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# There are 3 boxes A, B and C. In box A, there are 3 green marbles and

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There are 3 boxes A, B and C. In box A, there are 3 green marbles and  [#permalink]

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Updated on: 11 Dec 2016, 09:03
3
00:00

Difficulty:

55% (hard)

Question Stats:

73% (03:06) correct 27% (02:48) wrong based on 139 sessions

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There are 3 boxes A, B and C. In box A, there are 3 green marbles and 5 yellow marbles. In box B, there are 2 red marbles and 3 green marbles. In box C, there are 4 white marbles and 5 green marbles. Choose randomly a box then pick randomly a marble from that box. What is the probability that picked marble is green?

A. 1/8
B. 55/96
C. 2/15
D. 551/1080
E. 551/360

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Originally posted by broall on 11 Dec 2016, 05:21.
Last edited by broall on 11 Dec 2016, 09:03, edited 1 time in total.
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Re: There are 3 boxes A, B and C. In box A, there are 3 green marbles and  [#permalink]

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11 Dec 2016, 08:55
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Re: There are 3 boxes A, B and C. In box A, there are 3 green marbles and  [#permalink]

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11 Dec 2016, 08:59
fleamkt wrote:

I'll post the solution later Why dont you try this one?
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There are 3 boxes A, B and C. In box A, there are 3 green marbles and  [#permalink]

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Updated on: 12 Dec 2016, 01:21
1
$$P(A)*P(G)+P(B)*P(G)+P(C)*P(G)$$

$$\frac{1}{3} (\frac{3}{8} + \frac{3}{5} + \frac{5}{9}) = \frac{551}{1080}$$

First time I got 479, have you changed something?

Thanks for posting ones again.

Originally posted by vitaliyGMAT on 11 Dec 2016, 09:37.
Last edited by vitaliyGMAT on 12 Dec 2016, 01:21, edited 2 times in total.
Senior CR Moderator
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Re: There are 3 boxes A, B and C. In box A, there are 3 green marbles and  [#permalink]

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11 Dec 2016, 09:50
vitaliyGMAT wrote:
$$P(A)*P(G)+P(B)*P(G)+P(C)*P(G)$$

$$\frac{1}{3} (\frac{3}{8} + \frac{3}{5} + \frac{5}{9}) = \frac{551}{1080}$$

First time I got 479, have you changed something?

Thanks fo posting one again.

Yes, I've corrected some typos mistakes. Sorry about that.

Posted from my mobile device
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Re: There are 3 boxes A, B and C. In box A, there are 3 green marbles and  [#permalink]

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10 Jan 2017, 03:40
Vitality,,,,Why did u multiply by 1/3 of whole,,,please
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Re: There are 3 boxes A, B and C. In box A, there are 3 green marbles and  [#permalink]

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10 Jan 2017, 04:58
nguyendinhtuong wrote:
There are 3 boxes A, B and C. In box A, there are 3 green marbles and 5 yellow marbles. In box B, there are 2 red marbles and 3 green marbles. In box C, there are 4 white marbles and 5 green marbles. Choose randomly a box then pick randomly a marble from that box. What is the probability that picked marble is green?

A. 1/8
B. 55/96
C. 2/15
D. 551/1080
E. 551/360

The probability of picking a green marble from the different boxes is different.

Probability of picking box A = 1/3
Probability of picking a green marble from box A = 3/8

Probability of picking box B = 1/3
Probability of picking a green marble from box B = 3/5

Probability of picking box C = 1/3
Probability of picking a green marble from box C = 5/9

Probability of picking a green marble from box A, B or C = (1/3)*(3/8) + (1/3)*(3/5) + (1/3)*(5/9) = 1/8 + 1/5 + 5/27 = 551/1080
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Karishma
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Re: There are 3 boxes A, B and C. In box A, there are 3 green marbles and  [#permalink]

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13 Jan 2017, 00:13
I got the cake,,,actually I didn't observe the line,,,,choose randomly a box and then picking up marble,,,,,,Thank you Karishma mam,,,,,
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Re: There are 3 boxes A, B and C. In box A, there are 3 green marbles and  [#permalink]

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13 Jul 2017, 00:06
VeritasPrepKarishma wrote:
nguyendinhtuong wrote:
There are 3 boxes A, B and C. In box A, there are 3 green marbles and 5 yellow marbles. In box B, there are 2 red marbles and 3 green marbles. In box C, there are 4 white marbles and 5 green marbles. Choose randomly a box then pick randomly a marble from that box. What is the probability that picked marble is green?

A. 1/8
B. 55/96
C. 2/15
D. 551/1080
E. 551/360

The probability of picking a green marble from the different boxes is different.

Probability of picking box A = 1/3
Probability of picking a green marble from box A = 3/8

Probability of picking box B = 1/3
Probability of picking a green marble from box B = 3/5

Probability of picking box C = 1/3
Probability of picking a green marble from box C = 5/9

Probability of picking a green marble from box A, B or C = (1/3)*(3/8) + (1/3)*(3/5) + (1/3)*(5/9) = 1/8 + 1/5 + 5/27 = 551/1080

Hello VeritasPrepKarishma

I have solved this problem in the same way. I wonder how to do this using only combinatorics. Your inputs would be great
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Re: There are 3 boxes A, B and C. In box A, there are 3 green marbles and   [#permalink] 13 Jul 2017, 00:06
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