CBSE 2022 · Region 5 · Set 1 · Q3 · 2 marks
Let X be a random variable which assumes values $\displaystyle \mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}_{4}$ such that $\displaystyle 2 \mathrm{P}\left(\mathrm{X}=\mathrm{x}_{1}\right)=3 \mathrm{P}\left(\mathrm{X}=\mathrm{x}_{2}\right)=\mathrm{P}\left(\mathrm{X}=\mathrm{x}_{3}\right)=5 \mathrm{P}\left(\mathrm{X}=\mathrm{x}_{4}\right)$. Find the probability distribution of X.
Marking-scheme solution
\[\begin{aligned}
& \text { Let } \mathrm{P}\left(X=x_{3}\right)=k \\
& \mathrm{P}\left(X=x_{1}\right)=\frac{k}{2}, P\left(X=x_{2}\right)=\frac{k}{3}, P\left(X=x_{4}\right)=\frac{k}{5} \\
& \frac{k}{2}+\frac{k}{3}+k+\frac{k}{5}=1 \\
& k=\frac{30}{61}
\end{aligned}
\]
ProbabilityRandom Variable and its Probability DistributionApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.