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A black box contains eight marbles, of which three are red, and the re
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21 Jan 2015, 10:21
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76% (01:57) correct 24% (02:22) wrong based on 156 sessions
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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
Need help solving this using Combinations.
I tried solving it several times, however I always ended up with 3/7 as the final answer. Not sure where Im going wrong.
Re: A black box contains eight marbles, of which three are red, and the re
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21 Jan 2015, 20:10
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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
Re: A black box contains eight marbles, of which three are red, and the re
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21 Jul 2017, 21:55
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its an easy one 5C2/8C2 * 3C1/5C1 10/28 * 3/5 5/14 * 3/5 = 3/14--->A is answer
_________________
Push yourself again and again. Don't give an inch until the final buzzer sounds. -Larry Bird Success isn't something that just happens - success is learned, success is practiced and then it is shared. -Sparky Anderson -S
Re: A black box contains eight marbles, of which three are red, and the re
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13 Sep 2016, 13:23
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.
Re: A black box contains eight marbles, of which three are red, and the re
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13 Sep 2016, 16:36
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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:
Re: A black box contains eight marbles, of which three are red, and the re
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21 Jul 2017, 20:09
EMPOWERgmatRichC wrote:
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
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?
_________________
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!
_________________
You've got what it takes, but it will take everything you've got
Re: A black box contains eight marbles, of which three are red, and the re
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23 Mar 2019, 10:23
[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
gmatclubot
Re: A black box contains eight marbles, of which three are red, and the re
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23 Mar 2019, 10:23
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76% (01:57) correct 
