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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 :

1. $6.9 \times 10^6 \times e^{-20}$
2. $1.02 \times 10^6 \times e^{-20}$
3. $6.9 \times 10^3 \times e^{-20}$
4. $1.02 \times 10^3 \times e^{-20}$

$e^{-15}(\frac{15}{1!}+\frac{15^3}{3!}+\frac{15^5}{5!})=e^{-15}\times6.9\times10^{3}$

$Just$ for little more understanding:

The $Poisson\ distribution$ governs how many events happen in a given period of time. $Suppose$ that we have a process, in which events occur exactly every 10 seconds. Then the number of events that happen in a minute (i.e., 60 seconds) is 6.

https://math.stackexchange.com/questions/1536497/what-is-the-difference-between-a-poisson-and-an-exponential-distribution

Now coming to this question:

$\\1\ hr\rightarrow 20\ requests\\ 1\ min\rightarrow\ ?\\ 1\ min= \dfrac{1}{3}\ request\\ \\In\ 45\ minutes=45\times \dfrac{1}{3}=15\ requests$

Now for the rest of the explanation go for the $Best$ !!

$20$ request in $1$ hour. So we can expect $15$ request in $45$ minutes...

So, $\lambda = 15$ (expected value)

Poisson distribution formula$: f(x, \lambda) = p(X = x) = \dfrac{e^{-\lambda}*\lambda^x}{x!}$

$\text{Prob (1 request)} + \text{Prob (3 requests)} + \text{Prob (5 requests)}$
$\quad= p(1; 15) + p(3; 15) + p(5; 15)$
$\quad= {6.9} \times 10^{3} \times e ^ {-15}$
$\quad = {6.9}\times 10^{3}\times e^{5}\times e^{-20}$
$\quad= {1.02}\times {10^6}\times e^{-20}.$

How to get  mean if we use traditional method
@set2018 its not wrong just take out $_{e}-15$ as common and calculate, however the answer will not be exact we need to multiply and divide by $_{e}-15$ to make it $_{e}-20$ adjusting zeros on power of 10 we will get the exact answer which is option b.

@endurance1 can you please elaborate more on this?