Question 916261
Cage1: 5mice, 2W, 3B , p(w) = 2/5)
Cage2: 7mice: 4W, 3B, p(w) = 4/7
cage1 0r cage2 is chosen at random and three mice are selected, n = 3
expected number of white mice 
W
0 P(x =0) = {{{(3/5)(2/4)(1/3) + (4/7) (3/6)(2/5)}}} Choosing cage1 0r cage2
1 P(x=1) = {{{(2/5)(3/4)(2/3) + (3/5)(2/4)(2/3) + (3/5)(2/4)(2/3)}}} +{{{(4/7) (3/6)(2/5) + (3/7) (4/6)(3/5) +  (3/7) (2/6)(4/5) }}} 
2 P(x=2) = binompdf( 5, .4, 2) + binompdf(7, 4/7, 2) = similar to the above
3 P(x=3) = binompdf( 5, .4, 3) + binompdf(7, 4/7, 3) = 

expected = 0•P(x =0) + 1•P(x =1) + 2•P(x =2) + 3•P(x =3)