Evaluating is not the Key!

Suppose that a class of $40$ students contains $25$ boys and $15$ girls. A team of $15$ players is chosen at random, and $N$ is the number of boys it contains. Then $\mathbb{P}[N = x] \; = \; \frac{\binom{25}{x} \binom{15}{15-x}}{\binom{40}{15}} \hspace{2cm} 0 \le x \le 15$ so we are intetested in calculating $\mathbb{E}[N] \; = \; \sum_{x=0}^{15} x \frac{\binom{25}{x} \binom{15}{15-x}}{\binom{40}{15}}$ Evaluating this expectation can be done easily using indicator random variables. The team lines up for a photograph. Then $N = N_1 + N_2 + N_3 + \cdots + N_{15}$ where $N_j$ is the random variable that is equal to $1$ if the $j$ th player from the left is a boy, and $0$ if that player is a girl. Then $\mathbb{E}[N] \; = \; \sum_{j=1}^{15} \mathbb{E}[N_j]$ But $\mathbb{E}[N_j]$ is just the probability that (independent from all the other players chosen) the $j$ th player is a boy, and this is $\tfrac{25}{40}$ . Thus $\mathbb{E}[N] \; = \; 15 \times \tfrac{25}{40} \; = \; \tfrac{75}{8}$ making the answer $75+8=\boxed{83}$ .

Relevant wiki: Vandermonde's Identity

$\large {\text{We know that:} \ \binom{n}{k} = \cfrac{n!}{k!(n-k)!} \cdot \ \text{this can be transformed to:} \\ \binom{n}{k}=\cfrac{n}{k}\frac{(n-1)!}{(k-1)!(n-1-(k-1))!}=\cfrac{n}{k} \binom{n-1}{k-1} \cdot \\ \text{Now we can start with the definition of the expected value: } \ E[X]=\sum_{x=0}^{n} \cfrac{x{K\choose x}{M-K\choose n-x}}{M\choose n} \cdot \\ \text{Since for x=0, we add a 0 in this formula we can say that: } \ E[X]=\sum_{x=1}^{n}\cfrac{x{K\choose x}{M-K\choose n-x}}{M\choose n} \cdot}$

$\large { \text{Applying equation (1) we get: } \\ E[X]=\cfrac{nK}{M} \sum_{x=1}^{n} \cfrac{\binom{K-1}{x-1} \binom{M-1-(K-1)}{n-1-(x-1)}}{\binom{M-1}{n-1}} \cdot \\ \text{setting l = x - 1: } \implies E[X]=\cfrac{nK}{M} \sum_{l=0}^{n-1} \cfrac{\binom{K-1}{l} \binom{M-1-(K-1)}{n-1 -l}}{\binom{M-1}{n-1}} \cdot \\ \text{The sum of this equation:} \ \sum_{l=0}^{n-1} \cfrac{\binom{K-1}{l} \binom{M-1-(K-1)}{n-1 -l}}{\binom{M-1}{n-1}} \ \text{is 1 as it is the over all sum of the probabilities of a hypergeometric distribution (Vandermonde's Identity)} \cdot}$

Interpretation: Each term in the above sum can be interpreted as probability, that is the probability distribution of number of the number of blue balls in $n-1$ draws without replacement from a bag containing $K-1$ blue and $M-1-(K-1)$ green balls.

$\large {\text{Therefore we have:} \ E[X]=\cfrac{nK}{M} \cdot \\ \text{hence, the problem} \ \sum_{x = 0}^{15} \left( x \times \cfrac{\binom {25}{x} \times \binom {15}{15-x}}{\binom {40}{15}}\right ) \ \text{is just the same as:} \ E[X]=\sum_{{x=0}}^{n}\cfrac{x{K\choose x}{M-K\choose n-x}}{{M\choose n}} = \cfrac{nK}{M} \ \text{where n = 15, K = 25, M = 40.} \\ \implies \sum_{x = 0}^{15} \left( x \times \cfrac{\binom {25}{x} \times \binom {15}{15-x}}{\binom {40}{15}}\right ) = \cfrac{nK}{M} = \cfrac{15\times 25}{40} = \cfrac{75}{8} \\ \implies \boxed{a + b = 75 + 8 = 83.}}$

Its simply the co-efficient of $x^{14}$ in $25 (1+x)^{24} (x+1)^{15}$