Phone calls arrive at the rate of 48 per hour at the reservation desk for Indian Airlines. If no calls are currently being processed, what is the probability that the agent can take 3 minutes for personal time without being interrupted by a call?

0.091

The correct answer is Option (3) → 0.091

Given number of phone calls arrives per hour is 48.

Let random variable X be the number of phone calls in 3 minutes interval of time, then

$λ = 48 ×\frac{3}{60}⇒λ = 2.4$.

$P(X=0)=\frac{2.4^{0}e^{-2.4}}{0}=0.091$

Hence, the required probability is 0.091