$E_1$ and $E_2$ are two independent events of a random experiment such that $P\left(E_1\right)=\frac{1}{2}$…

$E_1$ and $E_2$ are two independent events of a random experiment such that $P\left(E_1\right)=\frac{1}{2}$ and $P\left(E_1 \cup E_2\right)=\frac{2}{3}$. Then match the items of List-I with the items of List-II.
The correct match is
  1. A-iii; B-iv; $\quad$ C-i; $\quad$ D-v
  2. A-iii; B-i; $\quad$ C-v; $\quad$ D-ii
  3. A-i; $\quad$ B-v; $\quad$ C-ii; $\quad$ D-iv
  4. A-v; $\quad$ B-i; $\quad$ C-iii; $\quad$ D-ii

Solution

$\mathrm{P}\left(\mathrm{E}_1\right)=\frac{1}{2}, \mathrm{P}\left(\mathrm{E}_1 \cup \mathrm{E}_2\right)=\frac{2}{3}$
Since, $E_1$ and $E_2$ are independent events $\therefore P\left(E_1 \cap E_2\right)=P\left(E_1\right) P\left(E_2\right)$ $\Rightarrow P\left(E_1 \cap E_2\right)=\frac{1}{2} P\left(E_2\right)$ ...(i) $\begin{aligned} & \text { Also, } P\left(E_1 \cup E_2\right)=P\left(E_1\right)+P\left(E_2\right)-P\left(E_1 \cap E_2\right) \\ & \Rightarrow \frac{2}{3}=\frac{1}{2}+P\left(E_2\right)-\frac{1}{2} P\left(E_2\right) [Using(i)]\\ & \Rightarrow \frac{1}{6}=\frac{1}{2} P\left(E_2\right) \Rightarrow \frac{1}{3}=P\left(E_2\right) \\ & \text { So, } P\left(E_1 \cap E_2\right)=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6} \\ & P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}=\frac{1}{6} \times 3=\frac{1}{2} \\ & P\left(\frac{\bar{E}_2}{E_1}\right)=\frac{P\left(\bar{E}_2 \cap E_1\right)}{P\left(E_1\right)}=\frac{P\left(E_1\right)-P\left(E_1 \cap E_2\right)}{P\left(E_1\right)}=\frac{\frac{1}{2}-\frac{1}{6}}{\frac{1}{2}}=\frac{2}{3} \\ & P\left(\bar{E}_1 \cup \bar{E}_2\right)=P\left(\overline{E_1 \cap E_2}\right)=1-P\left(E_1 \cap E_2\right)=1-\frac{1}{6}=\frac{5}{6} \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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