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Probability question

2025 · 4 Apr · Shift 1 · Q37

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2025 · 4 Apr · Shift 1 · Q37

Probability question

2025 · 4 Apr · Shift 1 · Q37

JEE MainMathematicsProbabilityMCQ+4 / −1

A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let

X

denote the number of defective pens. Then the variance of

X

is

A
$\frac{11}{15}$
B
$\frac{2}{15}$
C
$\frac{3}{5}$
D
$\frac{28}{75}$

View written solutionFree

Correct answer: D

Identify the distribution

A box has:

Total pens $N = 10$
Defective pens $K = 3$
Good pens $7$

A sample of $n = 2$ pens is drawn without replacement. Let $X$ = number of defective pens in the sample.

So, $X$ follows a hypergeometric distribution.

Use variance formula for hypergeometric distribution

For a hypergeometric random variable,

\operatorname{Var}(X)=n\cdot \frac{K}{N}\left(1-\frac{K}{N}\right)\cdot \frac{N-n}{N-1}

Substitute the values:

\operatorname{Var}(X)=2\cdot \frac{3}{10}\left(1-\frac{3}{10}\right)\cdot \frac{10-2}{10-1}

=2\cdot \frac{3}{10}\cdot \frac{7}{10}\cdot \frac{8}{9}

Now simplify:

2\cdot \frac{21}{100}\cdot \frac{8}{9} = \frac{42}{100}\cdot \frac{8}{9} = \frac{336}{900} = \frac{28}{75}

So,

\boxed{\operatorname{Var}(X)=\frac{28}{75}}

Option check

A: $\frac{11}{15}$ ❌
B: $\frac{2}{15}$ ❌
C: $\frac{3}{5}$ ❌
D: $\frac{28}{75}$ ✅

Thus, the correct option is:

\boxed{\text{D}}

Verification with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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