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A random variable $X$ has the probability distribution
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline \mathrm{p}(\mathrm{X}) & 0.15 & 0.23 & 0.12 & 0.10 & 0.20 & 0.08 & 0.07 & 0.05 \\
\hline
\end{array}
For the events $E=\{X$ is a prime number $\}$ and $\mathrm{F}=\{\mathrm{X} < 4\}$, then $\mathrm{P}(\mathrm{E} \cup \mathrm{F})$ is
Options:
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline \mathrm{p}(\mathrm{X}) & 0.15 & 0.23 & 0.12 & 0.10 & 0.20 & 0.08 & 0.07 & 0.05 \\
\hline
\end{array}
For the events $E=\{X$ is a prime number $\}$ and $\mathrm{F}=\{\mathrm{X} < 4\}$, then $\mathrm{P}(\mathrm{E} \cup \mathrm{F})$ is
Solution:
2094 Upvotes
Verified Answer
The correct answer is:
$0.77$
$$
\begin{array}{l}
\begin{array}{l}
P(E)=P(2 \text { or } 3 \text { or } 5 \text { or } 7) \\
=0.23+0.12+0.20+0.07=0.62 \\
P(F)=P(1 \text { or } 2 \text { or } 3) \\
=0.15+0.23+0.12=0.50 \\
P(E \cap F)=P(2 \text { or } 3) \\
=0.23+0.12=0.35 \\
\therefore P(E U F)=P(E)+P(F)-P(E \cap F) \\
=0.62+0.50-0.35=0.77
\end{array}
\end{array}
$$
\begin{array}{l}
\begin{array}{l}
P(E)=P(2 \text { or } 3 \text { or } 5 \text { or } 7) \\
=0.23+0.12+0.20+0.07=0.62 \\
P(F)=P(1 \text { or } 2 \text { or } 3) \\
=0.15+0.23+0.12=0.50 \\
P(E \cap F)=P(2 \text { or } 3) \\
=0.23+0.12=0.35 \\
\therefore P(E U F)=P(E)+P(F)-P(E \cap F) \\
=0.62+0.50-0.35=0.77
\end{array}
\end{array}
$$
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