Probability Question 196
Question: A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six, is
Options:
A) $ \frac{3}{8} $
B) $ \frac{1}{5} $
C) $ \frac{3}{4} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ E $ denote the event that a six occurs and $ A $ the event that the man reports that it is a 6,
we have $ P(E)=\frac{1}{6},P({E}’)=\frac{5}{6},P(A/E)=\frac{3}{4} $
and $ P(A/{E}’)=\frac{1}{4} $
From Baye’s theorem, $ P(E/A)=\frac{P(E).P(A/E)}{P(E).P(A/E)+P({E}’).P(A/{E}’)} $
$ =\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}. $
BETA
