Two witnesses both point to the suspect

Question

Before any testimony, a suspect is judged guilty with probability $10\%$ . Two witnesses view a lineup independently. Each witness correctly identifies a guilty suspect $70\%$ of the time, but also wrongly identifies an innocent suspect $20\%$ of the time. Both witnesses point to this suspect.

What is the probability the suspect is guilty?

Accepted Answer

Imagine 1{,}000 suspects in this situation: 100 guilty, 900 innocent. The two identifications are independent given the truth, so multiply. - Guilty, both identify: 100 	imes 0.70 	imes 0.70 = 49. - Innocent, both identify: 900 	imes 0.20 	imes 0.20 = 36. Of the 49 + 36 = 85 double-identifications, 49 are guilty: P(	ext{guilty} \mid 	ext{both point}) = \frac{49}{85} \approx 0.576 = 58\%. One identification alone gives only about 28\%, but a second independent match makes an all-innocent coincidence less likely, lifting belief past a coin flip. Takeaway. Independent corroboration multipl…

Two witnesses both point to the suspect

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