A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?

A. 9/100

B. 2/19

C. 1/8

D. 3/20

E. 3/10

(B) Use this link for an instructor video explanation:

http://www.youtube.com/watch?v=uG0wBeJCtLETo determine the probability that several events all take place, you must first determine the probability that each separate event occurs. In this case, to determine the probability of all 3 jellybeans being blue, you must first determine the separate probability of removing a blue jellybean on each trial.

The first time a jellybean is removed, there are 10 blue jellybeans out of a total of 20 jellybeans in the bag. The probability of removing a blue one is 10/20.

The second time a jellybean is removed, there are only 9 blue jellybeans out of a total of 19 (since we are only concerned with the case where the first was blue), so the probability of getting a blue one is 9/19.

The third time, there are 8 blue jellybeans out of a total of 18 in the bag (by the same reasoning), so the probability is 8/18.

To find the total probability, we must multiply the three probabilities together: 10/20 × 9/19 × 8/18. After reducing the fractions to 1/2 × 9/19 × 4/9, we get 2/19.

In the last step, we reduce numerators and denominators first: the 9 in 9/19 cancels out with the 9 in 4/9, and the 2 in 1/2 cancels out when the 4 in 4/9 is reduced to 2. Now, the multiplication is 1 × 1/19 × 2, or 2/19.

The correct answer is choice (B).

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Don't words "and are not replaced" mean that the probability of removing a blue jellybean on each trial is alway 1/2?