In a multi-user operating system on an average, $20$ requests are made to use a particular resource per hour. The arrival of requests follows a Poisson distribution. The probability that either one, three or five requests are made in $45$ minutes is given by : $6.9 \times 10^6 \times e^{-20}$ $1.02 \times 10^6 \times e^{-20}$ $6.9 \times 10^3 \times e^{-20}$ $1.02 \times 10^3 \times e^{-20}$

asked
Oct 30, 2014
in Probability
Ishrat Jahan
5k views