There are 3 blue disks, 5 green disks, and nothing else in a container

There are 3 blue disks, 5 green disks, and nothing else in a container. Disks will be taken one at a time, and at random without replacement, from the container, until all 3 blue disks have been chosen. What is the probability that the third blue disks will be the seventh disk chosen?

A) 3/28

B) 1/8

C) 5/28

D) 15/56

E) 3/8


Probability 3rd Blue is seventh chosen = (# combos with 3rd blue seventh)/(total possible combos of where 3rd blue can be)

1) The third blue can be in spots 3, 4, 5, 6, 7, or 8:


_ _ _B3


_ _ _ _B3


_ _ _ _ _B3


_ _ _ _ _ _B3


_ _ _ _ _ _ _B3


_ _ _ _ _ _ _ _B3




2) Number of arrangements for each scenario depends on what can be placed before:


_ _ _B3 -----------------> BB -------------> (2!)/(2!) -----------> 1


_ _ _ _B3 --------------> BBG -----------------> (3!)/(2! * 1!) ---------> 3


_ _ _ _ _B3 -------------> BBGG ----------------> (4!)/(2! * 2!) -------------> 6


_ _ _ _ _ _B3 -----------> BBGGG ----------------> (5!)/(3! * 2!) ---------------> 10


_ _ _ _ _ _ _B3 -------------> BBGGGG ----------------> (6!)/(4! * 2!) -----------> 15


_ _ _ _ _ _ _ _B3 -------------------> BBGGGGG ----------------> (7!)/(5! * 2!) -----> 21

Total Combos = 1 + 3 + 6 + 10 + 15 + 21 = 56

Total combos with B3 at seventh slot = 15

Prob B3 at seventh slot = 15/56

Answer: D

Since there are a total of 8 discs, 3 of them blue, there is 8C3 different ways of picking discs.
We are asked what is the probability that the third blue disks will be the seventh disk chosen, as a result we need to count the number of ways we can have G,G,G,G,B,B,B. Here the last blue disc is fixed but the order of selecting the other 6 are not. As a result, there is 6C2 different ways. As a result, answer is 6C2/8C3 = 15 / 56

3 blue, 5 green

Number of ways to arrange them in a row: 8!/3!5! = 56
Number of ways to arrange them in a row so that the 7th spot is occupied by the 3rd blue ---> means we have 2 blue and 4 green in the first 6 spots, 1 blue in the 7th spot, and 1 green in the 8th spot --> (# of ways to arrange 2 blue and 4 green) * (1 way to fill the 7th spot) * (1 way to fill the 8th spot) = 6!/2!4! = 15

Probability to have 3rd blue in the 7th spot = 15/56

Answer: D

We do not have to count the 8th disk as per question and 7th will always be Blue. We stop when all blue disks are chosen.
That means 8th disk is Green.
So we are left with 6 places from starting to work with. (2 Blue and 4 Green )
No of ways to place : 6!/(2!*4!) =15
Total No of ways :- 8!/(3!*5!)=56

Answer:- 15/56. Option D.

Approach:
Find the number of ways 6 disks can be selected * No of ways 7th disk can be selected as blue disk from left over disks (1 out of 2 disks)
\(\\
=> \frac{3C2 * 5C4}{8C6} * \frac{1}{2} = \frac{15}{56}\\
\)

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.