A munitions warehouse contains 50 bombs, of which 3 are defective (6%). A sample of 10 bombs is drawn and tested. What is the probability that the sample will contain at most 1 defective bomb?
P (Defective) = P(D) = 3/50 = 6% = 0.06
N= 10
Find P (at most 1 defective bomb) = Probability of 0 defective bomb + Probability of 1 defective bomb
So we need to find probability for x= 0 and x= 1
P(0)=0.5386151140949
P(1) = 0.34379688133717
P (at most 1 defective bomb) = P(0) + P(1) = 0.88241199543207
P (at most 1 defective bomb) = 0.8824
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