GATE IT 2007 | Question: 57

Question

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 :

• A. $6.9 \times 10^6 \times e^{-20}$
• B. $1.02 \times 10^6 \times e^{-20}$
• C. $6.9 \times 10^3 \times e^{-20}$
• D. $1.02 \times 10^3 \times e^{-20}$

Comments

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

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$Just$ 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 $

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

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What is the intended method to solve this question to get answer directley in the form as given in the question

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ans is option b .

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Ans

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could have done in less steps ig

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Note : Read the image carefully 

20 requests are made to use a particular resource per hour

 

This not λ = np here actually this average rate per unit interval and 45 is total time interval .In this case apply  , λ=rxt = 20 *(45/60) =15...

Concept 1 :

Source : (Wiki) : https://en.wikipedia.org/wiki/Poisson_distribution#Definitions:~:text=The%20equation%20can,%5B14%5D

Concept 2: Assumptions and validity of Poison distribution (It will give answer for " why 1, 3, or 5 requests are  mutually exclusive events " )

Sources : (wiki): https://en.wikipedia.org/wiki/Poisson_distribution#Definitions:~:text=The%20Poisson%20distribution%20is%20an,at%20exactly%20the%20same%20instant.

 

 

Answer 1

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

Comments

@endurance1 can you please elaborate more on this?

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probability that either one, three or five requests are made

Means

 Why not Prob (1 request or 3 requests or 5 requests)?

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@Chaithu555 Both are the same thing

$P$$($$X=1$ or $X=3$ or $X=5$$)$ = $P$$($$X=1$$)$+$P$$($$X=3$$)$+$P$$($$X=5$$)$
as all the intersection terms will be zero as the random variables are mutually exclusive or the sets will be disjoint.
 

Answer 2

 

ANSWER – (B)

Answer 3

 

Answer option (B)

Answer 4

Ans