Question

A telephone switch board receiving number of phone calls follows Poisson distribution with parameter $3$ per hour. The probability that it receives $5$ calls in $2$ hours duration is

• A. $\frac{e^{-3}{3}^5}{5!}$
• B. $1-\frac{e^{-3}{3}^5}{5!}$
• C. $\frac{e^{-6}{3}^5}{5!}$
• D. $1-\frac{e^{-6}{3}^5}{5!}$

Answer 1

The correct option is C $\frac{e^{-6}{3}^5}{5!}$

By using Poisson distribution,
$P(x;\mu )=\frac{e^{-\mu }{\mu }^x}{x!}$
$\mu$: The mean number of successes that occur in a specified region(parameter)
x: The actual number of successes that occur in a specified region.
P(x; $\mu$): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $\mu$
Poisson distribution with parameter 3 per hour. The probability that it receives 5 calls in 2 hours duration so, $\mu =3\times 2=6$ and $x=5$
$P(x;\mu )=\frac{e^{-6}{3}^5}{5!}$