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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Probability

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Probability

From a lot of 12 items containing 3 defectives, a sample of 5 items...

From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable X\mathrm{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 XX is mn\frac{m}{n}, where gcd⁡(m,n)=1\operatorname{gcd}(m, n)=1, then n−mn-m is equal to _________

Correct answer

0

Step-by-step solution

a=1−3C512C5 b=3⋅9C412C5c=3⋅9C312C5 d=1⋅9C212C5u=0.a+1.b+2⋅c+3⋅d=1.25σ2=0.a+1.b+4.c+9 d−u2σ2=105176\begin{aligned} & \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} \\ & \mathrm{u}=0 . \mathrm{a}+1 . b+2 \cdot c+3 \cdot \mathrm{d}=1.25 \\ & \sigma^2=0 . \mathrm{a}+1 . b+4 . c+9 \mathrm{~d}-\mathrm{u}^2 \\ & \sigma^2=\frac{105}{176} \end{aligned}
Ans. 176−105=71176-105=71
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