Question 1200745
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Answer: <font color=red size=4>0.535</font>
This value is exact without any rounding done to it.


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Explanation:


Let's give the judges code names: {J1,J2,J3}
We'll have them be presented in the order that was mentioned in the instructions.


J1 and J2 make their decisions independently of each other. 
The probability they reach a correct decision is p = 0.5 for each of those two judges.
If {J1,J2} reach the same outcome, then J3 rules the same way.


If {J1,J2} reach differing outcomes, then J3 thinks for her/himself to reach their own decision (making a mistake with probability q=0.43).
For this situation J3 makes the correct decision with probability 1-q=1-0.43=0.57


Here is a way to represent a two-way table of outcomes for the first two judges.
<table border = "1" cellpadding = "5"><tr><td></td><td>J2 right</td><td>J2 wrong</td></tr><tr><td>J1 right</td><td></td><td></td></tr><tr><td>J1 wrong</td><td></td><td></td></tr></table>
For example, the upper left corner has both J1,J2 give the correct ruling.


Here is a non-tabular format to represent those four outcomes:
J1 right, J2 right
J1 right, J2 wrong
J1 wrong, J2 right
J1 wrong, J2 wrong


Let's list those four outcomes in a new table. 
We'll have them as the 4 rows. The 2 columns will be the outcomes for J3.
<table border = "1" cellpadding = "5"><tr><td></td><td>J3 right</td><td>J3 wrong</td></tr><tr><td>J1 right,J2 right</td><td>A</td><td></td></tr><tr><td>J1 right,J2 wrong</td><td>B</td><td>C</td></tr><tr><td>J1 wrong,J2 right</td><td>D</td><td>E</td></tr><tr><td>J1 wrong,J2 wrong</td><td></td><td>F</td></tr></table>
I then added letters A through F to represent the various possible scenarios.
Scenario A is when all three judges make the correct decision.
The cell next to A is blank because J3 doesn't make the wrong decision when her/his two other colleagues make the correct decision (remember J3 mimics the other two judges if they rule the same way). This also explains why we have a blank cell in the bottom left corner.
Scenario B is when J1 gets it right, J2 gets it wrong, J3 gets it right. 
Scenario C is when J1 gets it right, J2 gets it wrong, J3 gets it wrong. 
And so on.


The letters A, B, and D correspond to situations where the court gets the correct outcome. 
This is because there are at least 2 judges that get the correct ruling. 
A = all 3 judges get it right
B = J1 and J3 get it right
D = J2 and J3 get it right
All other situations involve at least 2 judges getting the wrong verdict; hence the wrong final outcome.


To find the probability that the court gets the correct decision, we need to calculate the following: 
P(A)
P(B)
P(D)


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P(A) = P(J1 right,J2 right)
P(A) = P(J1 right)*P(J2 right)
P(A) = p*p
P(A) = p^2 
P(A) = 0.5^2 
P(A) = 0.25
This works because J1,J2 are independent of each other.
J3 does not alter the outcome here, so we ignore the probability associated with this judge.


P(B) = P(J1 right, J2 wrong, J3 right)
P(B) = P(J1 right)*P(J2 wrong)*P(J3 right)
P(B) = p*(1-p)*(1-q)
P(B) = 0.5*(1-0.5)*(1-0.43)
P(B) = 0.1425


P(D) = P(J1 wrong, J2 right, J3 right)
P(D) = P(J1 wrong)*P(J2 right)*P(J3 right)
P(D) = (1-p)*p*(1-q)
P(D) = (1-0.5)*0.5*(1-0.43)
P(D) = 0.1425


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Conclusion:
P(A) = 0.25
P(B) = 0.1425
P(D) = 0.1425


P(A)+P(B)+P(D) = 0.25+0.1425+0.1425 = <font color=red>0.535</font> represents the probability the court gets the right decision.
The court gets it right exactly 53.5% of the time.


This is around 50%, so it's about as good as a coin toss in my opinion. Those don't seem like good odds. 
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