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The probability distribution of a discrete r. v. $X$ is \begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2 & 3 & 4 \\
\hline \mathrm{P}(\mathrm{X}=x) & k & 2 k & 4 \mathrm{k} & 2 k & k \\
\hline
\end{array}
then value of $\mathrm{P}(\mathrm{X} \leq 2)$ is
Options:
\hline \mathrm{X}=x & 0 & 1 & 2 & 3 & 4 \\
\hline \mathrm{P}(\mathrm{X}=x) & k & 2 k & 4 \mathrm{k} & 2 k & k \\
\hline
\end{array}
then value of $\mathrm{P}(\mathrm{X} \leq 2)$ is
Solution:
1138 Upvotes
Verified Answer
The correct answer is:
$\frac{7}{10}$
Here $\mathrm{k}+2 \mathrm{k}+4 \mathrm{k}+2 \mathrm{k}+\mathrm{k}=10 \mathrm{k}=1 \Rightarrow \mathrm{k}=\frac{1}{10}$
Now $\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(0)+\mathrm{P}(1)+\mathrm{P}(2)=\frac{1}{10}+\frac{2}{10}+\frac{4}{10}=\frac{7}{10}$
Now $\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(0)+\mathrm{P}(1)+\mathrm{P}(2)=\frac{1}{10}+\frac{2}{10}+\frac{4}{10}=\frac{7}{10}$
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