Hi amidamani13,

You can actually treat this question as a straight Probability question (without using the Combination Formula at all). Here's how:

From the prompt we know about 2 boxes and the marbles within each:

The black box contains
3 red
5 blue
8 total

The white box contains
2 red
3 blue
5 total

We're told to pull 2 marbles from the black box and 1 marble from the white box. We're asked for the probability that ALL marbles pulled will be BLUE.

Let's start with the 1st marble you pull from the black box....
The probability that it's blue is 5/8

Once you pull that first marble, there's 1 fewer blue marble in the black box.
When you grab the 2nd marble from the black box....
The probability that it's blue is 4/7

For the 3rd marble, we pull from the white box.
The probability that it's blue is 3/5

Now, we multiply the results: (5/8)(4/7)(3/5) = 3/14

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Red Blue
Black box - 3 5
White box - 2 3

Total combinations to pick two balls from black box = 8C2
Total combinations to pick one ball from white box = 5C1

Total combinations to pick two blue balls from black box = 5C2
Total combinations to pick one blue ball from white box = 3C1

probability of picking blue balls = (5C2 * 3C1) / (8C2 * 5C1) = 3/14

Ans is A

its an easy one
5C2/8C2 * 3C1/5C1
10/28 * 3/5
5/14 * 3/5 = 3/14--->A is answer
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Moved. This is a PS question. Please post it in the correct forum.
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Hey,

Thanks for the reply.

Got it.

Hi,

Thank you for posting the question however I have a doubt . Why it can't be (5C2*3C1)/13C3. ?
Please help with your explanation,would really appreciate it.

Regards
Megha

Hi Megha,

You CAN do the math that way, but there's an error in your calculation. Since you're choosing marbles from 2 separate boxes, you CANNOT mathematically state that you're 'choosing 3 from 13' - you're actually 'choosing 2 from 8' AND '1 from 5.' You would have to change your calculation to the following:

(5C2)(3C1) / (8C2)(5C1) =

30 / 140 =
3/14

GMAT assassins aren't born, they're made,
Rich
_________________

This isn't a combinations problem, its a probability problem.

Probability of blue marble from black box 5/8*4/7
Probability of blue marble for white box 3/5
Multiple them all together 3/14

In your solution, marbles are pulled one by one. But, if all marbles are pulled at a time, what would be the case, since question does not say to pull one by one?
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Hi Mahmud6,

Even if all the balls are pulled out together, it will not make any difference.
The difference in probability problems occurs when there is replacement/not?
If there is a replacement, (i.e) if we need to put the marble back in the box,
there will be a difference in probability of pulling out the marble when 2 or marbles are to be picked.

In this case,
Probability of finding 2 blue marbles in the black box and 1 blue marble in the white box is as follow :
{as we have 5 blue marbles(out of 8 marbles) in the black box and 3 blue marbles(out of 5 marbles)}

\(\frac{5c2}{8c2} * \frac{3c1}{5c1}\) = \(\frac{10}{28} * \frac{3}{5}\) = \(\frac{3}{14}\)(Option A)

Hope that helps!
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[quote="amidamani13"]A black box contains eight marbles, of which three are red, and the rest are blue. A white box contains five marbles, of which two are red, and the rest are blue. If two marbles are chosen from the black box at random, and one marble is chosen from the white box at random, what is the probability that all marbles will be blue?

A. 3/14
B. 3/8
C. 5/13
D. 8/13
E. 17/20

Black box = 8 marbles ; red = 3 and blue - 5
white box = 5 marbles; red= 2 and blue = 3
so
two marbles from black box and 1 marble from white box
P that its blue marble
5/8* 4 / 7 * 3/5
solve we get
3/14
IMO A