Three friends (A, B, and C) will participate in a round-robin tournament in whic

Question

Three friends (A, B, and C) will participate in a round-robin tournament in which each one plays both of the others. Suppose P(A beats b) = 0.9. P(A beats C) = 0.2. P(b beats C) = 0.6. and that outcome of the three matches are independent of one another. What is the outcome of matches are independent of one another. What is the probability that A wins both her matches and that B beats? What is the probability that A wins both her matches? What is the probabilities that A loses both her matches? What is the probabilities that each person wins outcome one match? There are two different ways for this to happen.

Explanation / Answer

Solution:

P(A beats B) = 0.9, P(B beats A) = 1 - 0.9 = 0.1

P(A beats C) = 0.2, P(C beats A) = 1 - 0.2 = 0.8

P(B beats C) = 0.6, P(C beats B) = 1 - 0.6 = 0.4

a) The probability that A wins wins her matches and that beats C is  0.108

P(A win) = P(A beats B) × P(A beats C) × P(B beats C)

P(A win) = 0.9 × 0.2 × 0.6

P(A win) = 0.108

b) The probabilty that A wins both her matches is 0.18

P(A win) = P(A beats B) × P(A beats C)

P(A win) = 0.9 × 0.2

P(A win) = 0.18

c)The probabilty that A loses both her matches is

P(A win) = P(B beats A) × P(C beats A)

P(B beats A) = 1 - 0.9 = 0.1

P(C beats A) = 1 - 0.2 = 0.8

P(A loses) = 0.1 × 0.8

P(A win) = 0.08

d) The probability that each person wins one match is P(each win one match) = 0.44

Two different cases are:-

P(each wins 1 match) = P(A beats B) × P(B beats C) × P(C beats A) + P(A beats C) × P(B beats A) × P(C beats B)

P(each win one match) = (0.9 × 0.6 × 0.8) + ( 0.2 × 0.1 × 0.4)

P(each win one match) = 0.432 + 0.008

P(each win one match) = 0.44

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