Suppose X1 and X2 are independent binomial rvs and X1 B403 i

Suppose X1 and X2 are independent binomial r.v.s, and X1 B(4;0:3) (i.e., n = 4 trials
with success probability p = 0:3) and X2 B(3;0:3).
i) Find the conditional distribution of X1 given X1 +X2 = 6.
ii) Find E[X1|X1 +X2 = 6].

Solution

X1+X2 = 6

Cases

x1=3. x2=3

x1=4, x2=2

P(x1=3, x2=3)+P(x1=4, x2=2) = P(x1=3).P( x2=3)+P(x1=4).P(x2=2)

= 0.0756*0.027+0.0081*0.189 =0.00357

ii)

E(X1|X1+X2=6) =

(4*0.0756*0.027+3*0.0081*0.189)/0.00357

=3.57