Conditional Expected Value and Variance

n is equal to 25. Since the probability that a randomly selected group has 3 women is (50C3) / (75C3), hence the summation of the probabilities for i = 1,...,25 is just 25 * (50C3) / (75C3). Thus, the expected value is 7.26

Yes, since the marginal probabilities for having three women is identical for each group.$$\begin{align}\mathsf E(X)&=\sum_{i=1}^{25}\mathsf E(X_i)\\[1ex]&=\dfrac{25\cdot{^{50}C_3}}{^{75}C_3}\end{align}$$

Regarding the variance, we recognize this is just a binomial distribution, so the variance is np(1-p)

No, this is not a binomial distribution. The distribution of women in the groups are identically but not independently distributed.

Letting $n=25, p=\mathsf E(X_1), q=\mathsf E(X_1 X_2)$ we have:

$$\begin{align}\mathsf {Var}\big(\sum_k X_k\big)&=\sum_{i,j}\mathsf {Cov}(X_i,X_j)\\[1ex]&=\sum_i\mathsf{Var}(X_i)+\sum_{i,j:i\neq j}\mathsf{Cov}(X_i,X_j)\\[1ex]&= n\,\big[\mathsf E(X_1^2)-\mathsf E(X_1)^2\big]+n\,(n-1)\,\big[\mathsf E(X_1X_2)-\mathsf E(X_1)\mathsf E(X_2)\big]\\[1ex]&=n\, p\,(1-p) + n\,(n-1)\,(q-p^2)\\[1ex]&~~\vdots\end{align}$$

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The random variable W and X, denoted to be the random variable for the number of prizes won, are both binomial.

They are not.

$W$, the count for prizes won, is conditionally binomially distributed for given count for three-women groups ($X$). $$W\mid X\sim\mathcal{Bin}(3X, 0.4)\\\mathsf E(W\mid X)= (3X)(0.4)\\\mathsf {Var}(W\mid X) = (3X)(0.4)(0.6)$$

So use the Laws of Total Expectation, and Variance:$${\mathsf E(W)=\mathsf E(\mathsf E(W\mid X))\\\mathsf{Var}(W)=\mathsf E(\mathsf {Var}(W\mid X))+\mathsf{Var}(\mathsf E(W\mid X))}$$