Question 1015429
This problem uses the binomial probability formula,
Pr( k success in n trials) = nCk * p^k * q^(n-k)
:
We are given a sample size of 100 cars and told that 1% of the cars are defective and asked what is the probability of more than 2 cars being defective
:
let's look at a success as failure of a car, then p = .01 and q = 0.99
:
Pr( k > 2 ) = 1 - Pr(k=0) - Pr(k=1) - Pr(k=2)
:
Pr(k=0) = 100C0 * (.01)^0 * (.99)^(100-0) = 0.3660323412732289
Pr(k=1) = 100C1 * (.01)^1 * (.99)^(100-1) = 0.36972963764972616
Pr(k=2) = 100C2 * (.01)^2 * (.99)^(100-2) = 0.18486481882486308
:
Pr(k > 2 ) = 1 − 0.920626798 = 0.079373202 approx 0.0794