Exercise 2.27

Answers

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) = P(G)P(T|G) P(G)P(T|G) + P(Gc)P(T|Gc) = pp2 pp2 + (1 p)2p1p2 = p p + 2p1(1 p)

For P(G|T) to be larger than p, p1 has to be smaller than 1 2. This result makes sense, since if p1 = 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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2021-12-05 00:00
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