Question

The probability that a man hits a target is p = 0.1. He fires n = 100 times. The expected number n of times he will hit the target is :

• A. 33
• B. 30
• C. 20
• D. 10

Answer 1

Answer: (D)

10

Answer 2

Given : Probability of hitting the target = 0.1 i.e. p = 0.1 $\therefore \, q = 1 - p = 0.9$ Also given n = 100 $\therefore$ By Binomial distribution, we have Mean = $\mu$ = np, variance = npq $\therefore$ Expected number = $\mu$ = 100 $\times$ 0.1 = 10