Question
The probability that a part manufactured by a company will be defective is 0.05. If 15 such parts are selected randomly and inspected, then the probability that at least two parts will be defective is (round off to two decimal places).
Answer :
0.17
n = 15, p = 0.05, q = 0.95
x ~ B(n, p) ≈ x ~ P(λ)
λ = np = 15(0.05) = 0.75
P (Atleast two defective)
P(x ≥ 2) = 1 –P(x ≤ 1)
= 1−(P(0) + P (1)) = 1 - e-0.75 (1 + (0.75))
= 1 – 1.75 (e–0.75) = 0.1733 ≈ 0.17
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