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

Solution

Maths sir · 3 steps

Build the full probability table for X first; then variance is just E(X²) - [E(X)]².

1. Possible values. Two pens are drawn together, so X can be 0, 1 or 2. There are C(10, 2) = 45 equally likely samples. Of the 10 pens, 7 are good and 3 are defective.

   X	Ways	P(X)	X P(X)	X² P(X)
   0	C(7, 2) = 21	21/45	0	0
   1	C(3, 1)C(7, 1) = 21	21/45	21/45	21/45
   2	C(3, 2) = 3	3/45	6/45	12/45

2. Add the columns. E(X) = (21 + 6)/45 = 3/5 and E(X²) = (21 + 12)/45 = 11/15.

   Var(X) = 11/15 - (3/5)² = 55/75 - 27/75 = 28/75

3. So. The variance of X is 28/75, which is option (d).

The idea to keep. For any small discrete random variable, write out the table of X and P(X), then find E(X) and E(X²). Variance is E(X²) minus the square of the mean.

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