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A man is known to speak the truth in 75% cases. He throws a die and reports that it is a five. The probability that it is actually a five is

$\begin {array} {1 1} (1)\;\large\frac{3}{8} & \quad (2)\;\large\frac{3}{4} \\ (3)\;\large\frac{1}{5} & \quad (4)\;None\: of \: these \end {array}$

If E denote the event that five occurs and A the event
that the man reports that it is five.
$P(E)= \large\frac{1}{6}$
$P(E')= \large\frac{5}{6}$
$P(A/E)= \large\frac{3}{4}\: \: \: \: P(A/E')= \large\frac{1}{4}$
$\therefore$ By Baye's theorem
$P(E/A) = \large\frac{\large\frac{1}{6} \times \large\frac{3}{4}}{\large\frac{1}{6} \times \large\frac{3}{4} + \large\frac{5}{6} \times \large\frac{1}{4}}$
$= \large\frac{3}{8}$
Ans : (1)