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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 :
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
In a multi-user operating system on an average, 20 requests are made t...
20 request in 1 hour.. so we can expect 15 request in 45 minutes...
So, lemda = 15.. (expected value)
poission distribution formula: f(x, lemda) = p(X = x) = (lemda ^ x * e ^ - lemda)  / x!
Therefore p(one request) + p(3 request) + p(5 request)
= p(1; 15) + p(3; 15) + p(5; 15)
= 6.9 * 10^3 * e ^ -15..
= 6.9*103*e-15 = 6.9*103*e5*e-20 = 1.02*106*e-20..  Ans is (B)
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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 :a)b)c)d)Correct answer is option 'B'. Can you explain this answer?
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