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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. $1.02 \times 10^3 \times e^{-20}$
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48 votes

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

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