Question

A man is known to speak the truth in \(75\%\) of cases. If he throws an unbiased die and tells his friend that it is a six. Assuming that when he tells a lie, he specifically reports the result as a six, the probability that it is actually a six is...

• A. \(\frac{2}{3}\)
• B. \(\frac{5}{9}\)
• C. \(\frac{3}{8}\)
• D. \(\frac{8}{11}\)

Hint:

Apply Bayes' theorem with the two cases: the die shows six and he speaks the truth, or it does not and he lies.

Answer 1

The Correct Option is C

Step 1: Total probability of the report
$P(A)=\dfrac16\cdot\dfrac34+\dfrac56\cdot\dfrac14=\dfrac{3+5}{24}=\dfrac13$.

Step 2: Joint probability
$P(E_1\cap A)=\dfrac16\cdot\dfrac34=\dfrac18$.

Step 3: Divide
$P(E_1|A)=\dfrac{1/8}{1/3}=\dfrac38$.

Final Answer:
The probability is $\dfrac38$. \[ \boxed{\text{(C) }\dfrac{3}{8}} \]