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.