Discussion
Suppose that X and Y are i.i.d. random variables with c.d.f. F which is binomial with parameters n and p. Then, \(P(x = j \mid x + y = k)\) is equal to :
Answer: D
The conditional probability is \(\binom{n}{j}\binom{n}{k-j} / \binom{2n}{k}\), a hypergeometric distribution (printed as option D).
X and Y are independent Binomial(n, p), so \(X + Y\) is Binomial(2n, p). Then
\[ \begin{aligned} &P(X = j \mid X + Y = k) \\ &= \frac{P(X = j)\,P(Y = k - j)}{P(X + Y = k)} \\ &= \frac{\binom{n}{j}p^{j}q^{n-j} \cdot \binom{n}{k-j}p^{k-j}q^{n-k+j}}{\binom{2n}{k}p^{k}q^{2n-k}} \end{aligned} \]
All the powers of p and q cancel, leaving \(\binom{n}{j}\binom{n}{k-j}/\binom{2n}{k}\), free of p: given the total, it is like drawing k items from 2n of which n are 'X' items. The option prints the denominator as \(\binom{2n}{n}\); it is the only option with the hypergeometric numerator.
Option B still depends on p, which the conditioning removes. Option C, Binomial(k, ½), is the answer for two i.i.d. Poisson variables, not binomial ones, and \(j/k\) is not a probability distribution in j.
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