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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 :

- $6.9 \times 10^6 \times e^{-20}$
- $1.02 \times 10^6 \times e^{-20}$
- $6.9 \times 10^3 \times e^{-20}$
- $1.02 \times 10^3 \times e^{-20}$

+4

**$Just$ 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.

**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 $Best$ !!

+33 votes

Best answer

**Answer is (B)**

$20$ 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} $

$\quad= {1.02}\times {10^6}\times e^{-20}.$

0

why not can we take λ = 20 and then compute P(1)+P(2)+P(3) , which will give the probability of getting 1 or 3 or 5 request per hour and then take 3/4th of it.

Answer b is right only if we take λ = 15 and then compute P(1)+P(2)+P(3), which will be in requests per 45 mins.

Answer b is right only if we take λ = 15 and then compute P(1)+P(2)+P(3), which will be in requests per 45 mins.

0

λ = 15 and then compute P(1)+P(3)+P(5)

**e ^{-15}(15)** +

srestha where m wrong?

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