a boy is the same as the probability of a girl.

What is the probability that they will have three children of the same gender?

A- 0.5

B- 0.25

C- 0.375

D-0.125

E-none of these

Question 2 of 4

A person in a casino decides to play blackjack until he loses a game, but he will not play more than 3 games. Let L denote a loss and W denote a win.

What is the sample space for this random experiment?

A- S = {LLL, LLW, LWL, LWW, WLL, WLW, WWL, WWW}

B- S = {L, LW, LLW, LLL}

C- S = {L, LL, LLL}

D- S = {L, WL, WWL, WWW}

F- S = {L, WL, WWL}

Question 3 of 4

A person in a casino decides to play 3 games of blackjack. Let L denote a loss and W denote a win. Define the event A as "the person loses at least one game of blackjack."

What are the possible outcomes for this event?

A- {LLL, LLW, LWL, LWW, WLL, WLW, WWL}

B- {LLL, LLW, LWL, LWW, WLL, WLW, WWL, WWW}

C-{LWW, WLW, WWL}

D- {L, WL, WWL}

E-{L, LL, LLL}

Question 4 of 4

Four students attempt to register online at the same time for an Introductory Statistics class that is full. Two are on the football team and two are on the basketball team. They are put on a wait list. Prior to the start of the semester, two enrolled students drop the course, so the professor randomly selects two of the four wait list students and gives them seats in the class.

What is the probability that both students selected play the same sport?

A, 1/12

B, 1/6

C, 1/3

C, 1/2

D, It is impossible to tell because the outcomes are not equally likely.

Answer 1

1) P(boy) + P(girl) = 0.5

P(all three children are of same gender) = P(all 3 are boy) + P(all 3 are girl)

                                                                = (0.5)³ + (0.5³)

                                                                = 0.25

Option-B) 0.25

2) sample space = the person loses on the first try + the person wins the first and loses the second + the person wins the first two and loses the third + the person wins the first three

S = {L, WL, WWL, WWW}

Option-D) S = {L, WL, WWL, WWW}

3) Option-A) {LLL, LLW, LWL, LWW, WLL, WLW, WWL}

4) Total number of ways to select two students out of 4 = 4C2 = 4! / (2! * (4 - 2)!) = 6

P(both played the same sport) = P(both played football) + P(both played basketball)

                                                 = (2C2 / 6) + (2C2 / 6)

                                                 = 2/6

                                                 = 1/3

Option-C) 1/3