Answer Posted / kishore

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)

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