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Question
Assuming that a fatal accident in a factory during the year is 1/1200, calculate the probability that in a factory employing 300 workers there will be at least two fatal accidents in a year, (given e-0.25 = 0.7788).
Solution
Let p be the probability of a fatal accident in a factory during the year
p = `1/1200` and n = 300
λ = np = `300 xx 1/200 = 1/4`
λ = 0.25
x follows poison distribution with
P(x) = `("e"^(-lambda) lambda^x)/(x!) + ("e"^(-0.25)(0.25))/(x!)`
P(atleasttwo fatal accidents) = P(X ≥ 2)
= P(X = 2) + P(X = 3) + P(X = 3) + P(X = 4) + ……….
= 1 – P(X < 2)
= 1 – {P(X = 0) + P(X = 1)}
= `1 - {("e"^(-0.25) (0.25)^0)/(0!) + ("e"^(-0.25)(0.25)^0)/(1!)}`
= `1 - "e"^(-0.25) {(0.25)^0/(0!) + (0.25)^1/(1!)}`
= 1 – 0.7788 [1 + 0.25]
= 1 – 0.7788(1.25)
= 1 – 0.9735 = 0.0265
∴ P(X ≥ 2) = 0.0265
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