Question

In 1982 John Hinckley was on trial, accused of having attempted to kill Presiden Reagan. During Hinckley's trial, Dr. Daniel R. Weinberger told the court that when individuals diagnosed as schizophrenics were given computerized axial tomograph (CAT) scans, the scans showed brain atrophy in 30% of the cases compared with, only 2% of the scans done on normal people. Hinckley's defense attorney wanted to introduce as evidence Hinckley's CAT scan, which showed brain atrophy. The defense argued that the presence of atrophy strengthened the case that Hinckley suffered from mental illness.

We want to know the probability that Hinckley was schizophrenic given that he had brain atrophy. Suppose approximately 1.5% of people in the United States suffer from schizophrenia. Hence P(S) = 0.015, where S means schizophrenia. We also have P(A|S) 0.3 = (A means atrophy) and P(A|S complement) = 0.02.

Calculate P(S|A). Show your work

Answer 1

P(S/A) = 0.1859

Explanation:

The probability P(S/A) that Hinckley was schizophrenic given that he had brain atrophy is calculated as:

P(S/A) = P(S∩A)/P(A)

Where P(A) =P(S∩A) + P(S'∩A)

S means schizophrenia, A means atrophy and S' means S complement.

So, the probability P(S∩A) that a person has schizophrenia and atrophy is equal to:

P(S∩A) = P(S)*P(A/S)

P(S∩A) = 0.015*0.3 = 0.0045

Because 0.015 is the probability that a person suffer schizophrenia and 0.3 is the probability that a person had atrophy given that he has schizophrenia.

At the same way, the probability P(S'∩A) that a person doesn't has schizophrenia and atrophy is equal to:

P(S'∩A) = P(S')*P(A/S')

P(S'∩A) = (1-0.015)*0.02 = 0.0197

Finally, P(A) and P(S/A) are equal to:

P(A) = 0.0045 + 0.0197 = 0.0242

P(S/A) = 0.0045/0.242 = 0.1859