Question

A man is known to speak truth 3 out of 4 times. He throws a die and a number other than six comes up. Find the probability that he reports it is a six

Answer 1

Let E be the event that the man reports that six occurs in the throwing of the die and let S₁ be the event that six occurs and S₂ be the event that six does not occur.

P(S₁)=16

P(S₂)=1−1/6=5/6

P(E/S₁)=probability that the man reports that six occurs when 6 has actually occurred on the die.

P(E/S₁)=probability that the man speaks the truth=34

P(E/S₂)=probability that the man reports that six occurs when 6 has not actually occurred on the die.

P(E/S₂)=probability that the man does not speak the truth

⇒1−3/4=1/4

Hence by Baye's theorem we get,

P(S₁/E)=Probability that the report of the man that six has occurred is actually a six.

Answer 2

Let E be the event that the man reports that six occurs in the throwing of the die and let S1 be the event that six occurs and S2 be the event that six does not occur. P(S1)=16 P(S2)=1−1/6=5/6 P(E/S1)=probability that the man reports that six occurs when 6 has actually occurred on the die. P(E/S1)=probability that the man speaks the truth=34 P(E/S2)=probability that the man reports that six occurs when 6 has not actually occurred on the die. P(E/S2)=probability that the man does not speak the truth ⇒1−3/4=1/4 Hence by Baye's theorem we get,

Ans as 3/8