Question 1026011
Assume that each cage is equally likely to be chosen.
Let X = the number of white mice in the three mice chosen
Apparently, x = 0, 1, 2, 3.
Given the first cage is chosen, the following are the conditional probabilities for each value of X.
X -----> 0      1       2     3
p(x|1st cage)---> 0    3/10     6/10   1/10

Given the second cage is chosen, the following are the conditional probabilities for each value of X.
X -----> 0      1       2     3
p(x|2nd cage)---> 0    1/7      4/7   2/7

This means the combined (joint) probability mass function is as follows:

X -----> 0      1          2          3
p(x)---> 0    62/280     164/280     54/280

==> E(X) = {{{0*0 + 1*(62/280) + 2*(164/280) + 3*(54/280) = 1.97143}}}

Thus, the expected number of white mice in the selection is practically 2.