where the two terms correspond to whether or not the second queen sits next to the first couple.

By (3.39)–(3.41), E(N2) = 2 + n(n − 1) ·

2(2n − 3)

n(n − 1)2

, and hence var(N) = E(N2) − E(N)2 =

2(n − 2)

n − 1

. △

Exercise 3.42 Let N be the number of the events A1, A2, . . . , An which occur. Show that3 E(N) =

Xn i=1 P(Ai ).

3.6 Problems 1. Let X and Y be independent discrete random variables, each having mass function given by P(X = k) = P(Y = k) = pqk for k = 0, 1, 2 . . . , where 0 < p = 1 − q < 1. Show that P(X = k | X + Y = n) =

1 n + 1 for k = 0, 1, 2 . . . , n.

P3A similar fact is valid for an infinite sequence A1, A2, . . . , namely that the mean number of events that occur is

∞i =