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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 -

1. $e^-{15}$
2. $e^{-5}$
3. $1 - e^{-5}$
4. $1-e^{-10}$

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.
by

i didn't got how u got k=0???

@Arjun

@Jenny101
K=0,coz it is mentioned in the question that we have to find the probability of no requests  being made in 45 minutes.K is the number of requests.
edited by
@Arjun Sir can you please explain how subsituting k=0, will give the req. answer?

I think there is some modification re. in the formula i.e p(k)= (G^k * e^-G)/k!

Thanks

wrong formula @jenny101 . plz check again .

Poisson distribution p(k)= (G^k*e^-G)/k!

here k is no. of request in given time =0  bcz there is no request in 45 minutes

G is avg. request per 45 minutes = (45/60)*20=15

put eact value in formula: (15^0*e^-15)/0!

=e^-15