Four students (A, B, C, and D) are interviewing for an all-expenses-paid vacation to a country of their choice. Only one student will win a vacation. If A is twice as likely to win as B (i. e. P (A) = 2P (B)), B is 2/3 as likely to win as C, and C is one and a half times as likely to win as D, what are the probabilities that (a) A wins the vacation? (b) C does not win the vacation?

A. 0.36; B. 0.72

Step-by-step explanation:

Probability of A, B, C, D to win the interview = P (A), P (B), P (C), P (D)

Given : P (A) = 2 P (B) ∴ P (B) = P (A) / 2

P (B) = 2/3 P (C) ∴ P (C) = 3/2 P (B) ∴ P (C) = 3/2 [P (A) / 2 ]

So, P (C) = 3/4 P (A)

P (C) = 1.5 P (D) ∴ P (C) = 3/2 P (D) ∴ P (D) = 2/3 P (C)

∵ P (C) = 3/4 P (A) ∴ P (D) = 2/3 [3/4 P (A) ]

So, P (D) = 1/2 P (A)

Either of them will definitely win the interview. So probability of A or B or C or D winning = 1

So, P (A) + P (B) + P (C) + P (D) = 1

Putting above values : P (A) + P (A) / 2 + 3/4 P (A) + 1/2 P (A) = 1

P (A) [1 + 1/2 + 3/4 + 1/2] = [ (4+2+3+2) / 4] P (A)

∴ 2.75 P (A) = 1

So, P (A) = 1/2.75 = 0.36

P (B) = P (A) / 2 = 0.36 / 2 = 0.18

P (C) = 3/4 P (A) = 0.36 (3/4) = 0.27

P (D) = P (A) / 2 = 0.36 / 2 = 0.18

A. Probability A wins election = 0.36

B. Probability C doesn't win election = Pr A or B or D win election

= Pr (A) + Pr (B) + Pr (D) = 0.36 + 0.18 + 0.18 = 0.72