Probability question with black and white marbles

Question

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!

Answer 1

If it seems obviously correct, then it probably is. To be sure of yourself, you can use Bayes Theorem: $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

We have

$$P(\text{first is white}|\text{second is white})=\frac{P(\text{first is white}\cap\text{second is white})}{P(\text{second is white)}}$$

If this value is less than $0.5$ we would conclude that black is more likely.

Can you take it from here?

Answer 2

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.

Answer 3

Knowing the second marble is white means the first draw had to be between one white marble and one black marble, if it is done WITHOUT REPLACEMENT: that is, when the second draw is known, that selection is already spoken for, and the universe of possibilities for the first draw is CUT DOWN. So the p(white first) = p(black first) = 1/2, as the second draw has already been decided.

This can be shown using Bayes Theorem:

Let A = white 1st;

Let B = white 2nd:

p(A|B) = p(B|A)*p(A)/p(B), where p(A) is the prior probability before any data is observed, p(B) is the marginal probability.

In terms of A, B above:

p(white 1st | white 2nd) = p(white 2nd | white 1st)*p(white 1st before data) / [p(w1 | w2)p(w2) + p(w2 | black 1st)p(b1)]

p(white 1st) = p(w1) = 2/3 before any data is observed.

p(black 1st) = p(b1) = 1/3 before any data is observed

p(w2|w1) = prob that white is picked 2nd after white 1st = 1 white/(b+w left) = 1/2

p(w2|b1) = prob that white is picked 2nd after black 1st = 1 white/(w+w left) = 1

Hence, p(w1|w2) = p(w2|w1)p(w1)/[p(w2|w1)p(w1)+p(w2|b1)p(b1)] = (1/2)(2/3)/[(1/2)(2/3)+(1)(1/3)] = (1/3)/[1/3+1/3)] = (1/3)/(2/3) = 1/2.

Hence, p(b1|w2) = p(w2|b1)p(b1)/[p(w2|b1)p(b1)+p(w2|w1)p(w1)] = (1)(1/3)/[(1)(1/3)+(1/2)(2/3)] = (1/3)/[1/3+1/3)] = (1/3)/(2/3) = 1/2.

Answer 4

Knowing that one white marble was destined to be the second marble drawn, entails that there were three marbles in the bag which were equally likely to have been the first marble drawn (and set aside), and only one among them was white .

It is the same senario as drawing a white marble and asking what the probability that the next marble will be white ... just with the video played backwards.