Question

A telephone exchange receives on an average 5 calls per minute. The probability of receiving 3 or less calls per minute is :

• A. \(15e^{-5}\)
• B. \(\frac{118}{3}e^{-5}\)
• C. \(\frac{15}{4}e^{-5}\)
• D. \(\frac{13}{3}e^{-4}\)

Answer 1

The Correct Option is B

The problem involves finding the probability of receiving 3 or fewer calls per minute at a telephone exchange, given the arrival rate of 5 calls per minute. This problem can be solved using the Poisson distribution.

The Poisson probability mass function is given by:

\(P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!}\)

where \(\lambda\) is the average rate of occurrence, and \(k\) is the number of occurrences we want the probability for. Here, \(\lambda=5\) and \(k=0,1,2,3\).

The probability of receiving 3 or less calls per minute is the sum of probabilities for \(k=0\), \(k=1\), \(k=2\), and \(k=3\).

Calculate the individual probabilities:

\(P(X=0)=\frac{5^0 e^{-5}}{0!}=e^{-5}\)

\(P(X=1)=\frac{5^1 e^{-5}}{1!}=5e^{-5}\)

\(P(X=2)=\frac{5^2 e^{-5}}{2!}=\frac{25}{2}e^{-5}\)

\(P(X=3)=\frac{5^3 e^{-5}}{3!}=\frac{125}{6}e^{-5}\)

Add these probabilities to get the total probability for 3 or fewer calls:

\(P(X\leq 3)=e^{-5}+5e^{-5}+\frac{25}{2}e^{-5}+\frac{125}{6}e^{-5}\)

Factor out \(e^{-5}\) from the expression:

\(P(X\leq 3)=\left(1+5+\frac{25}{2}+\frac{125}{6}\right)e^{-5}\)

Convert each term to a common denominator, calculate the sum, and simplify:

\(1=\frac{6}{6},\,5=\frac{30}{6},\,\frac{25}{2}=\frac{75}{6},\,\frac{125}{6}=\frac{125}{6}\)

Add these fractions:

\(P(X\leq 3)=\left(\frac{6}{6}+\frac{30}{6}+\frac{75}{6}+\frac{125}{6}\right)e^{-5}= \left(\frac{236}{6}\right)e^{-5}\)

Simplify the fraction:

\(\frac{236}{6}=\frac{118}{3}\)

Therefore, the probability of receiving 3 or fewer calls per minute is:

\(\frac{118}{3}e^{-5}\)

This matches the correct option provided: \(\frac{118}{3}e^{-5}\).