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probability of selecting exactly one blue marble and two red marbles from the bowl after three successive marbles are withdrawn from the bowl?

A. 2/81
B. 3/28
C. 2/27
D. 1/28
E. 1/84

P(good outcome) = P(one way) * total possible ways.

Let B = blue and R = red.

P(one way):
One way to select exactly 1 blue marble and 2 red marbles is BRR.
P(B on the 1st pick) = 3/9. (Of the 9 marbles, 3 are blue.)
P(R on the 2nd pick) = 3/8. (Of the 8 remaining marbles, 3 are red.)
P(R on the 3rd pick) = 2/7. (Of the 7 remaining marbles, 2 are red.)
Since we want all of these events to happen, we MULTIPLY:
3/9 * 3/8 * 2/7 = 1/28.

Total possible ways:
RBB is only ONE WAY to select exactly 1 blue marble and 2 red marbles.
Now we must account for ALL OF THE WAYS to select exactly 1 blue marble and 2 red marbles.
Any arrangement of the letters BRR represents one way to select exactly 1 blue marble and 2 red marbles.
Thus, to account for ALL OF THE WAYS to select exactly 1 blue marble and 2 red marbles, the result above must be multiplied by the number of ways to arrange the letters BRR.
Number of ways to arrange 3 elements = 3!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical R's:
3!/2! = 3.

Multiplying the results above, we get:
P(exactly 1 blue marble and 2 red marbles) = 3 * 1/28 = 3/28.

The correct answer is B.

More practice:
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https://www.beatthegmat.com/probability-t227448.html[/quote]

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probability of selecting exactly one blue marble and two red marbles from the bowl after three successive marbles are withdrawn from the bowl?

A. 2/81
B. 3/28
C. 2/27
D. 1/28
E. 1/84

We are given there are 3 blue marbles, 3 red marbles, and 3 yellow marbles in a bowl. We must determine the probability of selecting one blue and two red marbles in 3 attempts.
On the first draw, since there are 3 red marbles and 9 total marbles, there is a 3/9 chance that a red marble will be selected. Next, since there are 2 red marbles and 8 total marbles left, there is a 2/8 chance a red marble will be selected on the second draw. Finally, when selecting the blue marble, since there are 3 blue marbles and 7 marbles left, there is a 3/7 chance a blue marble will be selected for the final marble. However, there are 3 different ways to select the 2 red and 1 blue marbles:

R - R - B:

R - B - R:

B - R - R:

Note that each of these 3 ways has the same probability of occurring, even though the individual probabilities appear in a different order. Thus, the total probability is:

3 x (3/9 x 2/8 x 3/7) = 3 x (1/3 x 1/4 x 3/7) = 3/28

Alternate Solution:

There are 9C3 = (9 x 8 x 7)/(3 x 2 x 1) = 84 ways to choose 3 marbles from a total of 9 marbles.

There are 3C1 = 3 ways to choose a blue marble and 3C2 = 3 ways to choose a red marble. Thus, there are 3 x 3 = 9 ways to make a selection that involves two red and one blue marble.

Thus, the probability that the selection consists of two red marbles and one blue marble is 9/84 = 3/28.

Answer: B