/sci/ - Science & Math

>>10558879
Not at all; just because two balls are identical doesn't mean its the same outcome. Choosing a box first is actually a red herring meant to confuse you, and it can be replaced by a model describing sequences like I did.

Lets ignore the "pull a second ball" part of the question and slightly modify it so I can explain this concept.
You have two boxes. One box has 2 gold balls. The second box had 1 gold ball and one silver ball.
You randomly choose a box (this is equivalent to the boxes being opaque) and pull out a gold ball.
What is the probability that you pulled the ball from box 1?

Formally, this is "What is the probability that you chose box 1, given you pulled a gold ball?"
Event A is you choose box 1 (and thus event A' being you choosing box 2)
Event B is you choosing a gold ball (and thus event B' being you choosing a silver ball)

P(A|B) = P(A intersection B) / P(B) = P(A ^ B) / P(B)

As we can see in the shitty probability tree to the right, you'll notice that the probability of A intersection B (choosing a gold ball from box 1) is 0.5
The probability of choosing a gold ball, however, is the probability of choosing a gold ball from box 1 plus the probability of choosing a gold ball from box 2, 0.5 + 0.25, or 0.75.

P(A ^ B) / P(B) = 0.5 / 0.75 = 2/3

It's left to the reader to show that this example is actually identical to the OP's original problem.