There are three blue marbles, three red marbles, and three yellow marbles in a bowl. What is the probability of selecting exactly one red marble and exactly one blue marble from the bowl after three successive marbles are withdrawn from the bowl?

(A) 1/27
(B) 3/56
(C) 3/28
(D) 9/56
(E) 9/28.

The OA is E

After looking up the answer, I used the following logic to reach the answer.
Please let me know if I have used the correct logic.
There are 9 ways of picking the first marble, 8 ways of picking the second and 7 ways of picking the third.
So, the denominator of the probability is 9 * 8 * 7.

To come up with exactly one red and exactly one blue also means that we need one red, one blue and one yellow after 3 successive draws.
There are 3 ways of picking a blue, 3 ways of picking a red and 3 ways of picking an yelllow. So far, the numerator has 3 * 3 * 3.
Additionally there are 6 ways of combining Blue, Red and yellow.
Here is the list: RBY, RYB, BYR, BRY, YBR, YRB. (Straight Permutation).
so, the total ways of picking the numerator is
3 * 3 * 3 * 6.
so, the probability is 3 * 3 * 3 * 6 / 9 * 8 * 7. This give us the answer.

Please add your comments and clarifications.
Paddy



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