Question 1192268: Three judges must vote for the decision. They vote independently of each other and from the previous experience it is known that probabilites of separate right decisions are 0.77, 0.58, 0.79.
a) what is the probability that the overall decision will be right?
b) if the joint decision was correct, what is the probability that the first judge made the decision correctly?
c) that second and third judges made the right decision?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
A = first judge
B = second judge
C = third judge
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) = 0.77
P(B) = 0.58
P(C) = 0.79
Their complements
P(~A) = 1 - P(A) = 1 - 0.77 = 0.23
P(~B) = 1 - P(B) = 1 - 0.58 = 0.42
P(~C) = 1 - P(C) = 1 - 0.79 = 0.21
which represent the probabilities of making the incorrect decision.
Let's calculate the probability the first two judges (A and B) get the correct decision, while judge C gets it wrong.
P(A and B only) = P(A, B, ~C)
P(A and B only) = P(A)*P(B)*P(~C) ... works because their decisions are independent of one another
P(A and B only) = 0.77*0.58*0.21
P(A and B only) = 0.093786
Let 
Similarly,
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) = 0.23*0.58*0.79
P(B and C only) = 0.105386
Let 
And,
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) = 0.77*0.42*0.79
P(A and C only) = 0.255486
Let 
Lastly,
P(A and B and C) = P(A)*P(B)*P(C)
P(A and B and C) = 0.77*0.58*0.79
P(A and B and C) = 0.352814
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 (i.e. two or more) 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 0.807472. This is the answer to part (a).
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
0.702086/0.807472 = 0.869486
which is approximate.
This is the probability of judge A being correct given the overall ruling was correct.
This takes care of part (b).
For part (c), I interpret the wording to mean ONLY judges B and C get the correct decision. This means judge A gets it wrong.
This will involve case  which has probability of 0.105386
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Answers:
a) 0.807472
b) 0.869486 (approximate)
c) 0.105386
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