GATE | Gate IT 2007 | Question 55

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 × 10⁶ × e^-20
(B) 1.02 × 10⁶ × e^-20
(C) 6.9 × 10³ × e^-20
(D) 1.02 × 10³ × e^-20


Answer: (B)

Explanation: In 1 hr. requests received =20
So in 45 min requests received = 15
So, lemda = 15

So, using poission distribution formula: P(x; μ) = (e^-μ) (μ^x) / x!

p(one request) + p(3 request) + p(5 request)

= p(1; 15) + p(3; 15) + p(5; 15)

= 6.9 * 10³ * e^-15

= 1.02 × 10⁶ × e^-20


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