If \(E_1\) and \(E_2\) are two events of a random experiment such that \(P\left(E_1\right)=\frac{1}{8},…
If \(E_1\) and \(E_2\) are two events of a random experiment such that \(P\left(E_1\right)=\frac{1}{8}, P\left(E_1 \mid E_2\right)=\frac{1}{3}\), \(P\left(E_2 \mid E_1\right)=\frac{1}{4}\), then match the items of List-I with the items of List-II.
\(\begin{array}{lllc}
\hline & \text { List-I } & & \text { List-II } \\
\hline \text { (A) } & P\left(E_2\right) & \text { I. } & \frac{3}{16} \\
\hline \text { (B) } & P\left(E_1 \cup E_2\right) & \text { II. } & \frac{3}{29} \\
\hline \text { (C) } & P\left(\bar{E}_1 \mid \bar{E}_2\right) & \text { III. } & \frac{3}{32} \\
\hline \text { (D) } & P\left(E_1 \mid \bar{E}_2\right) & \text { IV. } & \frac{26}{29} \\
\hline & & \text { V. } & \frac{13}{32} \\
\hline
\end{array}\)
The correct match is
- \(\begin{array}{cc} & A & B & C & D \\ & I & III & IV & II \end{array}\)
- \(\begin{array}{cc} & A & B & C & D \\ & III & I & IV & V \end{array}\)
- \(\begin{array}{cc} & A & B & C & D \\ & III & I & IV & II \end{array}\)
- \(\begin{array}{cc} & A & B & C & D \\ & I & II & V & IV \end{array}\)
Solution
For two given events \(E_1\) and \(E_2\), the given information are \(P\left(E_1\right)=\frac{1}{8}, P\left(E_1 \mid E_2\right)=\frac{1}{3}\) and
\(\begin{aligned}
& P\left(E_2 \mid E_1\right)=\frac{1}{4} \\
& \because P\left(E_2 \mid E_1\right)=\frac{1}{4} \Rightarrow \frac{P\left(E_1 \cap E_2\right)}{P\left(E_1\right)}=\frac{1}{4} \\
& \Rightarrow \quad P\left(E_1 \cap E_2\right)=\frac{1}{32} \\
& \therefore \quad P\left(E_2\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_1 \mid E_2\right)}=\frac{\frac{1}{32}}{\frac{1}{3}}=\frac{3}{32} \\
& \therefore P\left(E_1 \cup E_2\right)=P\left(E_1\right)+P\left(E_2\right)-P\left(E_1 \cap E_2\right) \\
& =\frac{1}{8}+\frac{3}{32}-\frac{1}{32}=\frac{3}{16}
\end{aligned}\)
\(\begin{aligned}
\because \quad P\left(\bar{E}_1\right) & =\frac{7}{8} \text { and } P\left(\bar{E}_2\right)=\frac{29}{32} \\
\text {and } P\left(\bar{E}_1 \cap \bar{E}_2\right) & =P\left(\overline{E_1 \cup E_2}\right) \\
& =1-P\left(E_1 \cup E_2\right)=\frac{13}{16} \\
\therefore \quad P\left(\bar{E}_1 \mid \bar{E}_2\right) & =\frac{P\left(\bar{E}_1 \cap \bar{E}_2\right)}{P\left(\bar{E}_2\right)}=\frac{\frac{13}{16}}{\frac{29}{32}}=\frac{26}{29} \\
\text {and } \quad P\left(E_1 \mid \bar{E}_2\right) & =1-P\left(\bar{E}_1 \mid \bar{E}_2\right)=1-\frac{26}{29}=\frac{3}{29}
\end{aligned}\)
Hence, option (c) is correct.
Asked in: AP EAMCET 2019 (23 Apr Shift 1)
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