Question

A traffic office imposes on an average 5 number of penalties daily on traffic violators. Assume that the number of penalties on different days is independent and follows a Poisson distribution. The probability that there will be less than 4 penalties in a day is _____ .

• A. 0.265

Answer 1

The correct option is A 0.265

Let X = {number of penalties per day}
Then average number of penalties $\left(\lambda \right)=5$ day (given)

$P(X<4)$

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

$=e^{-\lambda }+\lambda e^{-\lambda }+\frac{{\lambda }^2}{2!}{e}^{-\lambda }+\frac{{\lambda }^3}{3!}{e}^{-\lambda }$

$=e^{-\lambda }(1+\lambda +\frac{{\lambda }^2}{2}+\frac{{\lambda }^3}{6})$

$=e^{-5}(1+5+\frac{25}{2}+\frac{125}{6})$

$=\frac{e^{-5}}{6}(36+72+125)=0.265$