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

Question

Justice Glenn

Business

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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Answers (1)

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Arjun Bell · Answer 1

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