14
A jar contains jelly beans. There are
4
4 red jelly beans,
4
4 blue jelly beans and
6
6 green jelly beans. You randomly take two jelly beans out to eat. What is the probability that one jelly bean is red and the other jelly bean is green? Write your answer as a fraction. You do not need to simplify.
The total number of jelly beans in the jar is 4 + 4 + 6 = 14.
To find the probability of selecting a red jelly bean first and a green jelly bean second, we need to consider the number of ways this can happen.
There are 4 red jelly beans and 6 green jelly beans, so the number of ways to select a red jelly bean first and a green jelly bean second is 4 * 6 = 24.
However, we need to consider the order of the two jelly beans. Since we are taking two jelly beans out, the first one can be red and the second one green, but it could also be the other way around.
Therefore, the total number of successful outcomes is 24 * 2 = 48.
The total number of possible outcomes is the number of ways to select 2 jelly beans out of 14, which is given by the combination formula: C(14, 2) = 14! / (2! * (14 - 2)!) = (14 * 13) / (2 * 1) = 91.
Therefore, the probability that one jelly bean is red and the other jelly bean is green is 48/91.