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A man speaks truth \( 2 \) out of \( 3 \) times. He picks one of the natural numbers in the set \( S=\{1,2, \),
\( 3,4,5,6,7\} \) and reports that it is even. The probability that it is actually even is
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\( 3,4,5,6,7\} \) and reports that it is even. The probability that it is actually even is
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Verified Answer
The correct answer is:
\( \frac{3}{5} \)
$\mathrm{S}=\{1,2,3,4,5,6,7\}$
$E_{1}=$ An even number is picked, $E_{2}=$ An odd number is picked
$P\left(E_{1}\right)=\frac{3}{7}, P\left(E_{2}\right)=\frac{4}{7}$
$E: A$ man reports an even number
$P\left(E \mid E_{1}\right)=\frac{2}{3}$
$P\left(E \mid E_{2}\right)=\frac{1}{3}$
Required probability $=P\left(E_{1} \mid E\right)$
$=\frac{P\left(E \mid E_{1}\right) P\left(E_{1}\right)}{P\left(E \mid E_{1}\right) P\left(E_{1}\right)+P\left(E \mid E_{2}\right) P\left(E_{2}\right)}$
$=\frac{\left(\frac{2}{3}\right)\left(\frac{3}{7}\right)}{\left(\frac{2}{3}\right)\left(\frac{3}{7}\right)+\left(\frac{1}{3}\right)\left(\frac{4}{7}\right)}=\frac{6}{6+4}=\frac{3}{5}$
$E_{1}=$ An even number is picked, $E_{2}=$ An odd number is picked
$P\left(E_{1}\right)=\frac{3}{7}, P\left(E_{2}\right)=\frac{4}{7}$
$E: A$ man reports an even number
$P\left(E \mid E_{1}\right)=\frac{2}{3}$
$P\left(E \mid E_{2}\right)=\frac{1}{3}$
Required probability $=P\left(E_{1} \mid E\right)$
$=\frac{P\left(E \mid E_{1}\right) P\left(E_{1}\right)}{P\left(E \mid E_{1}\right) P\left(E_{1}\right)+P\left(E \mid E_{2}\right) P\left(E_{2}\right)}$
$=\frac{\left(\frac{2}{3}\right)\left(\frac{3}{7}\right)}{\left(\frac{2}{3}\right)\left(\frac{3}{7}\right)+\left(\frac{1}{3}\right)\left(\frac{4}{7}\right)}=\frac{6}{6+4}=\frac{3}{5}$
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