Twenty pairs of individuals playing in a bridge tournament have been seeded 1, . . . , 20. In the first part of the tournament, the 20 are randomly divided into 10 east–west pairs and 10 north–south pairs. a.? ?What is the probability that x of the top 10 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)?

Answer: Step1: Given, twenty pairs of individuals playing in a bridge tournament have been seeded 1, . . . , 20. In the first part of the tournament, the 20 are randomly divided into 10 east–west pairs and 10 north–south pairs. Here, M = 10, n = 10, N = 20 P(X = x) = h(x;n,M,N) ( x)( n x = N (n) Step2: a). The aim is what is the probability that x of the top 10 pairs end up playing east–west. 20 pairs of individuals were randomly divided into 10 east–west pairs and 10 north–south pairs. Let ‘x’ be the number of top 10 pairs from east-west group. Probability that ‘x’ of the top 10 pairs end up playing east-west is. 10 2010 h(x;10,10,20) = (x )( 10 x (10