Mean and variance. I think my book is wrong for variance in part A. Here's the problem:
At a local high school, 12 boys and 4 girls are applying to MIT. Suppose that
the boys have a 10% chance of getting in and the girls have a 20% chance. (a)
Find the mean and variance of the number of students accepted. (b) What is
more likely: 2 boys and no girls accepted or 1 boy and 1 girl?
For part A i am getting 1.72 for the variance but the book says 1.82.  Variance should be n(1-p)p. so, 4*(1-.2).2+12(1-.1)*.1=1.72  Is the book wrong?
Also not sure about part b, but the answer is 1 boy and 1 girl. 
 A: Let $X$ be the random number of girls accepted, and $Y$ be the random number of boys accepted.  Then $$X \sim \operatorname{Binomial}(n = 4, p = 0.2), \\ Y \sim \operatorname{Binomial}(n = 12, p = 0.1).$$  The total number of students accepted is $X+Y$ and its variance is $$\operatorname{Var}[X+Y] \overset{\text{ind}}{=} \operatorname{Var}[X] + \operatorname{Var}[Y] = 4(0.2)(1-0.2) + 12(0.1)(1-0.1) = 1.72,$$ as you claimed.
For the second part, we are asked to compute the probability $$\Pr[(X = 0) \cap (Y = 2)] \overset{\text{ind}}{=} \Pr[X = 0]\Pr[Y = 2],$$ as and compare that against $$\Pr[X = Y = 1] \overset{\text{ind}}{=} \Pr[X = 1] \Pr[Y = 1].$$  These are easily computed using the respective binomial probability mass functions.
A: Your variance computation is right, under appropriate independence assumptions.
As to the probabilities,  under the same independence assumptions, the probability that $2$ boys and no girls are accepted is $\binom{12}{2}(0.1)^2(0.9)^{10}\binom{4}{0}(0.2)^0(0.8)^4$. 
The probability $1$ boy and $1$ girl is accepted is $\binom{12}{1}(0.1)^1(0.9)^{11}\binom{4}{1}(0.2)^1(0.8)^3$. Compute.
