Help needed with a probability problem

[psriniva]

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of selecting exactly one red marble and exactly one blue marble from the bowl after three successive marbles are withdrawn from the bowl?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



[/spoiler]

[Stuart@KaplanGMAT]

Your method is good.

Here's a more formal way we could look at it.

Probability = # desired outcomes / total # of possibilities

When we want to find the probability of multiple events all occuring, we find the individual probabilities and multiply them together.

So, the probability of the 3 marbles being different colours is:

3/9 * 3/8 * 3/7

Further, we need to account for the different orders in which we can get 3 different colours. As you noted, we use the permutations formula to get 3! = 3*2*1 = 6

So, here's our formula for the probability of getting 3 different colours:

3! (3/9 * 3/8 * 3/7) = (6*3*3*3/9*8*7) = 9/28

[Ian Stewart]

The above approaches work, but I find this much simpler:

1st marble: doesn't matter what you pick. P = 1.
2nd marble: 8 marbles remain. You must pick one of the 6 that are different from the 1st selection. P = 6/8
3rd marble: 7 marbles remain. You must pick one of the 3 that are different from the first and second selections. P = 3/7.

so: 1*(6/8)*(3/7) = 9/28

[durgesh79]

The above approaches work, but I find this much simpler:

1st marble: doesn't matter what you pick. P = 1.
2nd marble: 8 marbles remain. You must pick one of the 6 that are different from the 1st selection. P = 6/8
3rd marble: 7 marbles remain. You must pick one of the 3 that are different from the first and second selections. P = 3/7.

so: 1*(6/8)*(3/7) = 9/28

Thanks a lot !!! I was thinking hard to try this approach but somehow wasnt able to put down on paper....

[beeparoo]

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?

OMG!! I completely mis-interpreted this question and understood boldened part to mean:

First select 3 marbles, THEN select one red marble, THEN select one blue marble, thus totalling 5 selected marbles.

I racked my brain over this one for an hour today trying to determine the 2-min (optimised) method of computing this without resorting to manual counting.

Was this a question prepared by the GMAC people?