A man is known to speak truth 2 out of 3 times. He throws a die and declares that it is one that appeared. The probability that it is actually one is

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$\frac{2}{3}$

$\frac{2}{5}$

$\frac{2}{6}$

$\begin{array}{l} 2 \\ \bar{7} \end{array}$

answer is D.

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Detailed Solution

$P (A) = \frac{1}{6}, P (\bar{A}) = \frac{5}{6}$

Let ‘E’ be the event that the man declares that it is one

$\begin{array}{l} P (\frac{E}{A}) = \frac{2}{3}, P (\frac{E}{\bar{A}}) = \frac{1}{3} \\ P (\frac{A}{E}) = \frac{P (A) . P (\frac{E}{A})}{P (A) . P (\frac{E}{A}) + P (\bar{A}) . P (\frac{E}{\bar{A}})} \end{array}$

$\begin{array}{l} = \frac{\frac{1}{6} \times \frac{2}{3}}{(\frac{1}{6} \times \frac{2}{3}) + (\frac{5}{6} \times \frac{1}{3})} \\ = \frac{2}{7} \end{array}$

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