JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Probability

Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random variable

X

to be the number of rotten apples in a draw of two apples, the variance of

X

Answer: (D)

Solution

Let

X

= number of rotten apples drawn (out of 3 rotten, 15 good, 18 total, choosing 2):

P (X = 0) = \frac{( 2 15 )}{( 2 18 )} = \frac{105}{153}, P (X = 1) = \frac{( 1 3 ) ( 1 15 )}{( 2 18 )} = \frac{45}{153}, P (X = 2) = \frac{( 2 3 )}{( 2 18 )} = \frac{3}{153}

E [X] = 0 \cdot \frac{105}{153} + 1 \cdot \frac{45}{153} + 2 \cdot \frac{3}{153} = \frac{51}{153} = \frac{1}{3}

E [X^{2}] = 0 + 1 \cdot \frac{45}{153} + 4 \cdot \frac{3}{153} = \frac{57}{153}

Var (X) = E [X^{2}] - (E [X])^{2} = \frac{57}{153} - \frac{1}{9} = \frac{57}{153} - \frac{17}{153} = \frac{40}{153}

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option