Question

The distribution of the number of road accidents per day in a city is Poisson with mean 4. Find the number of days out of 100 days when there will be 

(i) no accident 

(ii) at least 2 accidents and 

(iii) at most 3 accidents.

Answer 1

Let X be the Poisson variable denoting the number of accidents per day. 

Given that mean is 4 (i.e,) λ = 4. The p.m.f is given by P(X = x) = (e^-44^x) / x!

(i) P (no accident) = P(X = 0) = e = 0.0183 

For 100 days we have 100 × 0.0183 = 1.83 ~ 2 

Hence out of 100 days there will be no accident for 2 days.

(ii) P (atleast 2 accidents) = P (X ≥ 2) 

= 1 – P (X < 2) 

= 1 – [P(X = 1) + P(X = 0)] 

= 1 – [e^-4 (4) + e^-4 ] 

= 1 – (0.0183) (5) 

= 1 – 0.0915

 = 0.9085 

For 100 days we have 100 × 0.9085 ~ 91 

Hence out of 100 days there will be at least 2 accidents for 91 days.

(iii) P (atmost 3 accidents) = P (X ≤ 3) 

= P (X = 0 ) + P (X = 1 ) + P (X = 2) + P (X = 3) 

= e^-4 [ 1 + 4/1 + 16/2 + 64/6] 

= (0.0183) [23.6667] 

= 0.4331 

For 100 days we have 100 × 0.4331 ~ 43 

Hence out of 100 days, there will be atmost 3 accidents for 43 days.