# Question about conditional probability

1. Oct 26, 2012

### uasaki

1. The problem statement, all variables and given/known data

You start out with a bag that contains either a red marble or a green marble with equal probability. Then, a red marble is added to the bag. If I draw out a red marble the first time, what is the probability of drawing out a green marble the second time?

2. Relevant equations

3. The attempt at a solution

So, we know that P(R) = P(G) = 1/2, which implies that P(RR) = P(RG) = 1/2. I drew a decision tree

/ \
RG RR

/ \ / \

G R R R (the leaves correspond to the marbles left in the bag)

and it seems like the probability that I draw out a green marble the second time is 1 / 3 (because there are 3 ways in which I can draw out a red marble, and only one of those ways correspond to me drawing out a green marble). I'm not quite sure if this is the correct solution, because the two red marbles that might be in the bag would be indistinguishable, so I'm not really sure how it would affect the overall probability space.

I also tried calculating P(G | R) as follows. P(G | R) = P(G and R) / P(R). At first, I computed P(G and R) as 1/2, but I think counting the scenario where I draw out a green marble first and then I draw out a red marble might be overcounting the number of outcomes. P(R) is just 3/4.

2. Oct 26, 2012

### HallsofIvy

We add a red marble and then take out a red marble- we are in exactly the situation we were to start with! The probability is now 1/2 that the bag contains a red marble, 1/2 that it contains a green marble just as it was before.

(It does't matter whether the red marble removed was the red marble we just put in or the red marble that was there before we added a marble because they are identical.)

3. Oct 26, 2012

### Ray Vickson

This argument SEEMS OK at first, but it ultimately gives the wrong answer, as the following formal development will show. This seems related to puzzles like the Monte Hall problem, which can similarly lead one astray very easily.

Let R0 = {initial color = red}, G0 = {initial color = green}
R1 = {first draw is red}, R2 = {second draw is red}
Of course, we want the probability of green, but that is just the complement.

P(R1) = P(R1|R0)P(R0) + P(R1|G0)P(G0) = (1)(1/2) + (1/2)(1/2) = 3/4.
P(R2|R1) = P(R1&R2)/P(R1). The numerator is P(R1&R2|R0)P(R0)+P(R1&R2|G0)P(G0) = 1(1/2) + 0(1/2) = 1/2. Thus, P(R2|R1) = (1/2)/(3/4) = 2/3, as the OP claims.

After putting in a red and drawing out a red the _physical_ situation is the same as at the start. However, the _informational_ aspects are different, and that is why the probability changes from 1/2 to 2/3. In fact, if we repeat the actions n times (putting in a red, drawing out a red, putting it back, drawing out a ball again, etc, continuing as long as we draw red and we draw n reds in a row, that will certainly alter our assessment of the color of the companion ball. It might be a nice exercise to assume we draw n reds in a row and then get the conditional probability that the other ball is green.

RGV

Last edited: Oct 26, 2012