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# A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is 6. The probability that it is actually a 6 is

$\begin {array} {1 1} (A)\;\large\frac{1}{18} & \quad (B)\;\large\frac{1}{4} \\ (C)\;\large\frac{3}{8} & \quad (D)\;\large\frac{1}{2} \end {array}$

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## 1 Answer

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Let E denote the event that a six occurs and
A the event that the man reports that it is a 6.
$\therefore P(E)= \large\frac{1}{6}\: \: \: P(E) = \large\frac{5}{6}$
$P \bigg( \large\frac{A}{E} \bigg) = \large\frac{3}{4}\: \: \: P \bigg( \large\frac{A}{E} \bigg) = \large\frac{1}{4}$
$P \bigg( \large\frac{E}{A} \bigg) = \large\frac{P(E).P \bigg( \large\frac{A}{E} \bigg)}{P(E)P \bigg( \large\frac{A}{E} \bigg)+P(E).P \bigg( \large\frac{A}{E} \bigg) }$
$\Large\frac{\Large\frac{1}{6} \times \Large\frac{3}{4}}{\Large\frac{1}{6} \times \Large\frac{3}{4} + \Large\frac{5}{6} \times \Large\frac{1}{4}}$
$= \large\frac{3}{8}$
Ans : (C)

answered Jan 15, 2014

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