A drawer holds 4 red hats and 4 blue hats. What is the...

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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

The OA is C.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.

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Hi swerve,

We're told that a drawer holds 4 red hats and 4 blue hats. We're asked for 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. We can answer this question with a bit of note-taking and some basic arithmetic. To start, since we're putting each hat back into the drawer after pulling one out, the fact that there are 4 red hats and 4 blue hats is irrelevant. As long as there are an equal number of both colors, the math will be exactly the same (meaning that there could be 1 red hat and 1 blue hat).

Since there is an equal chance of pulling either of the two colors on each 'pull', there are (2)(2)(2)(2) = 16 possible outcomes

Given the 'goal', there are only a limited number of possible outcomes that 'fit' what we're looking for:
RRRB
RRBR
RBRR
BRRR

BBBR
BBRB
BRBB
RBBB

8 of the 16 = 1/2 of the outcomes give us what we're after.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

[GMATGuruNY]

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
[/quote

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:
https://www.beatthegmat.com/select-exact ... 88786.html
https://www.beatthegmat.com/probability- ... 14250.html
https://www.beatthegmat.com/a-single-par ... 28342.html
https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html

[Scott@TargetTestPrep]

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

The OA is C.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.

First, let's determine the probability of selecting exactly 3 red hats (and 1 blue hat).

P(R-R-R-B) = (1/2)^4 = 1/16

Since R-R-R-B can be arranged in 4!/3! = 4 ways, the overall probability is 4/16 = 1/4.

Since we have the same number of red and blue hats the probability of selecting exactly 3 blue hats (and 1 red hat) is also 1/4.

Thus, the probability of selecting exactly 3 red hats or 3 blue hats is 1/4 + 1/4 = 2/4 = 1/2.

Answer: C