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A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six, is
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The correct answer is:
${ }^{\frac{3}{8}}$
Let $E$ denote the event that a six occurs and $A$ the event that the man reports that it is a ' 6 ', we have
$P(E)=\frac{1}{6}, P\left(E^{\prime}\right)=\frac{5}{6}, P(A / E)=\frac{3}{4}$ and $P\left(A / E^{\prime}\right)=\frac{1}{4}$
$P(E / A)=\frac{P(E) \cdot P(A / E)}{P(E) \cdot P(A / E)+P\left(E^{\prime}\right) \cdot P\left(A / E^{\prime}\right)}$
$=\frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4}+\frac{5}{6} \times \frac{1}{4}}=\frac{3}{8}$.
$P(E)=\frac{1}{6}, P\left(E^{\prime}\right)=\frac{5}{6}, P(A / E)=\frac{3}{4}$ and $P\left(A / E^{\prime}\right)=\frac{1}{4}$
$P(E / A)=\frac{P(E) \cdot P(A / E)}{P(E) \cdot P(A / E)+P\left(E^{\prime}\right) \cdot P\left(A / E^{\prime}\right)}$
$=\frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4}+\frac{5}{6} \times \frac{1}{4}}=\frac{3}{8}$.
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