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21 July, 20:04

Eight equal-strength players, including Alice and Bob, are randomly split into 4 pairs, and each pair plays a game (i. e. 4 games in total), resulting in four winners. What is the probability that exactly one of Alice and Bob will be among the four winners

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  1. 21 July, 20:26
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    probability that exactly one of Alice and Bob will be among the winners is 0.5714

    Explanation:

    The question is to determine the probability that oneof Alice and Bob will be among the winners

    It is also important to note that Alice and Bob cannot win at the same time

    Therefore,

    Probability of Alice and Bob split up = 6/7

    Based on this probability, we make the case as follows

    First case: Alice wins and Bob loses

    Probability of Alice win = 1/2

    Probability of Bob lose = 1

    Probability of Alice winning and Bob losing = 1/2 x 1/2 = 1/4

    Second Case: Bob wins and Alice loses

    Probability of Bob win = 1/2

    Probability of Alice lose = 1

    Probability of Bob winning and Alice losing = 1/2 x 1/2 = 1/4

    Third Case

    The Probability that Alice and Bob play each other = 1/7

    Based on all the estalished Cases

    The required probability that exactly one of Alice and Bob will be among the winners is

    (6/7 x 1/4) + (6/7 x1/4) + 1/7

    = (2 x 6/7 x 1/4) + 1/7

    = 6/14 + 1/7

    = 3/7 + 1/7

    = 4/7 = 0.5714
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