# A probability question about Bayes'Rule,that is a little difficult for me.

#### zhangyu1534

80% of the murders committed in a certain town are committed by men. A dead body has been found.Two detectives examine the body.The first detective,who is right seven times out of ten,announces that the murderer was a male.But the other detective,who is right 3 times out of 4,says that the murderer was a woman.What is the probability that the murderer was a woman?

This is a question about Bayes'Rule.But I have no idea how the probability changes when there are two detectives.

#### romsek

MHF Helper
You have to assume the assertions of the two detectives are independent of one another. Then you can write

$P[\mbox{murderer was woman |D1 says man and D2 says woman}] = \dfrac{P[\mbox{D1 says man and D2 says woman | murderer was woman}]P[\mbox{murderer was woman}]}{P[\mbox{D1 says man and D2 says woman}]}=$

$\dfrac{\left((0.3)(0.75)\right)(0.2)}{\left((0.8)(0.7)+(0.2)(0.3)\right)\left((0.2)(0.75)+(0.8)(0.25)\right)}\approx 0.2074$

#### zhangyu1534

Thanks for giving a way to think about this problem.I reconsidered this problem.And I found the following may be clearer:
We can view the partition as the 80% of the crimes are committed by men and 20% by women.The question is P[murderer is woman|D1 is right and D2 is right].
So first consider P[D1 is right and D2 is right],which is calculated from two partitions of the problem:
P[D1 is right and D2 is right]=P[D1 is right and D2 is right|this murderer is a man]*P[this murderer is a man]+P[D1 is right and D2 is right|this murderer is a woman]*P[this murderer is a woman]
P[murderer is woman|D1 is right and D2 is right]=P[D1 is right and D2 is right|this murderer is a woman]*P[this murderer is a woman]/P[D1 is right and D2 is right]

the result is 0.243.
But i dont know how to type fraction.