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

$\frac{118}{3} e^{-5}$

The correct answer is Option (2) - $\frac{118}{3} e^{-5}$

$X \sim \text{Poisson}(5)$

$P(X \le 3) = \sum_{k=0}^{3} \frac{e^{-5} 5^k}{k!}$

$= e^{-5}\left(1 + 5 + \frac{25}{2} + \frac{125}{6}\right)$

$= e^{-5}\left(\frac{6 + 30 + 75 + 125}{6}\right)$

$= e^{-5}\left(\frac{236}{6}\right)$

$= \frac{118}{3}e^{-5}$

$P(X \le 3) = \frac{118}{3}e^{-5}$