Three friends A B and C will participate in a roundrobin tou

Three friends (A, B, and C) will participate in a round-robin tournament in which each one plays both of the others. Suppose that P(A beats B) = 0.6, P(A beats C) = 0.4, P(B beats C) = 0.5, and that the outcomes of the three matches are independent of one another.
(a) What is the probability that A wins both her matches and that B beats C?


(b) What is the probability that A wins both her matches?


(c) What is the probability that A loses both her matches?


(d) What is the probability that each person wins one match? (Hint: There are two different ways for this to happen.)

Solution

(a)

Since matches are independent from each other so the probability that A wins both her matches and that B beats C is

P(A beats B) *P(A beats C) *(P(B beats C) = 0.6*0.4*0.5=0.12

(b)

The  probability that A wins both her matches will be

P(A beats B) *P(A beats C) = 0.6*0.4=0.24

(c)

The probability that A loses both her matches will be

[1-P(A beats B)] *[1-P(A beats C)] = [1-0.6]*[1-0.4]=0.24

(d)

First way: A beats B, B beats C and C beats A. Probabiltiy of this will be

P(A beats B) * P(B beats C) *[1-P(A beats C)] =0.6* 0.5*[1-0.4]=0.18

Second way: A beats C, C beats B and B beats A. Probabiltiy of this will be

P(A beats C) * [1-P(B beats C)] *[1-P(A beats B)] =0.4* 0.5*[1-0.6]=0.08

Both thw above way are mututally exclusive so  the probability that each person wins one match is

0.18 + 0.08 = 0.26