. (3.39)

Now 12 Ai = 1Ai , since an indicator function takes only the values 0 and 1, and also 1Ai 1Aj =

1Ai∩Aj . Therefore, by symmetry, E(N2) = E





X i

1Ai + 2 X

i



 = nP(A1) + n(n − 1)P(A1 ∩ A2). (3.40)

Using conditional probability, P(A1 ∩ A2) = P(A1)P(A2 | A1)

=

2 n



1 n − 1 ·

1 n − 1 +

n − 2 n − 1 ·

2 n − 1



=

2(2n − 3)

n(n − 1)2

, (3.41)