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 poisson 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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