Question

If the function $P [x = x] =$

$$\begin{cases} \frac{K \cdot 2^x}{x!}, x = 0,1,2,3...... \\ 0 \text{ otherwise} \end{cases}$$

forms p.m.f. then value of K is

Answer 1

$K = e^{-2}$

Answer 2

For $P[X = x] = \frac{K \cdot 2^x}{x!}$ to be a p.m.f., we require

$$\sum_{x=0}^{\infty} \frac{K \cdot 2^x}{x!} = 1.$$

Since

$$\sum_{x=0}^{\infty} \frac{2^x}{x!} = e^2,$$

we have

$$K \cdot e^2 = 1 \quad \Rightarrow \quad K = e^{-2}.$$

Core Explanation:

• Sum over all $x$ gives $K \cdot e^2 = 1$.
• Solve to obtain $K = e^{-2}$.