In a multi-user operating system, 20 requests are made to use a particular resource per hour, on average. The probability that no requests are made in 45 minutes is -

Answer should be a) coz if in 60 min 20 req are made then in 45 minute 15 req are made.

By Poisson distribution $P (k) = \frac{ G^k \times e^{-G}}{k!}.$

Here value of $k$ is number of requests made in given time which is 0. $G$ is avg requests per 45 min. Substituting in formula and get $e^{-15}$ as result.