Question

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) = 0.2 P(A beats C) = 0.6 P(B beats C) = 0.8 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.)

Answer 1

# Given PCA beats B) = 0.2 PCA beats c) = 0.6 PCB beats c) = 0.8 (a) since matches are independent of one another Therefore Probabelity that a wens both her matches B beats C = PCA beats B) x PCA beats c) * and PCB beate c) 1) 0. 2x0.68 0.8 - 0.096 -) Probability that A wins both her matches and B beate c = 0.096 (6) Similarly Probability that A wins both her matchs. = PC A beats B) x P (A beats c) 0.2 x 0.6 0.12

(0) Probability that a looses both her matches = (1-PCA beats B)) (1-PCA beats) = (1-0.2) C1-0.6) 0.8xo.4 0.32 (d) There are two possibility when each wins one matches and cbeasA Elther A beats B , B beats a A beats c , cbeats B and B bensa Therefore Probability that each wins one match = P(A beats 6, 8 beate, ebeats A) +P (A beats e, c beats B, B beads A) Yoca beats B) x P (B beats c) x P cc beats A ) + (P (A beats clx PCC beats B) x P (B beats A) 10) (since matches are Independent)

Probability that each wins one match (0.2x0.8 x (1-0.6))+ (0.6x (1-0-0) *(1-0-2)) =(0.280.880.4) +(0.6x0.280.87 0.064 +0.096 o. 160 > Probability that each wins one match = 0.16