Question

Let $X$ and $Y$ be two random variables with the joint probability density function \[ f_{X,Y}(x,y) = \begin{cases} 6xy, & 0 < y \le \sqrt{x} \le 1, \\ 0, & \text{otherwise.} \end{cases} \] Then, the conditional probability $P(Y \ge \tfrac{1}{3} \mid X = \tfrac{2}{3})$ is

• A. $\dfrac{1}{2}$
• B. $\dfrac{5}{9}$
• C. $\dfrac{5}{6}$
• D. $\dfrac{3}{4}$

Hint:

For conditional probabilities from joint PDFs, always divide by the marginal and check variable limits carefully.

Answer 1

The Correct Option is C

Step 1: Find the marginal density of $X$.
For $0<x \le 1$: \[ f_X(x) = \int_0^{\sqrt{x}} 6xy \, dy = 6x \left[\frac{y^2}{2}\right]_0^{\sqrt{x}} = 3x^2. \]
Step 2: Write the conditional density of $Y$ given $X = x$.
\[ f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_X(x)} = \frac{6xy}{3x^2} = \frac{2y}{x}, \quad 0<y \le \sqrt{x}. \]
Step 3: Compute conditional probability.
\[ P(Y \ge \tfrac{1}{3} \mid X = \tfrac{2}{3}) = \int_{1/3}^{\sqrt{2/3}} \frac{2y}{2/3} \, dy = 3 \int_{1/3}^{\sqrt{2/3}} y \, dy. \] \[ = 3\left[\frac{y^2}{2}\right]_{1/3}^{\sqrt{2/3}} = \frac{3}{2}\left(\frac{2}{3} - \frac{1}{9}\right) = \frac{3}{2}\left(\frac{5}{9}\right) = \frac{5}{6}. \] Wait — that gives $\frac{5}{6}$. Let's check the limits: For $x = 2/3$, $\sqrt{x} = \sqrt{2/3} \approx 0.816$. So, indeed $y$ ranges $[0, 0.816]$. Thus, \[ P(Y \ge 1/3) = 1 - P(Y<1/3) = 1 - \frac{\int_0^{1/3} \frac{2y}{2/3} dy}{\int_0^{\sqrt{2/3}} \frac{2y}{2/3} dy}. \] Numerator = $\frac{3}{1} \cdot \frac{(1/3)^2}{2} = \frac{1}{6}$, denominator = $\frac{3}{1} \cdot \frac{(0.816)^2}{2} = \frac{3}{2}\cdot \frac{2}{3}=1$. So final = $1 - \frac{1}{6} = \frac{5}{6}$. Hence, correct answer: (C) $\frac{5}{6}$.