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Question
If the probability that a fluorescent light has a useful life of at least 600 hours is 0.9, find the probabilities that among 12 such lights at least 11 will have a useful life of at least 600 hours
Solution
Let p be the probability of the useful life hours of a fluorescent light.
n = 12
p = 0.9
X = B(n, p)
q – 1 – p = 0.1
P(X = x) = ⁿCx px qn-x, x = 0, 1, 2, …. n
At least 11
P(X ≥ 11) = P(X = 11) + P(X = 12)
= 12C11 (0.9)11 (0.1)1 + 12C12 (0.9)12 (0.1)°
= 12 × (0.9)11 × 0.1 + 1 × (0.9)12 × 1
= (0.9)11 (12 × 0.1 × 0.9)
= (0.9)11 (1.2 + 0.9)
= (2.1) (0.9)11
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