Q79 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics
A random variable $\mathrm{X}$ has the following probability distribution
\begin{array}{|l|c|c|c|c|c|c|c|c|c|}
\hline \mathbf{X}=\mathbf{x}_{\mathrm{i}}: & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline \mathbf{P}\left(\mathbf{X}=\mathbf{x}_{\mathrm{i}}\right): & 10 \mathrm{k} & 9 \mathrm{k} & 8 \mathrm{k} & 8 \mathrm{k} & 6 \mathrm{k} & 5 \mathrm{k} & 4 \mathrm{k} & 3 \mathrm{k} & \mathrm{k} \\
\hline
\end{array}
where $\mathrm{k}$ is a real number
If $\mathrm{A}=\left\{x_i / x_i\right.$ is a prime number $\}$ and
$\mathrm{B}=\left\{x_i / x_i>5\right\}$ are two events, then $P(A \cup B)=$
- Correct answerA$\frac{2}{3}$
- B$\frac{4}{9}$
- C$\frac{1}{27}$
- D$\frac{5}{6}$
Solution
Acc. to question
$$
\begin{aligned}
\mathrm{P}(\mathrm{A}) & =\mathrm{P}(2)+\mathrm{P}(3)+\mathrm{P}(5)+\mathrm{P}(7) \\
& =9 \mathrm{k}+8 \mathrm{k}+6 \mathrm{k}+4 \mathrm{k} \\
& =27 \mathrm{k} \\
\mathrm{P}(\mathrm{B}) & =\mathrm{P}(6)+\mathrm{P}(7)+\mathrm{P}(8)+\mathrm{P}(9) \\
& =5 \mathrm{k}+4 \mathrm{k}+3 \mathrm{k}+\mathrm{k}=13 \mathrm{k} \\
\mathrm{P}(\mathrm{A} & \cap \mathrm{B})=\mathrm{p}(7)=4 \mathrm{k}
\end{aligned}
$$
$\begin{aligned} & \begin{aligned} \mathrm{P}(\mathrm{A} \cup \mathrm{B}) & =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ & =27 \mathrm{k}+13 \mathrm{k}-4 \mathrm{k} \\ & =36 \mathrm{k}\end{aligned} \\ & \text { now } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{36 \mathrm{k}}{54 \mathrm{k}}=\frac{2}{3}\end{aligned}$
Hence, the correct answer is option: 'A'.