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Q.
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 :
Probability - Part 2
Solution:
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