Homework Answers
It is a problem of binomial probability distribution.
P (probability of getting defects) = 4.2% or .042
Q (probability of not getting defects) = 1 - .042 = .958
Required probability = 25C0 *P^0 * Q^25 + 25C1 *P^1 * Q^24 + 25C2 *P^2 * Q^23 + 25C3 *P^3 * Q^22
Required probability = 25C0 *.042^0 * .958^25 + 25C1 *.042^1 * .958^24 + 25C2 *.042^2 * .958^23 + 25C3 *.042^3 * .958^22
Required probability = 1*.042^0 * .958^25 + 25 *.042^1 * .958^24 + 300 *.042^2 * .958^23 + 2300 *.042^3 * .958^22
Required probability = .9806
So,
Probability to get 3 or less defects = .9806
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