Question

an urn contains 7 green and 2 red marbles. you reach in and grab 2 marbles. what is the probability there is one of each color in your hand?

Answer 1

The probability of drawing one green and one red marble is 7/18.

Step-by-step explanation:

To find the probability of drawing one green and one red marble, we need to calculate the probability for each case separately and then add them together.

First, let's calculate the probability of drawing a green marble on the first pick and a red marble on the second pick. There are 7 green marbles and 2 red marbles in the urn, so the probability of drawing a green marble on the first pick is 7/9. After drawing the green marble, there are 6 green marbles and 2 red marbles left, so the probability of drawing a red marble on the second pick is 2/8. Therefore, the probability of this case is (7/9) * (2/8).

Next, let's calculate the probability of drawing a red marble on the first pick and a green marble on the second pick. The probability of drawing a red marble on the first pick is 2/9. After drawing the red marble, there are 7 green marbles and 1 red marble left, so the probability of drawing a green marble on the second pick is 7/8. Therefore, the probability of this case is (2/9) * (7/8).

Now, we can add the probabilities of the two cases together to get the overall probability. (7/9) * (2/8) + (2/9) * (7/8) = 14/72 + 14/72 = 28/72 = 7/18. So the probability of drawing one green and one red marble is 7/18.