Ask Question
23 September, 17:45

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?

+5
Answers (1)
  1. 23 September, 20:20
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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 ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers