Fifty pairs of individuals playing in a bridge tournament have been seeded 1,, 50. In the first part of the tournament, the 50 are randomly divided into 25 east-west pairs and 25 north-south pairs.
(a) What is the probability that x of the top 25 pairs end up playing east-west?
(b) What is the probability that all of the top five pairs end up playing the same direction?
(c) If there are 2n pairs, what is the pmf of X = the number among the top n pairs who end up playing east-west? What are E(X) and V(X)?

168

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i am pretty sure that the answer is 168

what's the question?

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there is no question.

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

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hello umm i don't understand your question can you explain it more pls

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