Mathematics · Probability

JEE Main 2025 — 4 April, Morning Shift — Question 31

A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let $X$ denote the number of defective pens. Then the variance of $X$ is

Option A:
$\frac{11}{15}$
Option B:
$\frac{2}{15}$
Option C:
$\frac{28}{75}$
Correct
Option D:
$\frac{3}{5}$

Answer: C

Step-by-step solution

$X$	$P(X)$	$XP(X)$	$(X_i - \mu)^2$	$P_i(X_i - \mu)^2$
$X = 0$	$\frac{{}^7C_2}{{}^{10}C_2}$	$0$	$\left(0 - \frac{3}{5}\right)^2$	$\frac{7}{15} \left(\frac{9}{25}\right)$
$X = 1$	$\frac{{}^7C_1 \cdot {}^3C_1}{{}^{10}C_2}$	$\frac{7}{15}$	$\left(1 - \frac{3}{5}\right)^2$	$\frac{7}{15} \left(\frac{4}{25}\right)$
$X = 2$	$\frac{{}^7C_0 \cdot {}^3C_2}{{}^{10}C_2}$	$\frac{2}{15}$	$\left(2 - \frac{3}{5}\right)^2$	$\frac{2}{15} \left(\frac{49}{25}\right)$

$\mu=\sum X_{i} P\left(X_{i}\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5}$

Variance $(X)=$

$\sum P_{i}\left(X_{i}-\mu\right)^{2}=\frac{7}{15}\left(\frac{9}{25}\right)+\frac{7}{15}\left(\frac{4}{25}\right)+\frac{2}{15}\left(\frac{49}{25}\right)=\frac{28}{75}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Morning Shift
Subject: Mathematics
Chapter: Probability
Topic: Mean, variance, expected values of distributions

The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at…

Answer: C

Step-by-step solution

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