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 

The correct answer is e-15EXPLANATION:

To solve this problem, we can use the Poisson distribution, which is often used to model the number of events occurring in fixed intervals of time or space.

The Poisson distribution is given by the formula:

• \(P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}\)

Where:

• \(P(X = k) \) is the probability of observing  k events,
• e is the base of the natural logarithm (approximately 2.71828), 
• \(\lambda\) is the average rate of events per interval,
• k is the actual number of events observed.

In this case, the average rate of requests per hour (\( \lambda\)) is 20. We want to find the probability of no requests in 45 minutes. We can adjust the rate for the given time interval:

• \( \text{Adjusted rate for 45 minutes} = \frac{45}{60} \times \text{Average rate per hour}\)
• \( \text{Adjusted rate for 45 minutes} = \frac{3}{4} \times 20 = 15\)

Now, let \( \lambda = 15\) in the Poisson distribution formula to find the probability of no requests (\( P(X = 0)\)):

• \(P(X = 0) = \frac{e^{-15} \cdot 15^0}{0!} = e^{-15} \)

Therefore, the correct answer is: \( e^{-15}\)