Question

A man is know to speak the truth $3$ out if $4$ times. He throws a die and reports that it is a six. The probability that it is actually a six is:

• A. $\frac{3}{8}$
• B. $\frac{1}{5}$
• C. $\frac{3}{4}$
• D. None of these

Answer 1

The correct option is A $\frac{3}{8}$

Let E be the event that the man reports that six occurs in the throwing of the die and let $S_1$ be the event that six occurs and $S_2$ be the event that six does not occur.

$P\left(S_1\right)=\frac{1}{6},P\left(S_2\right)=1-\frac{1}{6}=\frac{5}{6}$

$P\left(E/S_1\right)$=probability that the man reports that six occurs when 6 has actually occurred on the die.

$P\left(E/S_1\right)$=probability that the man speaks the truth=$\frac{3}{4}$

$P\left(E/S_2\right)$=probability that the man reports that six occurs when 6 has not actually occurred on the die.

$P\left(E/S_2\right)$=probability that the man does not speak the truth

$⟹=1-\frac{3}{4}=\frac{1}{4}$

Hence by Baye's theorem, we get,

$P\left(S_1/E\right)$=Probability that the report of the man that six has occurred is actually a six.

$P\left(S_1/E\right)=\frac{P\left(S_1\right).P\left(E/S_1\right)}{P\left(S_1\right)P\left(E/S_1\right)+P\left(S_2\right).P\left(E/S_2\right)}⟹=\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}=\frac{1}{8}\times \frac{24}{8}=\frac{3}{8}$