Probability Without Replacement

Question

Bruno has 90 marbles. Of those, 45 are broken and 75 are painted red. If he randomly picks marbles one at a time, what is the probability that the first marble he gets is painted red, the second marble to be broken and the third to be both ( broken and red), without replacement? Show the complete solution.

Answer 1

There's definitely a 75/90 chance of drawing a red marble first, but it gets much more complicated after that.

We need to know how many are both broken and red. This information isn't given, and it needs to be. Unfortunately, we can't assume that these two properties are independent, because there are 75 marbles that are painted red, and half of the marbles are broken. Since we can't divide 75 by two evenly, we can't have exactly 75/s broken red marbles and 75/2 intact red marbles.

So let's look at the two extreme cases: that all the non-red marbles are broken, or none of the non-red marbles are broken.

First, if all of the non-red marbles are broken then we start with 15 non-red, broken marbles, 30 red broken marbles and 45 red intact marbles. Intact means "not broken". Our probability is then

30/90 (red, broken) * 29/89 (red, broken) * 28/88 (red, broken) +

30/90 (red, broken) * 15/89 (not red, broken) * 29/88 (red, broken) +

45/90 (red, intact) * 30/89 (red, broken) * 29/88 (red, broken) +

45/90 (red, intact) * 15/89 (not red, broken) * 30/88 (red, broken).

Second, if all the broken marbles are also red, then we start with 45 red, broken marbles, 15 intact red marbles, and 15 marbles that are intact and not red. Our probability is then:

45/90 (red, broken) * 44/89 (red, broken) * 43/88 (red,broken) +

15/90 (red, intact) * 45/89 (red, broken) * 44/88 (red, broken)

If it's somewhere in between, then the answer will be somewhere in between the first two scenarios given.