A bag of marbles contains 8 red marbles and 6 yellow marbles. Two marbles are drawn out of the bag at random without replacement. What is the probability that two red marbles are drawn out? Be sure to represent your answer as a fraction in simplest form.

prob(2reds) = (8/14)(7/13)

= ..

To solve this problem, we need to find the probability of drawing two red marbles out of the bag.

First, let's find the total number of marbles in the bag. Since there are 8 red marbles and 6 yellow marbles, the total number of marbles is 8 + 6 = 14.

Now, let's find the probability of drawing a red marble on the first draw. Since there are 8 red marbles out of 14 total marbles, the probability of drawing a red marble on the first draw is 8/14.

After the first draw, we need to adjust the likelihood for the second draw, as we did not replace the first marble. Since there is now one less marble in the bag, the total number of marbles is reduced to 14 - 1 = 13.

Next, let's find the probability of drawing a red marble on the second draw, given that a red marble was already drawn on the first draw. Since there are 7 red marbles left and 13 total marbles, the probability of drawing a red marble on the second draw is 7/13.

To find the probability of both events happening (drawing two red marbles), we multiply the probabilities of each event. So, the probability of drawing two red marbles is (8/14) * (7/13).

Now, let's simplify this fraction. The numerator can be simplified to 2 * 2 * 1 = 4, and the denominator can be simplified to 7 * 13 = 91. Therefore, the probability of drawing two red marbles is 4/91.

So, the answer to the question is that the probability of drawing two red marbles out of the bag is 4/91.