Mathematics · Probability

JEE Main 2024 — 6 April, Shift 2 — Question 25

From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable X denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$ , where $\operatorname{gcd}(m, n)=1$ , then $n-m$ is equal to $\qquad$

Answer: 71

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{a}=1-\frac{{ }^{3} \mathrm{C}_{5}}{{ }^{12} \mathrm{C}_{5}}$

$\mathrm{b}=3 \cdot \frac{{ }^{9} \mathrm{C}_{4}}{{ }^{12} \mathrm{C}_{5}}$

$\mathrm{c}=3 \cdot \frac{{ }^{9} \mathrm{C}_{3}}{{ }^{12} \mathrm{C}_{5}}$

$\mathrm{d}=1 \cdot \frac{{ }^{9} \mathrm{C}_{2}}{{ }^{12} \mathrm{C}_{5}}$

$u=0 . a+1 . b+2 . c+3 . d=1.25$

$\sigma^{2}=0 . a+1 . b+4 . c+9 d-u^{2}$

$\sigma^{2}=\frac{105}{176}$

Ans. 176 - $105=71$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Mathematics
Chapter: Probability
Topic: Random Variables, Binomial & Poission Distribution

If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly…

Answer: 71

Step-by-step solution

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