Q81 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics
An observer counts 240 vehicles per hour at a specific location on a highway. Assuming that the arrival of vehicles at the location follows Poisson distribution, the probability that more than two vehicles arrive over a 30 sec time interval is
- Correct answerA$\frac{e^2-5}{e^2}$
- B$\frac{e^2-2}{e^2}$
- C$\frac{1}{12 e^2}$
- D$\frac{12-e^2}{e^2}$
Solution
The average arrival rate, $\lambda$ is $240 \mathrm{veh} / \mathrm{h}$ 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,
$P(n>2)=1-P(n \leq 2)=1-\left\{e^{-2}+2 e^{-2}+2 e^{-2}\right\}$
$P(n>2)=\frac{e^2-5}{e^2}$
Hence, the correct answer is option: 'A'.