A probability of receiving k emergency calls in an hour is given by a function. Find m if you know that expected number of phone calls is 3

Question

Got stuck with this one. Please help

A probability of receiving k emergency calls in an hour is given by the probability function:

$$P(k)=\dfrac{\mathsf e^{-m} ~ m^k}{k!}$$

Find $m$ if you know that expected number of phone calls is $3$.

Do it by calculating the expected number of phone calls for specific m. For example try m=1, 1.5, 2, 2.5,… .

Do it as many times as you need to figure out formula.

Answer 1

The expected number of calls is $\mathsf E(K; m) = \sum\limits_{k=0}^\infty k~P(k)$

$$\begin{align}\mathsf E(K;m) = & ~\mathsf e^{-m} \sum_{k=0}^\infty \frac{k~m^k}{k!} \\[1ex] = & ~ m~\mathsf e^{-m}\sum_{h=0}^\infty \frac{m^h}{h!}\end{align}$$

Your instruction is to do this for particular values of $m$. $$\begin{align}\mathsf E(K;m=1) = & \mathsf e^{-1}\Big(0+1+\frac 2{2!}+\ldots\Big) \\[1ex] = & ~ 1~\mathsf e^{-1}\Big(0+1+\frac{1}{1!}+\frac 1{2!}+ ...\Big)\\[2ex] \mathsf E(K; m=2) = & ~ \mathsf e^{-2}\Big(0+2+\frac{4}{2!}+\frac{12}{3!}+\ldots\Big) \\[1ex] = & 2~\mathsf e^{-2}~\Big(0+1+\frac{2}{1!}+\frac {2^2}{2!}+\ldots\Big)\\[3ex] \textsf{et cetera}\end{align}$$

Carry on until you spot the obvious pattern. It emerges fairly soon.

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Hint: (spoiler warning)

A key point is to recall the Taylor Series expansion (around zero) of $\mathsf e^x$ ?