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A random variable $\mathrm{X}$ has the following probability distribution
\begin{array}{|r|l|l|l|l|l|l|l|l|l|}
\hline \mathrm{x} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & \mathrm{K} & 2 \mathrm{~K} & 3 \mathrm{~K} & 4 \mathrm{~K} & 4 \mathrm{~K} & 3 \mathrm{~K} & 2 \mathrm{~K} & \mathrm{~K} & \mathrm{~K} \\
\hline
\end{array}
Then $\mathrm{P}(3 < \mathrm{x} \leq 6)=$
Options:
\begin{array}{|r|l|l|l|l|l|l|l|l|l|}
\hline \mathrm{x} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & \mathrm{K} & 2 \mathrm{~K} & 3 \mathrm{~K} & 4 \mathrm{~K} & 4 \mathrm{~K} & 3 \mathrm{~K} & 2 \mathrm{~K} & \mathrm{~K} & \mathrm{~K} \\
\hline
\end{array}
Then $\mathrm{P}(3 < \mathrm{x} \leq 6)=$
Solution:
1928 Upvotes
Verified Answer
The correct answer is:
$\frac{3}{7}$
We know that
$$
\begin{aligned}
& \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+4 \mathrm{k}+4 \mathrm{k}+3 \mathrm{k}+2 \mathrm{k}+\mathrm{k}+\mathrm{k}=1 \Rightarrow 21 \mathrm{k}=1 \\
& \Rightarrow \mathrm{k}=\frac{1}{21}
\end{aligned}
$$
When $x=4, P=4 k=\frac{4}{21}$, When $x=5, P=3 k=\frac{3}{21}$,
When $\mathrm{x}=6, \mathrm{P}=2 \mathrm{k}=\frac{2}{21}$
$$
\therefore \mathrm{P}(3 < \mathrm{x} \leq 6)==\frac{4+3+2}{21}=\frac{9}{21}=\frac{3}{7}
$$
$$
\begin{aligned}
& \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+4 \mathrm{k}+4 \mathrm{k}+3 \mathrm{k}+2 \mathrm{k}+\mathrm{k}+\mathrm{k}=1 \Rightarrow 21 \mathrm{k}=1 \\
& \Rightarrow \mathrm{k}=\frac{1}{21}
\end{aligned}
$$
When $x=4, P=4 k=\frac{4}{21}$, When $x=5, P=3 k=\frac{3}{21}$,
When $\mathrm{x}=6, \mathrm{P}=2 \mathrm{k}=\frac{2}{21}$
$$
\therefore \mathrm{P}(3 < \mathrm{x} \leq 6)==\frac{4+3+2}{21}=\frac{9}{21}=\frac{3}{7}
$$
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