AAPL wrote:A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one?
A. 1/8
B. 1/4
C. 1/2
D. 3/8
E. 7/12
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P(exactly n times) = P(one way) * total possible ways.
Since 4 of the 8 hats are red, and 4 of the 8 hats are blue:
P(R) = 4/8 = 1/2.
P(B) = 4/8 = 1/2.
P(one way):
One way to select exactly 3 blue hats:
RBBB.
P(R on the 1st pick) = 1/2.
P(B on the 2nd pick) = 1/2.
P(B on the 3rd pick) = 1/2.
P(B on the 4th pick) = 1/2.
Since we want all of these events to happen, we MULTIPLY the fractions:
1/2 * 1/2 * 1/2 * 1/2 = 1/16.
Total possible ways:
RBBB is only ONE WAY to get exactly 3 blue hats.
Now we must account for ALL OF THE WAYS to get exactly 3 blue hats.
Any arrangement of the letters RBBB represents one way to get exactly 3 blue hats.
Thus, to account for ALL OF THE WAYS to get exactly 3 blue hats, the result above must be multiplied by the number of ways to arrange the letters RBBB.
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 3! to account for the three identical B's:
4!/(3!) = 4.
Multiplying the results above, we get:
P(exactly 3 blue hats) = 1/16 * 4 =
1/4.
By extension:
P(exactly 3 red hats) =
1/4.
Since a favorable outcome will be yielded by selecting exactly 3 blue hats OR exactly 3 red hats, we ADD the fractions in blue:
P(3 hats of the same color) = P(exactly 3 blue hats) + P(exactly 3 red hats) = 1/4 + 1/4 = 1/2.
The correct answer is
C.
More practice:
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