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 :

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

Tardigrade Admin · Accepted Answer

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

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 :

33

30

20

10

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

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 :

33

30

20

10

Given : Probability of hitting the target = 0.1 i.e. p = 0.1 ∴q=1−p=0.9 Also given n = 100 ∴ By Binomial distribution, we have Mean = μ = np, variance = npq ∴ Expected number = μ = 100 × 0.1 = 10

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