MCQ
If the function $p[X=x]=\left\{\begin{array}{cl}\frac{K \cdot 2^x}{x !}, & x=0,1,2,3 \\ 0, & \text { otherwise }\end{array}\right.$ forms p.m.f., then value of $K$ is
- A$\frac{5}{19}$
- B$\frac{1}{19}$
- C$\frac{3}{19}$
- D$\frac{2}{19}$
✓
Answer
(c) : Since the function is p.m.f so,
$
\begin{aligned}
& \Rightarrow \sum_{x=0}^3 \frac{K \cdot 2^x}{x !}=1 \\
& \Rightarrow K+2 K+2 K+\frac{4}{3} K=1 \Rightarrow \frac{19 K}{3}=1 \Rightarrow K=\frac{3}{19}
\end{aligned}
$
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