SOLUTION: Three judges A, B and C must make a decision by majority's vote. They make their individual decision independently. It is known that the judges make a correct decision with probabi

Question 1192064: Three judges A, B and C must make a decision by majority's vote. They make their individual decision independently. It is known that the judges make a correct decision with probabilities A - 0.79, B - 0.65 and C 0.79. What is the chance that their decision will be correct? If their collective decision was correct, what is the chance of judge's A decision being the right one? Or what is the probability that only B and C had the right decision?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

P(A) = probability that judge A makes the correct decision
P(B) and P(C) represent similar ideas for the other two judges.

Given probabilities
P(A) = 0.79
P(B) = 0.65
P(C) = 0.79

Their complements
P(~A) = 1 - P(A) = 1 - 0.79 = 0.21
P(~B) = 1 - P(B) = 1 - 0.65 = 0.35
P(~C) = 1 - P(C) = 1 - 0.79 = 0.21
which represent the probabilities of making the incorrect decision.

P(A and B only) = P(A, B, ~C)
P(A and B only) = P(A)*P(B)*P(~C)
P(A and B only) = 0.79*0.65*0.21
P(A and B only) = 0.107835
Let

P(B and C only) = P(~A, B, C)
P(B and C only) = P(~A)*P(B)*P(C)
P(B and C only) = 0.21*0.65*0.79
P(B and C only) = 0.107835
Let

P(A and C only) = P(A, ~B, C)
P(A and C only) = P(A)*P(~B)*P(C)
P(A and C only) = 0.79*0.35*0.79
P(A and C only) = 0.218435
Let

P(A and B and C) = P(A)*P(B)*P(C)
P(A and B and C) = 0.79*0.65*0.79
P(A and B and C) = 0.405665
Let

Cases through represent situations where exactly two judges get the right decision.
Case is when all three judges make the correct ruling.
All four represent when at least two judges get the correct ruling.

Add up the through

The probability that at least two judges reach the correct decision, and therefore get the correct overall ruling, is 0.83977

Cases , , and represent situations where judge A made the correct ruling that led to the overall ruling being correct.
The sum of these x values is

Dividing that second sum over the first sum calculated earlier will get us
0.731935/0.83977 = 0.871590
which is approximate.
This is the probability of judge A being correct given the overall ruling was correct.
This takes care of the second question mentioned.

For the third question, we go back to case which is when judges B and C are correct, but judge A is not correct.

------------------------------------------------------------------------
Summary:

What is the chance that their decision will be correct?
0.83977

If their collective decision was correct, what is the chance of judge's A decision being the right one?
0.871590

What is the probability that only B and C had the right decision?
0.107835

RELATED QUESTIONS
Three judges must vote for the decision. They vote independently of each other and from... (answered by math_tutor2020)

Three judges with a majority vote have to make a decision. The judges decide... (answered by ikleyn)

A court consists of 3 judges. Two of them, independently of each other, make correct... (answered by math_tutor2020,ikleyn)

A panel of 7 judges is to decide which of 2 final contestants (A and B) in a beauty... (answered by math_tutor2020)

I've completed some of the questions - however I am not sure if they are correct. Your... (answered by stanbon)

Multiple choice questions: A Type I error is committed if we make: a) a correct... (answered by stanbon)

Are the answers right? University of Phoenix Material Parametric and Nonparametric... (answered by stanbon)

I am having difficulty with the following homework assignment. Could someone please check (answered by stanbon)

Parametric and Nonparametric Data Identification Assignment Label each of the following... (answered by stanbon)