saurabhkamal1981 wrote:Hi Experts,

I downloaded permutation, combination and probability practice questions (54 questions) from here. In one of the questions i have founded that the following question's answer options are wrong or may be i am wrong. I did ask my tutor and he is also emphasizing on the answer options. The following question is

38. 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? (Tricky)

a) 1/8
b) ¼
c) ½
d) 3/8
e) 7/12

My answer is as follows:-

Ans. There are total 8 hats (4 red and 4 blue)
4 hats are randomly drawn. Out of these exactly 3 are red or exactly 3 are blue.

Case 1--> 3 red and 1 blue
Case 2--> 1 red and 3 blue

(4C3 * 4C1) + (4C3 * 4C1) = 1
8C1 * 8C1 * 8C1 * 8C1 128


or (4C3 * 4C1) + (4C3 * 4C1) / 8C1 * 8C1 * 8C1 * 8C1 = 1/128

Will be waiting for your reply.

Thanks & Regards
Saurabh Kamal

beat_gmat_09 wrote:

saurabhkamal1981 wrote:Hi Experts,

I downloaded permutation, combination and probability practice questions (54 questions) from here. In one of the questions i have founded that the following question's answer options are wrong or may be i am wrong. I did ask my tutor and he is also emphasizing on the answer options. The following question is

38. 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? (Tricky)

a) 1/8
b) ¼
c) ½
d) 3/8
e) 7/12

My answer is as follows:-

Ans. There are total 8 hats (4 red and 4 blue)
4 hats are randomly drawn. Out of these exactly 3 are red or exactly 3 are blue.

Case 1--> 3 red and 1 blue
Case 2--> 1 red and 3 blue

(4C3 * 4C1) + (4C3 * 4C1) = 1
8C1 * 8C1 * 8C1 * 8C1 128


or (4C3 * 4C1) + (4C3 * 4C1) / 8C1 * 8C1 * 8C1 * 8C1 = 1/128

Will be waiting for your reply.

Thanks & Regards
Saurabh Kamal

I am not a expert, but this is how i solved -

A hat drawn is replaced/put back in the drawer, hence there will be same number of hats each time on draw. There are 4 Red and 4 Blue, therefore the probability of picking same colored hat is 4/8 = 1/2. Now you should also consider the number of ways the 3 hats are chosen, it is given by = C(4,3) = 4 i.e. you may get hats in following ways - BBBR, RBBB, BRBB, BBRB, these are the number of ways of drawing the hats in which 3 hats have same color. Therefore the probability of getting only 3 Blue hats is 4*(1/2)^4 - as 4 hats are drawn and each has same probability.
Similarly the probability of getting only 3 Red (RRRB, RRBR, BRRR, RBRR) hats is 4*(1/2)^4
4*(1/2)^4 + 4*(1/2)^4 = 1/2.

Rahul@gurome wrote:The probability of having a blue hat in any draw is 4/8 = ½.
Similarly the probability of having a red hat in a draw is 4/8 = ½.

Let the four draws be denoted by d1, d2, d3 and d4.
So exactly 3 blue hats can be had in the following ways:
(1) d1 - blue, d2 - blue, d3 - blue, d4 - red. Probability of this is (1/2)^4.
(2) d1 - blue, d2 - blue, d3 - red, d4 - blue. Probability of this is (1/2)^4.
(3) d1 - blue, d2 - red, d3 - blue, d4 - blue. Probability of this is (1/2)^4.
(4) d1 - red, d2 - blue, d3 - blue, d4 - blue. Probability of this is (1/2)^4.

So the probability of having exactly 3 blue hats in 4 draws is the sum of all the above which is 4 * (1/2)^4 = ¼.

In a similar way, probability of having exactly 3 red hats in 4 draws is ¼.

So the required probability is ¼ + ¼ = ½.
The correct answer is c).