Probability with Marbles

Question

Suppose a bag contains 9 red marbles, 15 blue marbles, and 20 green marbles. If two marbles are drawn from the bag without replacement, what is the probability of drawing one blue and one green marble?

Answer 1

There are 44 total marbles. Thus, the probability of drawing one of 20 green marbles and one of 15 blue marbles is 20*15/C(44,2) = 20*15/(44*43/2!) = 10*15/11*43 = 150/473 = 0.317124735729387

Answer 2

To start, there are a total of 9+15+20 = 44 marbles. You have an equal probability of drawing each one.

Success will be defined as drawing B(lue), then G(reen) or by drawing G,B. Since those events don't affect each other, you can add their probabilities.

So first we need to find the probability of BG. Then you can find the probability GB.

We first draw a B. there were 15 blues out of 44 total marbles, so the probability of that was 9/44.

But now there's only 43 marbles left, and we want to draw a G. Well, there are 20 Gs out of 43 marbles, so that probability is 20/43.

The probability of both events B and then G occurring in exactly that order is found by multiplying them, so you get 9/44 * 20/43 = Probability of BG.

Now, find the probability of GB and add it to what you got for Prob(BG).