Question

In 1982 John Hinckley was on trial, accused of having attempted to kill the then US President Ronald Reagan. Hinckley was found not guilty by reason of insanity. The defense psychiatric reports found him to be insane while the prosecution reports declared him legally sane. In particular, the defense sought to establish that brain scan results like John Hinckley's appeared in a higher percentage of schizophrenics than of normal people his age.

During Hinckley's trial, an expert witness, Dr. Weinberger, told the court that when individuals suffering from schizophrenia were given computerized axial tomography (CAT) scans, the scans showed brain atrophy in 30% of cases compared with only 2% of scans done on normal people. Health records show that 1.5% of the people in the United States suffer from schizophrenia. Hinckleys CAT scan showed brain atrophy. Calculate the probability that Hinckley suffered from schizophrenia given his CAT scan showed brain atrophy.

Answer 1

O.186 or 18.6%

Explanation:

Let s = event of schizophrenia

B = brain atrophy

From our question we have:

P(s) = probability of those suffering from schizophrenia = 1.5% = 0.015

P(B|S) = probability of brain atrophy = 30% = 0.30

P(S') = 1 - 0.015 = 0.985

P(B|S') = 2% = 0.02

Probability Hinckley suffered from schizophrenia

= P(S|B) = P(B|S)P(S)/P(B|S)P(S)+P(B|S')P(S')

= (0.30x0.015)/(0.30)*(0.015)+(0.02)(0.985)

= 0.0045/0.0045+0.0197

= 0.0045/0.0242

= 0.185950

This is approximately 0.186 or 18.6%