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\(A\) speaks truth in \(20 \%\) of the cases and \(B\) in \(80 \%\) of the cases. Find the probability that their statements about an incident do not match.
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Verified Answer
The correct answer is:
\(\frac{17}{25}\)
Let \(E_1\) be the event of \(A\) speaking the truth
\(\begin{aligned}
\therefore \quad & P\left(E_1\right)=\frac{20}{100}=\frac{1}{5} \\
& P\left(\overline{E_1}\right)=1-\frac{1}{5}=\frac{4}{5}
\end{aligned}\)
\(E_2\) be the event of \(B\) speaking the truth.
\(\begin{aligned}
& P\left(E_2\right)=\frac{80}{100} \\
& P\left(E_2\right)=\frac{4}{5} \\
& P\left(\overline{E_2}\right)=1-\frac{4}{5}=\frac{1}{5}
\end{aligned}\)
Required, Probability \(=P\left(E_1\right) \cdot P\left(\bar{E}_2\right)+P\left(E_2\right) \cdot P\left(\bar{E}_1\right)\)
\(=\frac{1}{5} \times \frac{1}{5}+\frac{4}{5} \times \frac{4}{5}=\frac{1}{25}+\frac{16}{25}=\frac{17}{25}\)
\(\begin{aligned}
\therefore \quad & P\left(E_1\right)=\frac{20}{100}=\frac{1}{5} \\
& P\left(\overline{E_1}\right)=1-\frac{1}{5}=\frac{4}{5}
\end{aligned}\)
\(E_2\) be the event of \(B\) speaking the truth.
\(\begin{aligned}
& P\left(E_2\right)=\frac{80}{100} \\
& P\left(E_2\right)=\frac{4}{5} \\
& P\left(\overline{E_2}\right)=1-\frac{4}{5}=\frac{1}{5}
\end{aligned}\)
Required, Probability \(=P\left(E_1\right) \cdot P\left(\bar{E}_2\right)+P\left(E_2\right) \cdot P\left(\bar{E}_1\right)\)
\(=\frac{1}{5} \times \frac{1}{5}+\frac{4}{5} \times \frac{4}{5}=\frac{1}{25}+\frac{16}{25}=\frac{17}{25}\)
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