Dependence of posterior probability on parameters

Question

From page 88, Introduction to Probability (2019 2 edn) by Jessica Hwang and Joseph K. Blitzstein.

27. 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.

What I got so far is that P(guilty|evidence) = P(evidence|guilty)P(guilty)/(P(evidence|guilty)P(guilty)+P(evidence|not guilty)P(not guilty)).

P(guilty) = p

P(not guilty) $= 1 - p$

P(evidence|guilty) = 1/2 because one criminal has blood type 1

P(evidence|not guilty) = 1/5 because there are 5 blood types.

So I get P(guilty|evidence) $= \dfrac{0.5p}{0.5p + 0.2(1-p)}$ and I do not know if this is correct.

Answer 1

What I understand as the evidence is one of the criminals has type 1 and the other has type 2. So if our suspect having type 1 blood is guilty the probability of this evidence occurring is 1.

P(evidence|guilty)=1

And if not guilty the evidence has probability $2p_1p_2$ (both blood types occurring and order not important)

P(evidence|not guilty)=$2p_1p_2$

Hence we get $\frac{p}{p+2p_1p_2(1-p)}$

Answer 2

$\mathit G$: event that the suspect with blood type 1 is guilty
$\mathit E$: event that the suspect's blood type matches one of the blood types found at the crime scene

We want to calculate $\mathbf P(G\vert E)$

Using Baye's theorem,

$\mathbf P(G\vert E) = \frac{\mathbf P(G)\mathbf P(E\vert G)}{\mathbf P(E)} $

$\mathbf P(G) = p$

$\mathbf P(E) = \mathbf P(G)\mathbf P(E\vert G) + \mathbf P(G^c)\mathbf P(E\vert G^c) $

$\mathbf P(E\vert G)$ is the probability that the suspect's blood type matches one of the blood types found at the crime scene i.e. type 1 and 2, given that the suspect with blood type 1 is guilty. Or simply that the probability that the suspect's blood type was found at the crime scene given that he is guilty. If the suspect is guilty, his blood type with definitely be at the crime scene. So this probability amounts to one.
$\mathbf P(E\vert G) = 1$

$\mathbf P(E\vert G^c)$ is the probability that the suspects blood type matches the ones found in the crime scene, given that he is not guilty. In other words, the probability that the suspect has the same blood type as one of the criminals. This would amount to $\frac{p_1}{p_1 +\,p_2}$

This gives us the following result,

$\mathbf P(G\vert E) = \frac{p}{p\, +\,(1-p)\cdot\frac{p_1}{p_1 +\, p_2}} $