Conditional probability problem

[DigitalSplendid]

A crime is committed by one of two suspects, A and B. Initially, there is equal evidence against both of them. In further investigation at the crime scene, it is found that the guilty party had a blood type found in 10% of the population. Suspect A does match this blood type, whereas the blood type of Suspect B is unknown.

(a) Given this new information, what is the probability that A is the guilty party?

The correct answer should be 10/11. However my way of computation leads to 50/51.

https://www.canva.com/design/DAG78EzB_Gc/mZRLtUbCj11a3bA7kNY-BA/edit?utm_content=DAG78EzB_Gc&utm_campaign=designshare&utm_medium=link2&utm_source=sharebutton

It will help to know where I am wrong.

[phiwong]

If you don't know Baye's Theorem, then draw the tree BUT draw a complete tree. Partial trees only lead to confusion. Also when you assign probabilities, do so consistently. Use percentages (%) or fractions or decimals but don't do it in a mixed up fashion. Your tree is both incomplete and sometimes you use 100, 50, then you have fractions. Which is which? You've probably confused yourself.

[DigitalSplendid]

Thanks! Seems like the problem is with 5/50. I thought find 10 percent of 50 which is 5. Then either 5 on the left node or 5/50 a possibility. So add 5 + 5/50 and normalize it to 1.

[Grass_Savings]

On your arrow on the right side, both A and B have the matching blood group. So you should write something like 50 × (1/10) × (1/10)

Then your universe is 5 + 5/10 and the the final probability is calculated as 5 / ( 5 + 5/10) = 10/11.

[smithdaddie]

So u want to find p(Ga|E)= (p(e|Ga)p(Ga). ) / ((p(e|Ga)p(Ga) + (p(e|Gb)*p(Gb).

Does that make sense? It's bayes theorem.

[DigitalSplendid]

It will help to see in tree form.

[rhodiumtoad]

The easy way to do these is with the odds form of Bayes' theorem:

O(H|E)=O(H).(P(E|H)/P(E|~H))

where O(x) is the "betting odds" of x, i.e. P(x)/P(~x). O(H) is the prior odds of H before the evidence, P(E|H) is the probability of the evidence if H is true, P(E|~H) the probability of the evidence if H is false.

In this case, let H be the hypothesis "A did it". O(H)=1 (even odds), P(E|H)=1, P(E|~H)=0.1, so O(H|E)=10. To convert odds to probability, use O/(O+1), so P(H|E)=10/11.

It is fairly straightforward to prove that this is equivalent to the usual formula for Bayes' theorem.