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| 1. | 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» 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-44x) / 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. | |