probability

prob of takin particular hat is 4/8(1/2)
we will take 3 blue and 1 red
(BBBR)=(1/2)^4
(RBBB)=(1/2)^4
(BRBB)=(1/2)^4
(BBRB)=(1/2)^4
so total = 4*(1/2)^4=1/4


Similarly if we take 1 blue and 3 red
(RRRB)(BRRR)(RBRR)(RRBR) we will get 1/4


so total =(1/4)+(1/4)=1/2

C

Probability of choosing exactly three Blue hats which means the 4th hat has to be a Red.
So the Probability is same as choosing exactly one Red or one Blue Hat.
Choosing one Red 1/2 X (1/2)³
The choice of one Red Hat out of 4 draws happen in 4 ways . 1st ,2nd , 3rd or 4th draw
Prob of choosing Red = 4X1/16
Repeat for one Blue hat

Red or Blue so add the individual Prob

1/16 X 4 + 1/16 X 4 = [spoiler]1/2[/spoiler]

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.)

thank you all ....Though I still have one confusion ......

we are choosing 4 hats out of 8 so the total ways of doing this should be 8C4. However, all of you have chosen 2^4 which I understand is the right approach. But why doesnt 8C4 give us the right answer as we are choosing 4 hats out of 8. Think of the committee members questions where there are two commitees and we choose a certain number of people of out of them.

neoreaves wrote:thank you all ....Though I still have one confusion ......

we are choosing 4 hats out of 8 so the total ways of doing this should be 8C4. However, all of you have chosen 2^4 which I understand is the right approach. But why doesnt 8C4 give us the right answer as we are choosing 4 hats out of 8. Think of the committee members questions where there are two commitees and we choose a certain number of people of out of them.

If we weren't replacing the hat after each choice, then we could think of it as choosing 4 hats all at once. However, since this is a probability question "with replacement", the combinations formula is inapplicable.

So, if the question had been:

There are 8 hats in a drawer, 4 of which are red and 4 of which are blue. If Fred chooses 4 of the hats at random, what's the probability that exactly 3 of them are the same colour?

then we would have solved using the probability formula:

Probability = # of desired outcomes / total # of possibilities

and the denominator would be 8C4.

In fact, this is also an interesting question, so I challenge you: what would the final answer be?

There are 8 hats in a drawer, 4 of which are red and 4 of which are blue. If Fred chooses 4 of the hats at random, what's the probability that exactly 3 of them are the same colour?


Solving this...

For one color 4C3*4C1/8C4 = 16/70

For two = 2* 16/70 = 32/70

Ok this is a tough one for sure ...because the color is the same so 3C1 doesnt make sense because we always will be choosing the other color after choosing the first 3. However we have to look at the denominator which is 8C4. But anyways would love to get a clearer perspective on this. This has definitely become a very interesting question to brush up Combinations skills. Thanks specially to Stuart and others.