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3 years ago

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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?

1 answer:

Shkiper50 [21]3 years ago

4 0

Answer:

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

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