A man is known to speak truth 3 out of 4 times He throws a die reports - Maths - Probability

Question

A man is known to speak truth 3 out of 4 times He throws a die
& reports that it is a six Find the probability that it is
actually a six

Lalit Mehra · Accepted Answer

Let E1 : The event that six comes on the die E2 : The event that six does not comes on the die A : The event that man report it is a six $∴ P (E_{1}) = \frac{1}{6} and P (E_{2}) = 1 - P (E_{1}) = 1 - \frac{1}{6} = \frac{5}{6}$ Probability that the man report that there is a six on the die given that six comes on the die $= P (\frac{A}{E_{1}})$ = Probability that man speaks truth $= \frac{3}{4}$ Probability that the man reports that there is a six on the die given that six does not comes on the die $= P (\frac{A}{E_{2}})$ = Probability that man does not speak truth $= 1 - \frac{3}{4} = \frac{1}{4}$ By Baye's Theorem, we have $P (\frac{E_{1}}{A})$ = Probability that there is a six given that man reports that there is a six on die $= \frac{P (E_{1}) P (\frac{A}{E_{1}})}{P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}})} = \frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}} = \frac{3}{8}$

A man is known to speak truth 3 out of 4 times.He throws a die & reports that it is a six.Find the probability that it is actually a six.