Just looking for some help with this question. The answer seems obvious but I am questioning myself as it seems a little too obvious.
Max has a bag containing 2 black marbles and 2 white marbles. He draws a marble and sets it aside without looking at it. He then draws a second marble from the bag and it is white. Which is more likely?
A) The first marble drawn was equally likely to be black or white.
B) The first marble drawn was black.
C) The first marble drawn was white.
At first glance it would seem that A is the correct answer as the first draw is an independent event. But does knowing the result of the second draw affect the number of outcomes for the first draw?
Any help would be appreciated! Thank you!
Best Answer
Bayes' Theorem is one way to do this, but it isn't necessary.
Suppose the balls were labeled $W_1,W_2,B_1,B_2$ and that the second draw was $W_1$. Then we know the first ball must have been one of $W_2,B_1,B_2$. Nothing distinguishes these except for color so any one of them is equally likely to have been the first one chosen. Thus the probability that the first draw was Black is $\frac 23$ and the probability that it was White is $\frac 13$.
To help with intuition, here is another way to look at it. Imagine we draw out all $4$ balls (though in the end we only care about the first two draws). There are $\binom 42=6$ equally probable ways the balls might be ordered. We have $$WWBB,WBWB,WBBW,BBWW,BWBW,BWWB$$ Of these, only three have $W$ as the second draw, namely: $$WWBB,BWBW,BWWB$$ Thus, we know we are in one of these three states, each with probability $\frac 13$. We remark that in two of these three cases, the first ball is $B$.
To stress: I am assuming here that the first draw is NOT returned to the bag. It is drawn and then hidden away somewhere, not in the bag. That is critical. If, to the contrary, you show it to a friend (who notes the color) and then return it to the bag, then the two draws are entirely independent. In the context of my solution above, if you have replaced the first draw in the bag then there was no reason to exclude $W_1$ from the list of possibilities.