problem 27

2.1.20 problem 27

Let $G$ be the event that the suspect is guilty. Let $T$ be the event that one of the criminals has blood type $1$ and the other has blood type $2$ .

Thus,

P (G ∣ ∣ ∣ T) = \frac{P (G) P (T | G)}{P (G) P (T | G) + P (G^{c}) P (T | G^{c})} = \frac{p p_{2}}{p p_{2} + (1 - p) 2 p_{1} p_{2}} = \frac{p}{p + 2 p_{1} (1 - p)}

For $P (G | T)$ to be larger than $p$ , $p_{1}$ has to be smaller than $\frac{1}{2}$ . This result makes sense, since if $p_{1} = \frac{1}{2}$ , then half of the population has blood type $1$ , and finding it at the crime scene gives us no information as to whether the suspect is guilty.

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