Out of 1000 taxi drivers, find approximately the no. of drivers with:

(i) No accident in a year.

(ii) more than 3 accidents in a year.

Subject: Applied Mathematics 4

Topic: Probability

Difficulty: Medium

m = 3; N = 1000

$ P(X=r) = \frac{e^{-m}e^{r}}{r!} \\ P(X=r) = \frac{e^{-3}3^r}{r!} $

(i) $ P(X=0) = \frac{e^{-3}3^0}{0!} = e^{-3} = 0.049787 $

Therefore, number of drivers with no accident in a year = 1000 x 0.049787 = 49.787 = 50 drivers.

(ii) $ P(X\gt3) = P(X \geq 4) \\ = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)] \\ = 1 - [\frac{e^{-3}3^0}{0!} + \frac{e^{-3}3^1}{1!} + \frac{e^{-3}3^2}{2!} + \frac{e^{-3}3^3}{3!}] \\ = 1 - [0.049787 + 0.14936 + 0.22404 + 0.22404] \\ = 0.35276938 $

Expected number of drivers with more than 3 accidents = 1000 x 0.35276938 = 352.76938 = 353 drivers