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The multiuser operating system, $20$ requests are made to use a particular resource per hour, on an average the probability that no request are made in $45$ minutes is

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

Probability of events for a Poisson distribution

An event can occur 0, 1, 2, … times in an interval. The average number of events in an interval is designated $\lambda$ (lambda). Lambda is the event rate, also called the rate parameter. The probability of observing $k$ events in an interval is given by the equation

{\displaystyle P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}}

where

  • \lambda is the average number of events per interval
  • e is the number 2.71828... (Euler's number) the base of the natural logarithms
  • k takes values 0, 1, 2, …
  • k! = k × (k − 1) × (k − 2) × … × 2 × 1 is the factorial of k.

here $k=0 , \lambda =\frac{45\times 20}{60} =15$  (avg no of events in 45 minutes =15, since rate is 20 request per hr ) 

so probability of no request in 45 min is,

$P(0)=\frac{15^0 e^{-15}}{0!}=e^{-15}$

hence ans is A

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