>>12267935Not the guy you keep arguing with, but I'm here to defend 4/9. This will require a certain amount of brain cells.
We are looking for the probability that the 2nd ball is gold (B) given that the 1st ball is also gold (A). Without replacement. I'm gonna assume we can pick the 2nd ball from ANY box. The probability of picking a box at random {I,J,K} is 1/3. I have chosen to not incorporate this explicitely, to make for a lighter demonstration.
Given A, there are 3 gold balls we could have picked in the first round, so there are three different outcomes {X,Y,Z} after A, given that we don't replace the ball. The first two outcomes are actually identical. Given A, {X,Y,Z} represent the total set of events with each having an equal probability of 1/3.
P(B|A) is equal to the sum of P(B|X)P(X), P(B|Y)P(Y) and P(B|Z)P(Z)
P(B|X)P(X): Picked 1st gold ball in box 1
Box 1: P(I) = 1/3
One ball left in box 1, a gold ball, Probability of 1 given I and X
1/3 * 1 = 1/3
Box 2: P(J) = 1/3
One gold ball and one silver ball, Probability of 1/2 given J and X
1/3 * 1/2 = 1/6
Box 3: P(K) = 1/3
Two silver balls, Probability of 0 given K and X
P(B|X) = 1/3 + 1/6 + 0 = 1/2
P(B|X)P(X) = 1/2 * 1/3 = 1/6
P(B|Y)P(Y): Picked 1st gold ball in box 1
Box 1: P(I) = 1/3
One ball left in box 1, a gold ball, Probability of 1 given I and Y
1/3 * 1 = 1/3
Box 2: P(J) = 1/3
One gold ball and one silver ball, Probability of 1/2 given J and Y
1/3 * 1/2 = 1/6
Box 3: P(K) = 1/3
Two silver balls, Probability of 0 given K and Y
P(B|Y) = 1/3 + 1/6 + 0 = 1/2
P(B|Y)P(Y) = 1/2 * 1/3 = 1/6
P(B|Z)P(Z): Picked 1st gold ball in box 2
Box 1: P(I) = 1/3
Two gold balls. Probability of 1 given I and Z
1/3 * 1 = 1/3
Box 2: P(J) = 1/3
One ball left in box 2, a silver ball, Probability of 0 given J and Z
Box 3: P(K) = 1/3
Two silver balls, Probability of 0 given K and Z
P(B|Z) = 1/3
P(B|Z)P(Z) = 1/3 * 1/3 = 1/9
P(B|A) = P(B|X)P(X) + P(B|Y)P(Y) + P(B|Z)P(Z) =1/6 + 1/6 + 1/9 = 4/9
Seethe
