JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Probability

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Three defective oranges are accidentally mixed with seven good ones and 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)28/75✓ Correct
(B)14/25
(C)26/75
(D)18/25

Answer: (A)

Solution

Ten oranges contain three defective and seven good ones; two are drawn at random and

x

counts the defective ones drawn.

P (x = 0) = \frac{( 2 7 )}{( 2 10 )} = \frac{21}{45} = \frac{7}{15}

P (x = 1) = \frac{( 1 7 ) ( 1 3 )}{( 2 10 )} = \frac{21}{45} = \frac{7}{15}

P (x = 2) = \frac{( 2 3 )}{( 2 10 )} = \frac{3}{45} = \frac{1}{15}

Mean:

E (x) = 0 \cdot \frac{7}{15} + 1 \cdot \frac{7}{15} + 2 \cdot \frac{1}{15} = \frac{9}{15} = \frac{3}{5}

E (x^{2}) = 0 + 1 \cdot \frac{7}{15} + 4 \cdot \frac{1}{15} = \frac{11}{15}

Variance:

Var (x) = E (x^{2}) - [E (x)]^{2} = \frac{11}{15} - \frac{9}{25} = \frac{275 - 135}{375} = \frac{140}{375} = \frac{28}{75}

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Probability

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