Question

A person tried by a 3-judge panel is declared guilty if at least 2 judges cast votes of guilty. Suppose that when the defendant is, in fact, guilty, each judge will independently vote guilty with probability 0.7, whereas when the defendant is, in fact, innocent, this probability drops to 0.2. If 70 percent of defendants are guilty, compute the conditional probability that judge number 3 votes guilty given that judges 1 and 2 vote guilty.

Answer 1

Answer:

The probability is 0.6831

Step-by-step explanation:

Let's call G the event that the defendant is guilty, G' the event that the defendant is not guilty, A the event that the judge 1 vote guilty, B the event that the judge 2 vote guilty and C the event that the judge 3 vote guilty.

Then, the probability P(C/(A∩B)) that that judge number 3 votes guilty given that judges 1 and 2 vote guilty is calculated as:

P(C/(A∩B)) = P(C∩A∩B)/P(A∩B)

Then, P(A∩B) is calculated as:

P(A∩B) = P(G)*P(A/G)*P(B/G) + P(G')*P(A/G')*P(B/G')

P(A∩B) = 0.7*0.7*0.7 + 0.3*0.2*0.2

P(A∩B) = 0.343 + 0.012

P(A∩B) = 0.355

Where P(G) is the probability that the defendant is guilty, P(A/G) and P(B/G) are the probabilities that the judge 1 and 2 vote guilty given that the defendant is guilty, P(G') is the probability that the defendant is innocent and P(A/G') and P(B/G') are the probabilities that the judge 1 and 2 vote guilty given that the defendant is innocent.

At the same way, P(C∩A∩B) is calculated as:

P(C∩A∩B) = P(G)*P(A/G)*P(B/G)*P(C/G) + P(G')*P(A/G')*P(B/G')*P(C/G')

P(C∩A∩B) = 0.7*0.7*0.7*0.7 + 0.3*0.2*0.2*0.2

P(C∩A∩B) = 0.2401 + 0.0024

P(C∩A∩B) = 0.2425

Finally, P(C/(A∩B)) is:

P(C/(A∩B)) = 0.2425/0.355

P(C/(A∩B)) = 0.6831