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A man speaks the truth 3 out of 4 times. He throws a die
and reports it to be a 6. What is the probability of it
being a 6?
(a) 3/8
(b) 5/8
(c) 3/4
(d) None of the above
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Answer Posted / hynynu
prob. of getting 6 in a die p(e1)=1/6
prob. of not getting 6 in a die p(e2)=1-(1/6)=5/6
prob. of the man getting 6 in a die and reporting it as 6
i.e. truth p(e/e1)=3/4
prob. of the man NOT getting 6 in a die and reporting it as
6 i.e. lying p(e/e2)=1-(3/4)=1/4
P= p(e1)p(e/e1)
------------------------
p(e1)p(e/e1)+p(e2)p(e/e2)
p= 1/6*3/4
--------------------
(1/6*3/4)+(5/6*1/4)
p= 3/8
Answer is (a)
| Is This Answer Correct ? | 7 Yes | 1 No |
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