A random variable \(X\) has the probability distribution as given below. Let \(E=[X \mid X\) is prime…
A random variable \(X\) has the probability distribution as given below. Let \(E=[X \mid X\) is prime number] and \(F=\{X \mid X < 4\}\), then \(P(E \cup F)=\)
\(\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline P(X) & K & 2 K & K^2 & 2 K^2 & 5 K^2 & K & K & 2 K \\
\hline
\end{array}\)
\(\frac{38}{64}\)
\(\frac{39}{64}\)
\(\frac{42}{64}\)
\(\frac{17}{64}\)
Solution
Given probability distribution is
\(\begin{array}{c|c|c|c|c|c|c|c|c}
\hline \boldsymbol{X} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline P(X) & K & 2 K & K^2 & 2 K^2 & 5 K^2 & K & K & 2 K \\
\hline
\end{array}\)
\(\begin{aligned}
\because & \Sigma P(X) & =1 \\
\Rightarrow & 8 K^2+7 K & =1 \\
\Rightarrow & 8 K^2+7 K-1 & =0 \\
\Rightarrow & 8 K^2+8 K-K-1 & =0 \\
\Rightarrow & 8 K(K+1)-1(K+1) & =0 \\
\Rightarrow & K=\frac{1}{8} \text { as } K & > 0 .
\end{aligned}\)
\(\because\) The events (given) \(E=\{x \mid x\) is a prime}
\(=\{2,3,5,7\}\)
and \(F=\{X \mid X < 4\}=\{1,2,3\}\)
\(\begin{aligned}
& \therefore \quad P(E \cup F)=P(E)+P(F)-P(E \cap F) \\
& =\left(2 K+K^2+5 K^2+K\right)+\left(K+2 K+K^2\right)-\left(2 K+K^2\right) \\
& =6 K^2+4 K=6 \frac{1}{64}+\frac{4}{8}=\frac{38}{64}
\end{aligned}\)
Hence, option (a) is correct.