JEE Main 2025 Jan 28 Shift 1, Mathematics Q19: Introduction To Probability

JEE Main2025Jan 28, Shift 1Mathematics

Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If $x$ denotes the number of defective oranges, then the variance of $x$ is :

A
$\frac{28}{75}$
B
$\frac{14}{25}$
C
$\frac{26}{75}$
D
$\frac{18}{25}$

Solution

Sample 2 oranges from 10 (3 defective, 7 good). Total ways $= (2 10) = 45$ .

$P (X = 0) = \frac{( 2 7 )}{45} = \frac{21}{45} = \frac{42}{90}$
$P (X = 1) = \frac{( 1 7 ) ( 1 3 )}{45} = \frac{21}{45} = \frac{42}{90}$
$P (X = 2) = \frac{( 2 3 )}{45} = \frac{3}{45} = \frac{6}{90}$

Mean: $E [X] = 0 \cdot \frac{42}{90} + 1 \cdot \frac{42}{90} + 2 \cdot \frac{6}{90} = \frac{54}{90} = \frac{3}{5}$ .

$E [X^{2}] = 0 + 1 \cdot \frac{42}{90} + 4 \cdot \frac{6}{90} = \frac{66}{90} = \frac{11}{15}$ .

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

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q19: Introduction To Probability

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