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

Answer 1

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