A large shipment of batteries (several million) contains 4% that are defecttive. It 60 batteries are selected randomly, What is the probability that at most two are defective?

0.5676

Step-by-step explanation:

We have been given these informations from this question ,

probability of 4% = 0.04

q = 1 - p = 1 - 0.04 = 0.96

Sample size n = 60

We solve further by Using binomial probability formula,

P(X = x) = (nCx)*px*(1 - p)^n - x

We have:

(60C0)x 0.04⁰x(0.96⁶⁰+(60C1)x0.04¹x(0.96)⁵⁹+(60C2)x0.04²x (0.96)⁵⁸

From this we would get the probability to be = 0.5676