VJesus12 wrote:5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn without replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?
A. 4/33
B. (5/36)^2
C. 1/2
D. (31/36)^2
E. 29/33
P(not 2 blue and 2 purple) = 1 - P(exactly 2 blue and exactly 2 purple).
P(exactly n times) = P(one way) * total possible ways.
P(one way):
One way to get exactly 2 blue and exactly 2 purple:
BBPP.
P(B on the 1st pick) = 5/12. (Of the 12 marbles, 5 are blue.)
P(B on the 2nd pick) = 4/11. (Of the 11 remaining marbles, 4 are blue.)
P(P on the 3rd pick) = 4/10. (Of the 10 remaining marbles, 4 are purple.)
P(P on the last pick) = 3/9. (Of the 9 remaining marbles, 3 are purple.)
Since we want all of these events to happen, we MULTIPLY:
5/12 * 4/11 * 4/10 * 3/9 = 2/99.
Total possible ways:
BBPP is only ONE WAY to get exactly 2 blue and exactly 2 purple.
Now we must account for ALL OF THE WAYS to get exactly 2 blue and exactly 2 purple.
Any arrangement of the letters BBPP represents one way to get exactly 2 blue and exactly 2 purple.
Thus, to account for ALL OF THE WAYS to get exactly 2 blue and exactly 2 purple, the result above must be multiplied by the number of ways to arrange the letters BBPP.
Number of ways to arrange 4 elements = 4!.
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 2 identical B's and by another 2! to account for the 2 identical P's:
4!/(2!2!) = 6.
Multiplying the results above, we get:
P(exactly 2 blue and exactly 2 purple) = 2/99 * 6 = 4/33.
Thus:
P(not 2 blue and 2 purple) = 1 - 4/33 = 29/33.
The correct answer is
E.
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