Question

PROBLEM 1. A hospital switch board receives an average of 4 emergency calls in a10 minute interval. What is the probability that (i) there are atmost 2 emergency calls(ii) there are exactly 3 emergency calls in a 10 minutes interval.​

Answer 1

The answers are 0.238 and 0.195.

Given:

The average of emergency calls is 4

To Find:

The probability that there are at most 2 emergency calls

The probability that there are exactly 3 emergency calls

Solution:

This type of situation will follow Poisson's distribution as the probability depends on the frequency of the outcome.
The probability in Poisson's distribution is given as

$p=\frac{\lambda^ke^{-\lambda}}{k!}$

where λ is the average given, k is the frequency of output, and e is Euler's number.

We have been given λ = 4.

Therefore for at most 2 emergency calls, we can have 0,1, or 2 calls.

The probability of having zero calls is

$p=\frac{4^0e^{-4}}{0!} =0.018$

The probability of having one call is

$p=\frac{4^1e^{-4}}{1!} =0.073$

The probability of having two calls is

$p=\frac{4^2e^{-4}}{2!} =0.147$

Therefore the probability of having at most 2 calls is the sum of the three probabilities

$p=0.018+0.073+0.147=0.238$

The probability of having exactly three calls is

$p=\frac{4^3e^{-4}}{3!} =0.195$

Hence the probability of having at most 2 emergency calls is 0.238 and the probability of having exactly 3 emergency calls is 0.195.

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