IIT JEE , AIEEE Forum - Mathematics , Algebra

saket_brutus

A man speaks the truth in 75% cases. He throws a die(unbiased) and tells his friend that it is a 6. What is the probability that it is a 6? (try using Baye's theorem plz)

edison

Here,

the probability of 6 is = 1/6

probability that a man speaks truth = 3/4

so, for a friend s.t. throw results in 6 is Probability = 3/24

saket_brutus

Sorry Edison Sir, but the answer is 3/8 (maybe its (1/2) * (3/4) ).

Cheers.

rahul_c

THIS IS A PROBLEM USING THE BAYE'S THEOREM

P(A1)=PROBABILITY OF GETTING 6 IN THE THROW OF DICE=1/6

P(A2)=" " " NOT " " " " " "=1-1/6=5/6

P(A/A1)=PROBABILITY THAT THE MAN SPEAKS THE TRUTH THAT IS REPORTS 6 WHEN 6 HAS OCCURED =75/100=3/4

P(A/A2)=PROBABILITY THAT THE MAN DOES NOT SPEAK THE TRUTH THAT IS REPORTS 6 WHEN 6 HAS NOT OCCURED =1-(75/100)=1/4

BY BAYE'S THEOREM

P(A1/A)=P(A1)*P(A/A2) / {P(A1)*P(A/A1)+P(A2)*P(A/A2)}

= (1/6*3/4) / { (1/6*3/4) +(5/6*1/4) }

=3/8

with regards

shakirshafi12

Let h1 be the event of a six coming
Let h2 be the event of a six not coming
P(h1)=1/6 which is probability of six coming
P(h2)=5/6 which is probability of six not coming
Let E be the event of him telling a six has come
P(E/h1)=75/100=3/4 which is the probabilty of him reporting a six when a six has come
P(E/h2)=25/100=1/4 which is the probabilty of him reporting a six when it has not come
{NOTE:THIS IS THE PROBABILITY OF HIM TELLING THE TRUTH BECAUSE IF HE TELLS A SIX HAS COME WHEN IT HAS ACTUALLY COME IT MEANS HE IS TELLING THE TRUTH.)
USING BAYES THEOREM
P(h1/E)= P(E/h1).P(h1) / ( P(E/h1).P(h1)+P(E/h2).P(h2) )
where P(h1/E) is the probabilty of him tellinh that a six has come when it actually has come
===>(3/4)*(1/6) / { ((3/4)*(1/6))+((1/4)*(5/6)) }
=3/8
which is the required answer

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