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UGC NET CSE | July 2018 | Part 2 | Question: 53

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In a multi-user operating system, 30 requests are made to use a particular resource per hour, on an average. The probability that no requests are made in 40 minutes, when arrival pattern is a poisson distribution, is ______

  1. $e^{-15}$
  2. $1-e^{-15}$
  3. $1-e^{-20}$
  4. $e^{-20}$
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Poison’s Distribution = $\frac{(e^{-p} * p^{x})}{x!}$

No. of request in 60 minutes = 30

No. of request in 1 minute = 0.5

p = No. of request should be made in 40 minutes = 40 * 0.5 = 20

x = But In Reality How many Request in 40 minutes = 0

Poison’s Distribution = $\frac{(e^{-20} * 20^{0})}{0!}$ = $e^{-20}$

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