A random variable \(X\) has the probability distribution \(\begin{array}{lllllll} \hline X=x_i & 1 & 2 & 3 &…
A random variable \(X\) has the probability distribution
\(\begin{array}{lllllll} \hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P\left(X=x_i\right) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\ \hline \end{array}\)
If \(A=\left\{x_i / x_i\right.\) is a prime number}, \(B=\left\{x_i / x_i < 4\right\}\) are two events, then \(P(A \cup B)=\)
0.31
0.62
0.82
0.41
Solution
The given probability distribution for random variable \(x\)
\(\begin{array}{lllllll}
\hline x=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline P\left(X=x_i\right) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\
\hline
\end{array}\)
and two events \(A=\left\{x_i \mid x_i\right.\) is a prime number \(\}\) \(=\{2,3,5\}\)
and \(B=\left\{x_i \mid x_i < 4\right\}=\{1,2,3\}\)
So,
\(\begin{aligned}
& P(A \cup B)=P(X=1)+P(X=2)+P(X=3)+P(X=5) \\
& =0.2+0.3+0.12+0.2=0.82
\end{aligned}\)
Hence, option (3) is correct.