Question

3 students {A,B,C} tackle a puzzle together and offers a solution upon which majority of the three agrees.Probability of A solving the puzzle correctly is p.Probability of B solving the puzzle correctly is also p.C is a dumb student who randomly supports the solution of either A or B.There is one more student D,whose probability of solving the puzzle correctly is once again p.Out of the three member team {A,B,C} and one member team {D} which one is more likely to solve the puzzle correctly?

Answer 1

@ganeshie8

Answer 2

Well i started off like this:-
probability that the puzzle is solved by three member team is
P(AUBUC) ..
That can be written as P(A)+P(B)+P(C)-P(AintB)-P(BintC)-P(AintC)+P(AintBintC) ..
Int is for intersection.. Sorry bt no symbol for intersection as such in my cell..

Answer 3

P(C) =P(AUB)

Answer 4

.

Answer 5

@parthkohli

Answer 6

Just solve it in an elementary way that covers all cases.
As for the one-student team, the probability of getting a correct answer is obviously \(p\).
For the three-student team, let's see how it goes.

Case 1: \(P_1\): Student A gets it right, student B gets it right.
(i) C agrees with A - \(\frac{P_1}{2}\) - right answer.
(ii) C agrees with B - \(\frac{P_1}2\) - right answer.
Total probability of getting right answer in case 1: \(p^2\)

Case 2: \(P_2\): Student A gets it right, student B gets it wrong.
(i) C agrees with A - \(\frac{P_2}2\) - right answer.
(ii) C agrees with B - \(\frac{P_2}{2}\) - wrong answer.
Total probability of getting right answer in case 2: \(\frac{p(1-p)}2\)

By an argument of symmetry, we can say that it'd be the same in case 3.

Finally in case 4, this probability is completely zero.

So total probability\[= p^2 + \frac{p-p^2}{2}+\frac{p-p^2}{2}\]\[=p\]

Answer 7

And that makes sense, right? I mean the situation doesn't really change in the extreme cases.
When you have one right, one wrong then while the probability of choosing the right one is halved down, it also means that you are twice as likely to have a right answer (if student A gets it wrong, then B will get it right).

Answer 8

You halved the probability for C to go with either of the two ....
Like p1/2 in the very first case for C to go with A and p1/2 for C to go eith the ans of B ..(by symmetry) .
Isn't it?