From page 88, Introduction to Probability (2019 2 edn) by Jessica Hwang and Joseph K. Blitzstein.
- Suppose that there are $5$ blood types in the population, named type $1$ through type $5$, with probabilities $p_1, p_2,\cdots ,p_5$. A crime was committed by two individuals. A suspect,who has blood type $1$, has prior probability $p$ of being guilty. At the crime scene, blood evidence is collected, which shows that one of the criminals has type $1$ and the other has type $2$.
Find the posterior probability that the suspect is guilty, given the evidence. [Rest of question is omitted.]
I am fully aware of these two other posts
- Dependence of posterior probability on parameters
- The conditional probability of the evidence on a crime scene?
where the solution is discussed, however, I don't understand what is going on. In fact, there seem to be $3$ different answers. Hence, I wish to repost the problem.
My attempt:
Let $E$ be the event that the suspect is guilty. Let $B$ be the event that the blood type of suspect matches blood type $1$, i.e. one of the samples found in the crime scene. Let $E^c$ be the event where the suspect is not guilty. Let $p_1$ be the $p$ (blood type $1$).
Therefore, by invoking Bayes, we have: $P(E \mid B) = \frac{P(E)P(B \mid E)}{P(E)P(B\mid E))+ (P(E^c)P(B\mid E^c)}$
This gives: $P(E\mid B) = \frac{p}{p+(1-p)p_1}$
Both posts cited above, however, states a different answer. Where am I going wrong in my thinking? In my opinion, given that the suspect is not guilty, then $P(B\mid E^c)$ reduces to just being $p_1$, the naturally occurring probability of the blood type.