Question

At a telephone exchange, telephone calls arrive independently at an average rate of 1 call per minute, and the number of telephone calls follows a Poisson distribution. Five time intervals, each of duration 2 minutes, are chosen at random. Let \( p \) denote the probability that in each of the five time intervals at most 1 call arrives at the telephone exchange. Then \( e^{10} p \) (in integer) is equal to ________

Hint:

For Poisson distributions, the probability of at most 1 event is the sum of the probabilities of 0 and 1 events. Use this when working with Poisson processes.

Answer 1

Correct Answer: 243

Since the average rate of calls is 1 call per minute, in each 2-minute interval, the expected number of calls is 2. The number of calls in each time interval follows a Poisson distribution with mean 2. We need to find the probability that at most 1 call arrives in each of the five intervals. The probability of at most 1 call in a Poisson distribution is given by: \[ P(X \leq 1) = P(X = 0) + P(X = 1) = e^{-2} \left( \frac{2^0}{0!} + \frac{2^1}{1!} \right) = e^{-2} \left( 1 + 2 \right) = 3e^{-2}. \] Thus, the probability \( p \) that at most 1 call arrives in each of the five intervals is: \[ p = \left( 3e^{-2} \right)^5 = 3^5 e^{-10}. \] Now, we need to find \( e^{10} p \): \[ e^{10} p = e^{10} \times 3^5 e^{-10} = 3^5 = 243. \] Thus, \( e^{10} p \) is \( \boxed{243} \).