>>269406882p = 0.75
Why? Let b1 = 2 gold, b2 = 1 gold 1 silver, b3 = 2 silver. Now we have a set of three random variables: X = value of first draw, Y = value of second draw, Z = the choice of box {1,2,3}
So in this context, the question is asking the value of P(X = gold | Y = gold). But, Y = gold is equivalent to saying that Z != b3. Ergo P(X = gold | Y = gold) = P(X = gold | Z != b3) = [P(Z != b3 | X = gold) * P(X = gold)]/P( Z != b3).
Now P(X = gold) = 0.5, P(Z != 3) = 2/3. So we have:
p = [P(Z != b3 | X = gold) * 3]/4.
Now P(Z != b3 | X = gold) = 1. This should be obvious but for the brainlets among you, P(Z != b2| X = gold) is the probability that the box is not b3 given that the first draw is gold, since b3 contains no gold balls, then P(X = gold & Z = b3) = 0.
