Customers arrive at a bus station at the rate of 5 per minute following Poisson distribution. What is the probability of 6 arrivals in a two-minute interval?
A) $\frac{e^{-5} 5^{3}}{3 !}$
B) $\frac{e^{-10} 10^{6}}{6 !}$
C) $\frac{e^{-5} 5^{3}}{5 !}$
D) $\frac{e^{-6} 10^{-10}}{10 !}$

Question

Quizplus · Accepted Answer

The probability of $k$ arrivals in a time interval in a Poisson distribution is given by $P(k) = \frac{e^{-\lambda} \lambda^{k}}{k !}$, where $\lambda$ is the average rate of arrivals multiplied by the time interval. Here, $\lambda = 5 \text{ customers/minute} \times 2 \text{ minutes} = 10$, and $k = 6$. Thus, the correct formula is $P(6) = \frac{e^{-10} 10^{6}}{6 !}$, which matches option B.

Customers Arrive at a Bus Station at the Rate of 5