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 that

P(A beats B) = 0.3
P(A beats C) = 0.8
P(B beats C) = 0.9

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.)

Explanation / Answer

consider the events

E1-A beats B, E2-A beats C and E3-B beats C

E1, E2 and E3 are independent.

a) p1= the probability that A wins both her matches and that B beats C=P(E1 E2 E3)

p1=P(E1) P(E2) P(E3) = 0.3*0.8*0.9= 0.216 (since E1, E2 and E3 are independent.)

b) p2=the probability that A wins both her matches=P(E1 E2)

p1=P(E1) P(E2)=0.3*0.8=0.24

c) p3=the probability that A loses both her matches=P(E1' E2')

where E1'- B beats A and E2'- C beats A.

p3=P(E1') P( E2')=(1-P(E1)) (1-P(E2))= 0.7*0.2=0.14

d) there are two ways for each person wins one match 1) A win A lost and C win 2) A lost B win C lost.

p4= probability that each person wins one match=

P(E1 E2' E3)+P(E1' E2 E3')=P(E1) (1-P(E2)) P(E3)+(1-P(E1)) P(E2) (1-P(E3))

p4=0.054+0.056= 0.11

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