UGC NET CSE | June 2023 | Part 2: 8

Question

In a multiuser operating system, $20$ requests are made to use a particular resource per hour, on an average. The probability that no request is made in $45$ minutes is

• A. $\mathrm{e}^{-15}$
• B. $\mathrm{e}^{-5}$
• C. $1-\mathrm{e}^{-5}$
• D. $1-\mathrm{e}^{-10}$

Answer 1

source for this – GO CLASSES

Answer 2

• We need to find the probability of no events (i.e., x = 0) in 45 minutes, which is 3/4th of an hour. 

• The average rate of events per hour is given as 20. - So, λ = 20 events per hour.

• We need to find the probability of 0 events in 3/4th of an hour. - So, the time interval is 3/4th of an hour, and the average rate of events per interval is λ = 20 * 3/4 = 15.

• Substituting these values in the Poisson distribution formula, we get: P(0) = (e^-15 * 15^0) / 0! = e^-15

• Therefore, the probability that no requests are made in 45 minutes is e^-15, which is approximately 3.06 * 10^-7. 

• Answer: - The correct answer is option 'A' (e^-15).