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Probability question

2024 · 5 Apr · Shift 1 · Q52
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  5. /2024 · 5 Apr · Shift 1 · Q52

Probability question

2024 · 5 Apr · Shift 1 · Q52

JEE MainMathematicsProbabilityNumerical+4 / −1
From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable XXX denote the number of defective items in the sample. If the variance of XXX is σ2\sigma^2σ2, then 96σ296 \sigma^296σ2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 56

  1. Identify the distribution

Since 5 items are drawn without replacement from 10 items, of which 3 are defective, the random variable XXX = number of defective items in the sample follows a hypergeometric distribution with:

  • Population size: N=10N=10N=10
  • Number of defective items: K=3K=3K=3
  • Sample size: n=5n=5n=5
  1. Use the variance formula for hypergeometric distribution

For a hypergeometric random variable,

Var⁡(X)=n⋅KN(1−KN)⋅N−nN−1\operatorname{Var}(X)=n\cdot \frac{K}{N}\left(1-\frac{K}{N}\right)\cdot \frac{N-n}{N-1}Var(X)=n⋅NK​(1−NK​)⋅N−1N−n​

Substitute the values:

σ2=5⋅310(1−310)⋅10−510−1\sigma^2=5\cdot \frac{3}{10}\left(1-\frac{3}{10}\right)\cdot \frac{10-5}{10-1}σ2=5⋅103​(1−103​)⋅10−110−5​ =5⋅310⋅710⋅59=5\cdot \frac{3}{10}\cdot \frac{7}{10}\cdot \frac{5}{9}=5⋅103​⋅107​⋅95​
  1. Simplify
5⋅310=325\cdot \frac{3}{10}=\frac{3}{2}5⋅103​=23​

So,

σ2=32⋅710⋅59\sigma^2=\frac{3}{2}\cdot \frac{7}{10}\cdot \frac{5}{9}σ2=23​⋅107​⋅95​ =2120⋅59=\frac{21}{20}\cdot \frac{5}{9}=2021​⋅95​ =105180=712=\frac{105}{180}=\frac{7}{12}=180105​=127​

Thus,

σ2=712\sigma^2=\frac{7}{12}σ2=127​
  1. Compute 96σ296\sigma^296σ2
96σ2=96⋅712=8⋅7=5696\sigma^2=96\cdot \frac{7}{12}=8\cdot 7=5696σ2=96⋅127​=8⋅7=56
  1. Final answer
56\boxed{56}56​
  1. Comparison with stored answer

Stored correct answer = 565656.

Our derived answer matches the stored answer.

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