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probability of selecting exactly one blue marble and two red marbles from the bowl after three successive marbles are withdrawn from the bowl?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.
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https://www.beatthegmat.com/probability-t227448.html[/quote]
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