A bag contains 8 red marbles, 7 blue marbles and 5 green marbles. If two marbles are drawn out of the bag, what is the exact probability that both marbles drawn will be red?

Answer 1

There are a couple of ways to approach this problem, but one common method is to use the formula for probability of independent events:

P(A and B) = P(A) × P(B)

In this case, we want to find the probability of drawing two red marbles, which we can represent as the event A followed by the event B. Since we're drawing without replacement (i.e. we don't put the first marble back in the bag before drawing the second marble), the probability of the second event depends on the outcome of the first event. Specifically, if we draw a red marble on the first try, there will be one less red marble and one less total marble in the bag, which will affect the probability of drawing another red marble on the second try. So we need to calculate the probability of each event separately and adjust the second probability accordingly.

First, let's find the probability of drawing a red marble on the first try. There are 8 red marbles out of a total of 20 marbles in the bag, so the probability of drawing a red marble on the first try is:

P(A) = 8/20 = 2/5

Next, let's find the probability of drawing a red marble on the second try, assuming we did not replace the first marble. Now there are only 7 red marbles left out of a total of 19 marbles in the bag, so the probability of drawing a red marble on the second try is:

P(B|A) = 7/19

The vertical bar | in P(B|A) means "given that A occurred", so we're calculating the probability of drawing a red marble on the second try, given that we already drew a red marble on the first try.

Finally, we can plug these probabilities into the formula:

P(A and B) = P(A) × P(B|A) = (2/5) × (7/19) = 14/95

So the exact probability of drawing two red marbles is 14/95 or about 0.147, or about a 14.7% chance.