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Probability question

2022 · 28 Jul · Shift 2 · Q38
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Probability question

2022 · 28 Jul · Shift 2 · Q38

JEE MainMathematicsProbabilityNumerical+4 / −1
A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X\mathrm{X}X be the number of white balls, among the drawn balls. If σ2\sigma^{2}σ2 is the variance of X\mathrm{X}X, then 100σ2100 \sigma^{2}100σ2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 56

  1. Identify the distribution

Let XXX be the number of white balls in 333 draws without replacement from a bag containing:

  • 444 white balls
  • 666 black balls

So, XXX follows a hypergeometric distribution with parameters: N=10, K=4, n=3N=10,\, K=4,\, n=3N=10,K=4,n=3 where:

  • NNN = total number of balls
  • KKK = number of white balls
  • nnn = number of draws
  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=3⋅410(1−410)⋅10−310−1\sigma^2=3\cdot \frac{4}{10}\left(1-\frac{4}{10}\right)\cdot \frac{10-3}{10-1}σ2=3⋅104​(1−104​)⋅10−110−3​

  1. Simplify step-by-step

1−410=6101-\frac{4}{10}=\frac{6}{10}1−104​=106​

So, σ2=3⋅410⋅610⋅79\sigma^2=3\cdot \frac{4}{10}\cdot \frac{6}{10}\cdot \frac{7}{9}σ2=3⋅104​⋅106​⋅97​

Now, 3⋅79=733\cdot \frac{7}{9}=\frac{7}{3}3⋅97​=37​

Thus, σ2=73⋅24100\sigma^2=\frac{7}{3}\cdot \frac{24}{100}σ2=37​⋅10024​

σ2=168300=1425\sigma^2=\frac{168}{300}=\frac{14}{25}σ2=300168​=2514​

  1. Compute 100σ2100\sigma^2100σ2

100σ2=100⋅1425=56100\sigma^2=100\cdot \frac{14}{25}=56100σ2=100⋅2514​=56

  1. Final answer

100σ2=56100\sigma^2=56100σ2=56

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