Abag contains 6 red jelly beans, 4 green jelly beans, and 4blue jelly beans. if we choose a jelly bean, then another jelly bean without putting the first one back in the bag, what is the probability that the first jelly bean will be green and the second will be red?

The probability of picking a green jelly bean with the first pick is 4/14 = 2/7, because there are 14 jelly bean in total (6 red + 4 green + 4 blue) and 4 of them are green.

If you pick a green jelly bean at the beginning, you have 13 jelly beans remaining, of which 6 are red. So, the probability of picking a red jelly bean is now 6/13.

You want these two events to happen one after the other, to be more precise you want to pick a green jelly bean with the first pick AND a red jelly bean with the second pick. We know the probabilities of the two events, so we have to multiply them to get the probability of them happening both:

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)