A telephone exchange receives on an average 5 calls per minute. The probability of receiving 3 or less calls per minute is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) - $\frac{118}{3} e^{-5}$
$X \sim \text{Poisson}(5)$
$P(X \le 3) = \sum_{k=0}^{3} \frac{e^{-5} 5^k}{k!}$
$= e^{-5}\left(1 + 5 + \frac{25}{2} + \frac{125}{6}\right)$
$= e^{-5}\left(\frac{6 + 30 + 75 + 125}{6}\right)$
$= e^{-5}\left(\frac{236}{6}\right)$
$= \frac{118}{3}e^{-5}$
$P(X \le 3) = \frac{118}{3}e^{-5}$