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If the function $P[X=x]=\left\{\begin{array}{cc}\frac{K \cdot 2^x}{x !}, & x=0,1,2,3 \\ 0, & \text { otherwise }\end{array}\right.$ Forms p.m.f., then value of $K$ is
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The correct answer is:
$\frac{3}{19}$
$\begin{aligned} & \sum P(x)=1 \\ & \Rightarrow k \cdot \frac{2^0}{0 !}+k \cdot \frac{2^1}{1 !}+k \cdot \frac{2^2}{2 !}+k \cdot \frac{2^3}{3 !}=1 \\ & \Rightarrow \frac{19}{3} k=1 \\ & \Rightarrow k=\frac{3}{19}\end{aligned}$
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