Question

The arrival rate of customers arriving at a bank counter follows a Poisson distribution with a mean rate of 4 per 10 minutes interval. Find the probability that i. Exactly 0 customer will arrive in 10 minutes interval ii. Exactly 2 customers will arrive in 10 minutes interval iii. At most 2 customers will arrive in 10 minutes interval iv. At least 3 customers will arrive in 10 minutes interval 

Answer 1

Let $N$ be the number of customers arriving in 10 minutes interval.

$N$ has $Poisson(4)$ distribution, hence $P(N=n)=\frac{4^ne^{-4}}{n!}$.

i. $P(N=0)=\frac{4^0e^{-4}}{0!}=e^{-4}$.

ii. $P(N=2)=\frac{4^2e^{-4}}{2!}=8e^{-4}$.

iii.

$P(N\leq2)=\\ P(N=0)+P(N=1)+P(N=2)=\\ e^{-4}+\frac{4^1e^{-4}}{1!}+8e^{-4}=13e^{-4}.$

iv.

$P(N\geq3)=P(N>2)=1-P(N\leq2)=1-13e^{-4}.$

Answer: i. $e^{-4}\approx0.0183$, ii. $8e^{-4}\approx 0.1465$, iii. $10e^{-4}\approx 0.1831$, iv. $1-13e^{-4}\approx 0.7619$.