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. Does the evidence make it more likely or less likely that the suspect is guilty, or does this depend on the values of the parameters $p, p_1,. . . , p_5$? If it depends, give a simple criterion for when the evidence makes it more likely that the suspect is guilty.
Proposed solution: Let $A_1$ be the event that the suspect (with blood type 1) is guilty. Let $X$ be the event that one of the criminals have blood type 1 and the other has type 2. We can then use Bayes' rule to define
$$ \begin{aligned}P(A_1|X) &=\frac{P(X|A_1)P(A_1)}{P(X)} \\ &=\frac{P(X|A_1)P(A_1)}{P(X|A_1)P(A_1)+P(X|A_1^c)P(A_1^c)} \\ &=\frac{P(X|A_1)p}{P(X|A_1)p+P(X|A_1^c)(1-p)} \\ &=\frac{P(X|A_1)p}{P(X|A_1)p+P(X|A_1^c)(1-p)} \\ &=\frac{1p}{1p_2p+2p_1p_2(1-p)} \\ \end{aligned} $$ I have two questions on what happens on the last line? The last line I got from the following source here.
First, where does this equality come from $P(X|A_1) = 1p_2$ ? Let $X_1$ be the event that the first criminal's blood type is blood type 1, and so $P(X_1)=p_1$ if the criminal is unknown. And, let $X_2$ be the event that the second criminal's blood type is blood type 2, so $P(X_2)=p_2$ if that criminal is still unknown. Assume also that the two criminal's blood type is independent. Is it then that
$P(X|A_1) = P(X_1\cap X_2|A_1) = P(X_1|A_1)P(X_2|A_1)=1p_2$? In other words, $P(X_1|A_1)=1$ since the suspect is now assumed to be the criminal, and $ P(X_2|A_1)=p_2$ since the second criminal is unknown and thus we equate this with the population parameter.
Second, where does this other equality come from $P(X|A_1^c) = 2p_1p_2$ ? Here, with the notation and logic in my first question, I regard the following to be true $P(X|A_1^c) = P(X_1\cap X_2|A_1^c) = P(X_1|A_1^c)P(X_2|A_1^c) = p_1p_2$. Hence, since the two criminals are unknown, the probability of the evidence is just the population parameters occurring simultaneously (the blood of the two criminals being spilled on the crime scene). In other words, where does the 2 come from?
For your first question, given the knowledge that $A_1$ is guilty, we know that one criminal has blood type $1$, namely $A_1$. Under these conditions, the probability that $X$ holds is the probability that the other criminal has blood type $2$, which is $p_2$. Hence, $\Bbb P(X|A_1) = p_2$.
For your second question, the $2$ comes from the ordering or labeling of the criminals. The first criminal has blood type $1$ with probability $p_1$ and the second criminal has blood type $2$ with probability $p_2$, but we could change the order and still obtain a valid result, so we add them up.
In the link provided, this $2$ is accounted for in $\binom21$. This is a hint at how one would generalize this for other variations of the problem, using multinomial coefficient; note that $\binom21 = \binom2{1,1}$.
Notice that in the first case there is no such freedom with labels, as in the second case. This is because one of the criminals has the 'label' $A_1$ and there is only one other criminal.