MHT CET · Maths · Probability
If the p. m. f. of a random variable \(\mathrm{X}\) is
\(\begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{X} & 1 & 2 & 3 & 4 & 5\mathrm{P}(\mathrm{X}=x) & k & \frac{k}{3} & \frac{k}{4} & \frac{k}{2} & \frac{k}{2} \\ \hline\end{array}\)
then \(k=\)
- A \(\frac{15}{31}\)
- B \(\frac{1}{12}\)
- C \(\frac{11}{12}\)
- D \(\frac{12}{31}\)
Answer & Solution
Correct Answer
(D) \(\frac{12}{31}\)
Step-by-step Solution
Detailed explanation
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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