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 first a red marble then a yellow marble are drawn? Represent your answer as a fraction in simplest form.
prob(red, then yellow) = (8/14)(6/13_
= ...
To find the probability that first a red marble and then a yellow marble are drawn, we can use the concept of conditional probability.
First, let's find the probability of drawing a red marble on the first draw.
The bag contains a total of 8 red marbles and 6 yellow marbles, so the probability of drawing a red marble on the first draw is:
P(R) = (number of favorable outcomes)/(total number of possible outcomes)
= 8/14
= 4/7
After drawing a red marble on the first draw, there are now 7 red marbles and 6 yellow marbles left in the bag.
Now, let's find the probability of drawing a yellow marble on the second draw, given that a red marble was drawn on the first draw.
The bag contains a total of 7 red marbles and 6 yellow marbles, so the probability of drawing a yellow marble on the second draw is:
P(Y|R) = (number of favorable outcomes)/(total number of possible outcomes)
= 6/13
Now, to find the probability of both events happening (drawing a red marble first and then drawing a yellow marble), we can multiply the probabilities of the two events:
P(R then Y) = P(R) * P(Y|R)
= (4/7) * (6/13)
= 24/91
Therefore, the probability of first drawing a red marble and then drawing a yellow marble is 24/91.