Verified Solution • 2025–26 Session moderate

QJEEM24Ma4735: JEE Main 2024 Mathematics - Class 12 JEEM24Ma4735 Solution: From a lot of 10 items, which include 3 defective items, a…

From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable XX denote the number of defective items in the sample. If the variance of XX is σ2\sigma^2 , then 96σ296 \sigma^2 is equal to __________.

Step-by-Step Verified Solution

xx 0 1 2 3
P(x)P(x) 7C510C5=112 C4⋅3C110C5=512 7C3⋅3C210C5=512 7C2⋅3C310C5=112
xP(x)xP(x) 0 512\frac{5}{12} 1012\frac{10}{12} 312\frac{3}{12}

amp;μ=∑xP(x)=0+512+1012+312=32amp;σ2=∑(x−μ)P(x)=∑(x−32)2P(x)amp;=94×112+14×512+14×512+94×112=712\begin{aligned}& \mu=\sum x P(x)=0+\frac{5}{12}+\frac{10}{12}+\frac{3}{12}=\frac{3}{2} \\& \sigma^2=\sum(x-\mu) P(x)=\sum\left(x-\frac{3}{2}\right)^2 P(x) \\& =\frac{9}{4} \times \frac{1}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{9}{4} \times \frac{1}{12}=\frac{7}{12}\end{aligned}

⇒σ2⋅96=8×7=56\Rightarrow \sigma^2 \cdot 96=8 \times 7=56

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