Question

5. The probability of a man hitting a target is ¼.

1. If he fires 7 times, what is the probability of his hitting the target at least twice?
2. How many times must he fire so that the probability of his hitting the target at least one is greater than 2/3?

Answer 1

P(hitting a target) = 1/4 = 0.25

a) n = 7

P(X > 2) = 1 - (P(X = 0) + P(X = 1))

              = 1 - (7C0 * 0.25⁰ * (1 - 0.25)^7-0 + 7C1 * 0.25¹ * (1 - 0.25)^7-1)

              = 1 - 0.4449

              = 0.5551

b) P(at least one) > 2/3

or, 1 - P(hitting the target zero times) > 2/3

or, 1 - (1 - 0.25)ⁿ > 2/3

or, (1 - 0.25)ⁿ < 1/3

or, n = 4