Discussion

Let \(X\) and \(Y\) be two random variables having the joint probability density function

\[ f(x,y) = \begin{cases} 2 & \text{if } 0 \lt x \lt y \lt 1 \\ 0 & \text{otherwise.} \end{cases} \]

Then the conditional probability \(P\left(X \le \frac{2}{3} \mid Y = \frac{3}{4}\right)\) is equal to

Answer: D

The conditional probability is \(\frac{8}{9}\).

The marginal density of Y is \(f_{Y}(y) = \int_{0}^{y} 2\,dx = 2y\) for \(0 \lt y \lt 1\). So the conditional density of X given \(Y = y\) is

\[ \begin{aligned} f_{X \mid Y}(x \mid y) &= \frac{2}{2y} = \frac{1}{y}, \\ &0 \lt x \lt y \end{aligned} \]

That is, given \(Y = \frac{3}{4}\), X is uniform on \((0, \frac{3}{4})\):

\[ \begin{aligned} P\left(X \le \tfrac{2}{3} \mid Y = \tfrac{3}{4}\right) &= \frac{2/3}{3/4} \\ &= \frac{8}{9} \end{aligned} \]

\(\frac{2}{3}\) would be the answer if X were uniform on \((0, 1)\), ignoring the conditioning.

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