JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Probability

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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) $\frac{28}{75}$ ✓ Correct
(C) $\frac{2}{15}$
(D) $\frac{3}{5}$

Answer: (B)

Solution

Drawing

2

pens from

10

with

3

defective,

X

takes values

0, 1, 2

with

P (0) = \frac{( 2 7 )}{( 2 10 )} = \frac{7}{15}

P (1) = \frac{( 1 7 ) ( 1 3 )}{( 2 10 )} = \frac{7}{15}

and

P (2) = \frac{( 2 3 )}{( 2 10 )} = \frac{1}{15}

. The mean is

E (X) = \frac{7}{15} + \frac{2}{15} = \frac{3}{5}

, and

E (X^{2}) = \frac{7}{15} + \frac{4}{15} = \frac{11}{15}

. The variance is

\frac{11}{15} - (\frac{3}{5})^{2} = \frac{55 - 27}{75} = \frac{28}{75}

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option

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