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Question
If the probability that a fluorescent light has a useful life of at least 800 hours is 0.9, find the probabilities that among 20 such lights at least 2 will not have a useful life of at least 800 hours. [Given : (0⋅9)19 = 0⋅1348]
Solution
Let X be the number of fluorescent lights that have a useful life of at least 800 hours.
P(a light has useful life of at least 800 hours) = p = 0.9, q = 1 - 0.9 = 0.1
Given n = 20
∴ X ~ B (20, 0.9)
The p.m.f. of X is given by
P(X = x) = p(x) = 20Cx (0.9)x (0.1)20-x , x = 0,1,2, ……,20
P(at least 2 lights will not have a useful life) = P(at most 18 will have useful life)
= P(X ≤ 18) = 1 - P(X > 18)
= 1 - [P(X = 19) + P(X = 20)]
= 1 - [20C19 (0.9)19 (0.1) + 20C20 (0.9)20]
`= 1-[20xx9^19/10^20+9^20/10^20]=1-[9^19/10^20(20+9)]`
`=1-((9^19xx29)/10^20)`
Let M=`(29xx9^19)/10^20`
log M=log 29 + 19 log 9 - 20 log 10
= 1.4624 + 19 × 0.9542 - 20 × 1
= 1.4624 + 18.1298 - 20
= 19.5922 - 20
= 19.5922 - 19 - 1
`= bar1 .5922`
∴ M = Antilog ( `bar1 .5922` ) = 0.3910
∴ P(at least two lights will not have a useful life) = 1 - 0.3910 = 0.6090
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