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Telephone calls come into an exchange according to a Poisson process with 5 calls per minute on the average. The probability that no call will come in during the two-minute period 10 A.M. to 10:02 A.M. on a particular day is
  • a)
    0
  • b)
    e-10
  • c)
    2/5 e-5
  • d)
    1 - e-10
Correct answer is option 'B'. Can you explain this answer?
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Telephone calls come into an exchange according to a Poisson process w...
λ = 5 x 2 = 10
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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).
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Telephone calls come into an exchange according to a Poisson process with 5 calls per minute on the average. The probability that no call will come in during the two-minute period 10 A.M. to 10:02 A.M. on a particular day isa)0b)e-10c)2/5 e-5d)1 - e-10Correct answer is option 'B'. Can you explain this answer?
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