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Q.

A man is known to speak truth 3 out of 4 times. He throws a die and report that it is a six. Then the probability that it is actually six is

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a

$\frac{3}{4}$

b

$\frac{3}{8}$

c

$\frac{3}{16}$

d

$\frac{1}{3}$

answer is B.

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Detailed Solution

(2)
A = man reports six
E₁= six appears $\Rightarrow P (E_{1}) = \frac{1}{6}$
E₂ = six does not appears $\Rightarrow P (E_{2}) = \frac{5}{6}$
$P (E_{1} | A) = \frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}} = \frac{3}{8}$

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