Q78 / 160 · TS EAMCET 2020 (09 Sep Shift 2) · Mathematics

$E_1$ and $E_2$ are two independent events of a random experiment with $P\left(E_1\right)=\frac{1}{2}$ and $P\left(E_1 \cup E_2\right)=\frac{2}{3}$. Then match the items of List-1 with those of List-II. Figure
The correct match is $\begin{array}{lllllllll}A & \text { B } & \text { C } & \text { D } & \text { A } & \text { B } & \text { C } & \text { D }\end{array}$

  • A
    $\begin{array}{llll}\text { III } & \text { II } & \text { I } & \text { IV }\end{array}$
  • Correct answer
    B
    III IV I II
  • C
    $\begin{array}{llll} & \text { III } & \text { II } & \text { I }\end{array}$
  • D
    $\begin{array}{llll}\text { IV } & \text { III } & \text { I } & \text { II }\end{array}$

Solution

We have, $P\left(E_1\right)=\frac{1}{2}$ and $P\left(E_1 \cup E_2\right)=\frac{2}{3}$. Since, $E_1$ and $E_2$ are independent events $\therefore \quad P\left(E_1 \cap E_2\right)=P\left(E_1\right) \times P\left(E_2\right)$ Let $P\left(E_2\right)$ be $x$. Then, $\quad P\left(E_1 \cap E_2\right)=\frac{x}{2}$ $\begin{array}{rlrl} & & P\left(E_1 \cup E_2\right) & =P\left(E_1\right)+P\left(E_2\right)-P\left(E_1 \cap E_2\right) \\ & \Rightarrow & \frac{2}{3}=\frac{1}{2}+x-\frac{x}{2} \\ \Rightarrow & \frac{2}{3}=\frac{1}{2}+\frac{x}{2} \\ \Rightarrow & \frac{x}{2}=\frac{2}{3}-\frac{1}{2}=\frac{4-3}{6}=\frac{1}{6} \\ \Rightarrow & \frac{x}{2}=\frac{1}{6} \Rightarrow x=\frac{1}{3} \\ \therefore & P\left(E_2\right)=\frac{1}{3} \\ \text { and } & P\left(E_1 \cap E_2\right)=\frac{1 / 3}{2}=\frac{1}{6}\end{array}$ $\begin{aligned} & \text { Now, } \begin{aligned} & P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}=\frac{1 / 6}{1 / 3}=\frac{1}{6} \times 3=\frac{1}{2} \\ & P\left(\frac{\bar{E}_2}{E_1}\right)=\frac{P\left(\bar{E}_2 \cap E_1\right)}{P\left(E_1\right)}=\frac{P\left(E_1\right)-P\left(E_1 \cap E_2\right)}{P\left(E_1\right)} \\ &=\frac{\frac{1}{2}-\frac{1}{6}}{\frac{1}{2}}=\frac{1 / 3}{1 / 2}=\frac{1}{3} \times 2=\frac{2}{3} \\ & P\left(\bar{E}_1 \cup \bar{E}_2\right)=1-P\left(E_1 \cap E_2\right)=1-\frac{1}{6}=\frac{5}{6}\end{aligned} \\ & \therefore A \rightarrow \mathrm{III}, B \rightarrow \mathrm{IV}, C \rightarrow \mathrm{I}, D \rightarrow \mathrm{II}\end{aligned}$

Hence, the correct answer is option: 'B'.

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