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A random variable $\mathrm{X}$ has the following probability distribution
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline \mathrm{X}=\mathrm{x} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & 0.15 & 0.23 & \mathrm{k} & 0.10 & 0.20 & 0.08 & 0.07 & 0.05 \\
\hline
\end{array}
For the events $E=\{x / x$ is a prime number $\}$ and $F=\{x / x < 4\}$ then $P(E \cup F)=$
Options:
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline \mathrm{X}=\mathrm{x} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & 0.15 & 0.23 & \mathrm{k} & 0.10 & 0.20 & 0.08 & 0.07 & 0.05 \\
\hline
\end{array}
For the events $E=\{x / x$ is a prime number $\}$ and $F=\{x / x < 4\}$ then $P(E \cup F)=$
Solution:
2729 Upvotes
Verified Answer
The correct answer is:
$0.77$
$\Sigma \mathrm{P}=1$
$0.15+0.23+\mathrm{K}+0.10+0.20+0.08+0.07+0.05=1$
$\begin{aligned} & 0.88+\mathrm{K}=1 \\ & \mathrm{~K}=0.12 \\ & \mathrm{E}=\{x: x \text { is a prime number }\} \\ & \mathrm{E}=\{2,3,5,7\} \\ & \mathrm{F}=\{1,2,3\} \\ & \mathrm{E} \cup \mathrm{F}=\{1,2,3,5,7\} \\ & \mathrm{P}(\mathrm{E} \cup \mathrm{F})=\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=5) \\ & +\mathrm{P}(\mathrm{X}=7)\end{aligned}$
$=0.15+0.23+0.12+0.20+0.07=0.77$
$0.15+0.23+\mathrm{K}+0.10+0.20+0.08+0.07+0.05=1$
$\begin{aligned} & 0.88+\mathrm{K}=1 \\ & \mathrm{~K}=0.12 \\ & \mathrm{E}=\{x: x \text { is a prime number }\} \\ & \mathrm{E}=\{2,3,5,7\} \\ & \mathrm{F}=\{1,2,3\} \\ & \mathrm{E} \cup \mathrm{F}=\{1,2,3,5,7\} \\ & \mathrm{P}(\mathrm{E} \cup \mathrm{F})=\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=5) \\ & +\mathrm{P}(\mathrm{X}=7)\end{aligned}$
$=0.15+0.23+0.12+0.20+0.07=0.77$
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