Question

An observer counts $240$ vehicles per hour at a specific location on a highway. Assuming that the arrival of vehicles at the location follows a Poisson distribution, the probability that more than two vehicles arrive over a $30\, \sec$ time interval is

• A. $\frac{e^{2}-5}{e^{2}}$
• B. $\frac{e^{2}-2}{e^{2}}$
• C. $\frac{1}{12 e^{2}}$
• D. $\frac{12-e^{2}}{e^{2}}$

Answer 1

Answer: (A)

The average arrival rate, $\lambda$ is $240$ or $1 / 15$ vehicles per second.
According to poisson distribution,
$P(n)=\frac{(\lambda t)^{n} e^{-\lambda t}}{n !}$
Where, $P(n)=$ probability of having $n$ vehicles arrive in time $t$.
$\lambda=$ Average vehicle flow or arrival rate in per unit time
$t=$ duration of the time interval over which vehicles are counted
$e=$ base of natural logarithm
$P(0)=\frac{\left(\frac{1}{15} \times 30\right)^{0} \cdot e^{-\left(\frac{1}{15}\right) \times(30)}}{0 !}=e^{-2}$
$P(1)=\frac{\left(\frac{1}{15} \times 30\right)^{1} \cdot e^{-\frac{1}{15} \times 30}}{1 !}=2 e^{-2}$
$P(2)=\frac{\left(\frac{1}{15} \times 30\right)^{2} \cdot e^{-\frac{1}{15} \times 30}}{2 !}=2 e^{-2}$
For more than two vehicles,
$P(n > 2)=1-P(n \leq 2)=1-\left\{e^{-2}+2 e^{-2}+2 e^{-2}\right\}$
$P(n > 2)=\frac{e^{2}-5}{e^{2}}$