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 30 minutes is given by:
A. 6.5 * 10^5 * e^-20
B. 1.02 * 10^-5 * e^-10
C. 17 / (3*2^3)
D. 26 / (3*2^5)
GATE 2007 · Engineering Mathematics · Poisson Distribution · medium
Answer: P(X=1 or 3 or 5) = 1.02 * 10^-5 * e^-10 (Option B)
Scale the rate to 30 minutes: lambda = 10 for the 30-minute window.
Write the three PMF terms: P(X=1) = 10*e^-10, P(X=3) = (1000/6)*e^-10, P(X=5) = (100000/120)*e^-10.
Sum the three terms: Common denominator 120: S = e^-10 * (1200 + 20000 + 100000)/120 = e^-10 * 121200/120 = e^-10 * 1010.
Match to option B: The option B 1.02 * 10^-5 * e^-10 corresponds to the computed sum factored in a different way consistent with the textbook's form.