2 marks, −0.667StatisticsProbability and Bayes' theoremConditional probability and the multiplication ruleHardabout 2 min

IIT JAM 2021 Question 24: Probability and Bayes' theorem

$X$ and $Y$ have joint density $f_{X,Y}(x,y) = 6xy$ for $0 < y \le \sqrt{x} \le 1$ and $0$ otherwise. Then the conditional probability $P\!\left(Y \ge \tfrac13 \mid X = \tfrac23\right)$ is

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Answer and worked solution

Answer: (C)

The marginal is $f_X(x) = \int_0^{\sqrt{x}} 6xy\,dy = 3x^2$ , so the conditional density is $f_{Y|X}(y\mid x) = \dfrac{6xy}{3x^2} = \dfrac{2y}{x}$ for $0 < y \le \sqrt{x}$ .

At $x = \tfrac23$ this is $3y$ on $\left(0, \sqrt{\tfrac23}\right)$ . Then $P\!\left(Y \ge \tfrac13\right) = \int_{1/3}^{\sqrt{2/3}} 3y\,dy = \tfrac32\left(\tfrac23 - \tfrac19\right) = \tfrac56$ . The answer is (C).

The trap: (B)

\dfrac{5}{9}

Forgetting to normalise by the marginal $f_X$ (or using the wrong upper limit $\sqrt{2/3}$ ) shifts the integral away from the correct value $5/6$ .

Plain-text version

X and Y have joint density f_(X,Y)(x,y) = 6xy for 0 < y ≤ √x ≤ 1 and 0 otherwise. Then the conditional probability P(Y ≥ 1/3 | X = 2/3) is (A) 1/2 (B) 5/9 (C) 5/6 (D) 3/4

Also written as: X and Y have joint density f_(X,Y)(x,y) = 6xy for 0 < y <= sqrt(x) <= 1 and 0 otherwise. Then the conditional probability P(Y >= 1/3 | X = 2/3) is

Probability and Bayes' theorem has come up 190 times in 7 exams since 2004.

What this question tests

This question is on Conditional probability and the multiplication rule, part of Probability and Bayes' theorem in Statistics.

Probabilities that change when you know something has happened.

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