MHT CET · Maths · Probability
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
- A \(\frac{5}{19}\)
- B \(\frac{2}{19}\)
- C \(\frac{3}{19}\)
- D \(\frac{1}{19}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{19}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sum P(x)=1 \\ & \Rightarrow k \cdot \frac{2^0}{0 !}+k \cdot \frac{2^1}{1 !}+k \cdot \frac{2^2}{2 !}+k \cdot \frac{2^3}{3 !}=1 \\ & \Rightarrow \frac{19}{3} k=1 \\ & \Rightarrow k=\frac{3}{19}\end{aligned}\)
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