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Question:
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If the p. m. f. of a random variable $\mathrm{X}$ is
\begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 \\
\hline \mathrm{P}(\mathrm{X}=x) & k & \frac{k}{3} & \frac{k}{4} & \frac{k}{2} & \frac{k}{2} \\
\hline
\end{array}
then $k=$
Options:
\begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 \\
\hline \mathrm{P}(\mathrm{X}=x) & k & \frac{k}{3} & \frac{k}{4} & \frac{k}{2} & \frac{k}{2} \\
\hline
\end{array}
then $k=$
Solution:
1473 Upvotes
Verified Answer
The correct answer is:
$\frac{12}{31}$
Here $\mathrm{k}+\frac{\mathrm{k}}{3}+\frac{\mathrm{k}}{4}+\frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{2}=1$
$\therefore \mathrm{k}\left(\frac{12+4+3+6+6}{12}\right)=1 \Rightarrow \mathrm{k}\left(\frac{31}{12}\right)=1 \Rightarrow \mathrm{k}=\frac{12}{13}$
$\therefore \mathrm{k}\left(\frac{12+4+3+6+6}{12}\right)=1 \Rightarrow \mathrm{k}\left(\frac{31}{12}\right)=1 \Rightarrow \mathrm{k}=\frac{12}{13}$
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