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A random variable $X$ has the probability distribution
\begin{array}{l|c|c|c|c|c|c|c|c}
\hlinex & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hlineP(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 prime number $\}$ and $F=\{x < 4\}$ the probability of $P(E \cup F)$ is
Options:
\begin{array}{l|c|c|c|c|c|c|c|c}
\hlinex & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hlineP(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 prime number $\}$ and $F=\{x < 4\}$ the probability of $P(E \cup F)$ is
Solution:
1218 Upvotes
Verified Answer
The correct answer is:
$0.77$
Given, $E=\{x$ is a prime number $\}$
$$
\begin{aligned}
P(E) &=P(2)+P(3)+P(5)+P(7) \\
&=0.23+0.12+0.20+0.07=0.62 \\
\text { and } \quad F &=\{x < 4\} \\
P(F) &=P(1)+P(2)+P(3) \\
&=0.15+0.23+0.12=0.50 \\
\text { and } \quad P(E \cap F) &=P(2)+P(3) \\
=& 0.23+0.12=0.35 \\
\therefore \quad P(E \cup F) &=P(E)+P(F)-P(E \cap F) \\
&=0.62+0.50-0.35 \\
&=0.77
\end{aligned}
$$
$$
\begin{aligned}
P(E) &=P(2)+P(3)+P(5)+P(7) \\
&=0.23+0.12+0.20+0.07=0.62 \\
\text { and } \quad F &=\{x < 4\} \\
P(F) &=P(1)+P(2)+P(3) \\
&=0.15+0.23+0.12=0.50 \\
\text { and } \quad P(E \cap F) &=P(2)+P(3) \\
=& 0.23+0.12=0.35 \\
\therefore \quad P(E \cup F) &=P(E)+P(F)-P(E \cap F) \\
&=0.62+0.50-0.35 \\
&=0.77
\end{aligned}
$$
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