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QJEEM24Ma90f3: JEE Main 2024 Mathematics - Class 12 JEEM24Ma90f3 Solution: Three rotten apples are accidently mixed with fifteen good…

Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable xx to be the number of rotten apples in a draw of two apples, the variance of xx is

Multiple Choice Options

  • A 57153\frac{57}{153}
  • B 40153\frac{40}{153}
  • C 37153\frac{37}{153}
  • D 47153\frac{47}{153}

Step-by-Step Verified Solution

3 bad apples, 15 good apples.

Let X\mathrm{X} be no of bad apples

amp; Then P(X=0)=15C218C2=105153amp;P(X=1)=3C1×15C118C2=45153amp;P(X=2)=3C218C2=3153amp;E(X)=0×105153+1×45153+2×3153=51153amp;=13amp;Var⁡(X)=E(X2)−(E(X))2amp;=0×105153+1×45153+4×3153−(13)2amp;=57153−19=40153\begin{aligned}& \text { Then } \mathrm{P}(\mathrm{X}=0)=\frac{{ }^{15} \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{105}{153} \\& \mathrm{P}(\mathrm{X}=1)=\frac{{ }^3 \mathrm{C}_1 \times{ }^{15} \mathrm{C}_1}{{ }^{18} \mathrm{C}_2}=\frac{45}{153} \\& \mathrm{P}(\mathrm{X}=2)=\frac{{ }^3 \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{3}{153} \\& \mathrm{E}(\mathrm{X})=0 \times \frac{105}{153}+1 \times \frac{45}{153}+2 \times \frac{3}{153}=\frac{51}{153} \\& =\frac{1}{3} \\& \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-(\mathrm{E}(\mathrm{X}))^2 \\& =0 \times \frac{105}{153}+1 \times \frac{45}{153}+4 \times \frac{3}{153}-\left(\frac{1}{3}\right)^2 \\& =\frac{57}{153}-\frac{1}{9}=\frac{40}{153}\end{aligned}

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