Question

In a town, 10 accidents take place in the span of 50 days. Assuming that the number of accidents follows Poisson distribution, find the probability that there will be 3 or more accidents on a day.

(Given that e^-0.2 = 0.8187)

Answer 1

Here m `= 10/50 = 0.2`, and hence X ∼ P (m) with m = 0.2.

the p.m.f of X is P(X=x) `= (e^-m m^x)/(x!)`; x = 0,1,2,...

P(x ≥ 3) = 1 - P(X < 3)

`= 1 - [P(x = 0) + P(x = 1) + P(x = 2)]`

`=1 - [(e^-0.2 (0.2)^o)/(0!) + (e^-0.2 (0.2)^1)/(1!) + (e^-0.2 (0.2)^2)/(2!)]`

`= 1- [(0.8187xx1)/1 + (0.8187xx0.2)/1+ (0.8187xx0.04)/1]`

`= 1 - [0.8187 + 0.16374 + 0.016374]`

`= 1-0.9988`

`= 0.0012`