Question

A firm has three telephone numbers. They all receive numbers of calls that follows Poisson distributions, the first having a mean of 8, the second 4, and the third 3 in a period of half an hour. Find the probability that:
a. The second and the third will receive a total of exactly 6 calls in half an hour b. The firm will receive at least 12 calls in half an hour
c. Line one will receive at most six calls in an hour. d. Line one will receive no calls in 15 minutes

Answer 1

The probability that the second and the third will receive a total of exactly 6 calls in half an hour will be 0.1490.

The probability that the firm will receive at least 12 calls in half an hour is 0.8152.

The probability that line one will receive at most six calls in an hour will be 0.3134

How to calculate the probability?

The probability that the second and the third will receive a total of exactly 6 calls in half an hour will be:.= (e^-6 × 7^6)/6!

= 0.1490

The probability that the firm will receive at least 12 calls in half an hour will be:

= 1 - (0.1848)

= 0.8152

The probability that line one will receive at most six calls in an hour will be:

= e^-8 × 83/3!

= 0.3134

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