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
- 1115
- 215
- 35
- 2875
Answer
(4) 2875
Solution
X can be 0, 1, 2; total ways = 10C2 = 45
P(X = 0) = 7C245 = 2145, P(X = 1) = 3C1 7C145 = 2145, P(X = 2) = 3C245 = 345
E(X) = 0 + 21 + 645 = 35
E(X2) = 0 + 21 + 1245 = 1115
Var(X) = E(X2) - [E(X)]2
= 1115 - 925 = 55 - 2775 = 2875
Answer: (4)