Three rotten apples are accidently 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 is

Solution

Maths sir · 3 steps

Build the probability table for 0, 1 and 2 rotten apples, then use E(x²) - [E(x)]².

1. Count the draws. There are 18 apples, 3 rotten and 15 good. Two apples can be drawn in C(18, 2) = 153 ways, so every probability has denominator 153.

   x	P(x)	x P(x)	x² P(x)
   0	C(15, 2)/153 = 105/153	0	0
   1	(3 × 15)/153 = 45/153	45/153	45/153
   2	C(3, 2)/153 = 3/153	6/153	12/153
   Total	1	51/153 = 1/3	57/153

2. Mean and square. The mean is E(x) = 1/3, so [E(x)]² = 1/9 = 17/153. The table also gives E(x²) = 57/153.

   Var(x) = E(x²) - [E(x)]² = 57/153 - 17/153 = 40/153

3. So. The variance of x is 40/153, which is option (b).

The idea to keep. For a small discrete random variable, list each value with its probability, find E(x) and E(x²) from the table, and always subtract [E(x)]² at the end.

Answer from the NTA final answer key. Solution by PrepWiser.