In a multiuser operating system, 20 requests are made to use a particular…

• Requests follow a Poisson process.

• Average rate = 20 requests/hour.

• Time = 45 minutes = 0.75 hour.

Mean number of requests in 45 minutes:

λ=20×0.75=15\lambda = 20 \times 0.75 = 15λ=20×0.75=15

For a Poisson process, probability of zero requests in time ttt is:

P(0)=e−λP(0) = e^{-\lambda}P(0)=e−λ

So,

P(no request in 45 min)=e−15P(\text{no request in 45 min}) = e^{-15}P(no request in 45 min)=e−15