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Question
If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week
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Solution
In a poisson distribution
Mean (λ) = `3/20` = 0.15
P(X = x) = `("e"^(-lambda) lambda^x)/(x!)`
P(not be more than one failure) = P(X ≤ 1)
P(X = 0) + P(X = 1)
= `("e"^(-0.15) (0.15)^0)/(0!) + ("e"^(-0.15) (0.15)^1)/(1!)`
= `"e"^(-0.15) [(0.15)^0/(0!) + (0.15)^1/(1!)]`
= `"e"^(-0.15) [1 + 0.15]`
= `"e"^(-0.15) [1.15]`
= 0.86074 × (1.15)
= 0.98981
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