AP EAMCET · Maths · Probability
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
- A \(\begin{array}{cc} & A & B & C & D \\ & I & III & IV & II \end{array}\)
- B \(\begin{array}{cc} & A & B & C & D \\ & III & I & IV & V \end{array}\)
- C \(\begin{array}{cc} & A & B & C & D \\ & III & I & IV & II \end{array}\)
- D \(\begin{array}{cc} & A & B & C & D \\ & I & II & V & IV \end{array}\)
Answer & Solution
Correct Answer
(C) \(\begin{array}{cc} & A & B & C & D \\ & III & I & IV & II \end{array}\)
Step-by-step Solution
Detailed explanation
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…
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