neoreaves wrote:A drawer holds 4 red hats and 4 blue hats.What is the probability of getting exactly 3 red hats or exactly 3 blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?
A) 1/8
B) 1/4
C) 1/2
D) 3/8
E) 7/12
Whenever we have a 50/50 probability question, we can think of it as a "pseudo-coin flip" question.
Here, since we're replacing the hat after each draw, we have a pure 50/50 probability. We could rephrase the question as:
If a fair coin is flipped 4 times, what's the probability of getting either exactly 3 heads or exactly 3 tails?
First, to save time, we note that the probability of exactly 3 heads is identical to the probability of exactly 3 tails; so, we can just solve for one of them, then double it.
We can then apply the coin flip formula:
Prob of k results out of n flips = nCk/2^2
(nCk is the combinations formula, n!/k!(n-k)!).
In this question, we have n=4 and k=3, so:
Prob 3 heads out of 4 flips =
4C3/2^4 = 4!/3!1! / 16 = 4/16 = 1/4
Now we remember to double it (since 3 tails is also acceptable):
2 * 1/4 = 1/2... choose (C).
(The other approaches posted also work, I just wanted to provide an alternative solution - the more different approaches you see in practice, the more likely you'll apply the approach that works best for you on Test Day.)
