Statistics And Probability Question 41
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) $ \frac{1}{2} $
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Answer:
Correct Answer: A
Solution:
- [a] P (six occurs) $ =P(S_2)=\frac{1}{6} $ P (Six does not occur) = $ P(S_2)=\frac{5}{6} $ $ P(E|S_1)=P $ (Man speaks truth) $ =\frac{3}{4} $ $ P( E|E_2 )=P $ (Man does not speak the truth) $ =\frac{1}{4} $
$ \therefore $ By Baye?s theorem, $ P(S_1|E)=P $ (he reports that six has occurred is actually a six) $ =\frac{P(S_1)P(E|S_1)}{P(S_1)P(E|S_1)+P(S_2)P(E|S_2)} $ $ =\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}\times \frac{5}{6}\times \frac{1}{4}}=\frac{3}{8} $
BETA
