What is the Probability that Each Person Wins One Match?

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In a round-robin tournament among three friends A, B, and C, the probabilities of winning matches are given: P(A beats B) = 0.7, P(A beats C) = 0.3, and P(B beats C) = 0.6. The task is to calculate the probability that each person wins one match, noting that there are two distinct scenarios for this outcome. The initial calculation of P=(0.7)(0.7)(0.6)=0.294 is incorrect as it only considers one of the two scenarios. A complete solution requires evaluating both possible winning combinations for A, B, and C to determine the overall probability.
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Homework Statement



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) = 0.7, P(A beats C) = 0.3, P(B beats C) = 0.6, and that the outcomes of the three matches are independent of one another.

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

Homework Equations



not sure if there is a relevant equation to this

The Attempt at a Solution



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

so A, B and C would need to win a match

P(A beats B) = 0.7 (A wins)
P(A beats C) = 0.3, so we would take 1-.3=0.7 to figure out probability of C winning
P(B beats C) = 0.6 (B wins)

P=(0.7)(0.7)(0.6)=0.294 THIS IS INCORRECT
 
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hi needstatshelp! welcome to pf! :smile:
needstatshelp said:
(d) What is the probability that each person wins one match? (Hint: There are two different ways for this to happen.)

P=(0.7)(0.7)(0.6)=0.294 THIS IS INCORRECT

you've only done the probability for one of the two different ways :wink:
 
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