Telephone calls come into an exchange according to a Poisson process w...
To solve this problem, we can use the Poisson distribution formula. The Poisson distribution is commonly used to model the number of events occurring in a fixed interval of time when the events occur with a known constant mean rate and are independent of the time since the last event.
Let's break down the problem step by step:
Given:
- The average number of calls per minute is 5.
- We need to find the probability that no call will come in during the two-minute period from 10 A.M. to 10:02 A.M.
Step 1: Determine the average number of calls in a two-minute period.
Since the average number of calls per minute is 5, we can multiply it by 2 to get the average number of calls in a two-minute period:
Average number of calls in a two-minute period = 5 * 2 = 10
Step 2: Calculate the probability of no calls during the two-minute period.
The Poisson distribution formula is given by:
P(x; λ) = (e^(-λ) * λ^x) / x!
where P(x; λ) is the probability of x events occurring in the interval, λ is the average rate of events, e is the base of the natural logarithm, and x! is the factorial of x.
In our case, x = 0 (no calls) and λ = 10 (average number of calls in a two-minute period). Plugging these values into the formula, we get:
P(0; 10) = (e^(-10) * 10^0) / 0!
Step 3: Simplify the equation.
Anything raised to the power of 0 is 1, and 0! is equal to 1. Therefore, we can simplify the equation to:
P(0; 10) = e^(-10)
Step 4: Calculate the probability.
Using a calculator, we can find that e^(-10) is approximately 0.00004539992.
Therefore, the probability that no call will come in during the two-minute period from 10 A.M. to 10:02 A.M. is approximately 0.00004539992, which is approximately equal to e^(-10).
Hence, the correct answer is option 'B': e^(-10).