Question

A jar contains 6 red jelly beans, 6 green jelly beans, and 6 blue jelly beans.
If we choose a jelly bean, then another jelly bean without putting the first one back in the jar, what is the probability that the first jelly bean will be green and the second will be green as well?

Answer 1

the probability is 1/9

Answer 2

Given:

No. of red jelly beans:6

No. of green jelly beans:6

No. of blue jelly beans:6

To Find:

Probability that the :

1. 1st jelly bean drawn out will be green.
2. 2nd jelly bean drawn out will be green as well.

Solution:

Total no.of jelly beans in the bag=(No. of red jelly beans+No. of green jelly beans+No. of blue jelly beans)=(6+6+6)=18.

Total no. of green jelly beans=6

∴Probability that the drawn jelly bean will be green=$\dfrac{total no. of green jelly beans}{total no. of jelly beans}$=$\dfrac{6}{18}$=$\dfrac{1}{3}$

Total number of jelly  beans left in the bag=17.

If the 1st jelly bean drawn from the bag is green, then the total number of green jelly beans left in the bag =5.

∴Probability that the 2nd jelly bean drawn is green=$\dfrac{5}{17}$

If the 1st jelly bean drawn is not green, then the total number of green jelly beans in the bag=6

∴Probability that the 2nd jelly bean drawn from the bag is green=$\dfrac{6}{17}$.