JEE Main 2025 Apr 4 Shift 1, Mathematics Q19: Introduction To Probability

JEE Main2025Apr 4, Shift 1Mathematics

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

Solution

$X$ follows a hypergeometric distribution with $N = 10$ , $K = 3$ defective, $n = 2$ drawn. Using $(2 10) = 45$ :

$x$	$P (X = x)$
$0$	$(2 7) /45 = 21/45 = 7/15$
$1$	$(1 7) (1 3) /45 = 21/45 = 7/15$
$2$	$(2 3) /45 = 3/45 = 1/15$

$E (X) = 0 + \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5}$ .

$E (X^{2}) = 0 + \frac{7}{15} + \frac{4}{15} = \frac{11}{15}$ .

$Var (X) = E (X^{2}) - [E (X)]^{2} = \frac{11}{15} - \frac{9}{25} = \frac{55 - 27}{75} = \frac{28}{75}$ .

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JEE Main 2025 Apr 4 Shift 1, Mathematics Q19: Introduction To Probability

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