There are three blue marbles, three red marbles, and three yellow marb

Please tell me what is unclear here: there-are-three-blue-marbles-three-red-marbles-and-three-yellow-marb-231053.html#p1780047 Thank you.

Hey Bunuel,

I don't quite understand this part of the question "after three successive marbles are withdrawn from the bowl?"

I also don't understand the reasoning behind vitaliyGMAT answer....

Thanks!

It means that 3 marbles were taken from the bowl one by one.

This is quite an easy probability question, so if you have problems solving it you should go through the basics again. Please check the link s below:

Combinatorics Made Easy!

Theory on Combinations

DS questions on Combinations
PS questions on Combinations

Tough and tricky questions on Combinations

Probability Made Easy!

Theory on probability problems

Data Sufficiency Questions on Probability
Problem Solving Questions on Probability

Tough Probability Questions

Hope it helps.

We are given there are 3 blue marbles, 3 red marbles, and 3 yellow marbles in a bowl. We must determine the probability of selecting one blue and two red marbles in 3 attempts.
On the first draw, since there are 3 red marbles and 9 total marbles, there is a 3/9 chance that a red marble will be selected. Next, since there are 2 red marbles and 8 total marbles left, there is a 2/8 chance a red marble will be selected on the second draw. Finally, when selecting the blue marble, since there are 3 blue marbles and 7 marbles left, there is a 3/7 chance a blue marble will be selected for the final marble. However, there are 3 different ways to select the 2 red and 1 blue marbles:

R - R - B:

R - B - R:

B - R - R:

Note that each of these 3 ways has the same probability of occurring, even though the individual probabilities appear in a different order. Thus, the total probability is:

3 x (3/9 x 2/8 x 3/7) = 3 x (1/3 x 1/4 x 3/7) = 3/28

Alternate Solution:

There are 9C3 = (9 x 8 x 7)/(3 x 2 x 1) = 84 ways to choose 3 marbles from a total of 9 marbles.

There are 3C1 = 3 ways to choose a blue marble and 3C2 = 3 ways to choose a red marble. Thus, there are 3 x 3 = 9 ways to make a selection that involves two red and one blue marble.

Thus, the probability that the selection consists of two red marbles and one blue marble is 9/84 = 3/28.

Answer: B

Approach 1: Probability of drawing B R R = (3/9)*(3/8)*(2/7)*(3!/2!) = 3/28

Approach 2: Total # of ways of drawing 3 marbles out of 9 marbles = 9C3 =(9*8*7)/(1*2*3) = 3*4*7

# of ways of choosing 1 blue marble out of 3 blue marbles = 3C1 = 3

# of ways of choosing 2 red marbles out of 3 red marbles = 3C2 = 3

Favorable # of ways = 3*3

hence required Probability = 3*3/3*4*7 = 3/28


Answer B.


Thanks,
GyM

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.