Question

A marksman hits a target 0.91 of the time. He fires 10 shots at the target. Find the probability that he make no more than 8 hits.

Answer 1

Solution:

We have to find P(X≤8)

n = 10

p = 0.91

q = 1 – 0.91 = 0.09

P(X≤8) = 1 – P(X≥9)

P(X≥9) = P(X=9) + P(X=10)

P(X=x) = nCx*p^x*q^(n – x)

P(X=9) = 10C9*0.91^9*0.09^(10 – 9) = 0.385137

P(X=10) = 10C10*0.91^10*0.09^(10 – 10) = 0.389416

P(X≥9) = P(X=9) + P(X=10)

P(X≥9) = 0.385137 + 0.389416

P(X≥9) = 0.774553

P(X≤8) = 1 – P(X≥9)

P(X≤8) = 1 – 0.774553

P(X≤8) = 0.225447

Required probability = 0.225447