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There are 5 bags three of which contains 5 white and 2 black

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There are 5 bags three of which contains 5 white and 2 black  [#permalink]

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New post Updated on: 21 Oct 2014, 00:05
7
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  95% (hard)

Question Stats:

22% (02:44) correct 78% (02:29) wrong based on 392 sessions

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There are 5 bags three of which each contains 5 white and 2 black balls, and remaining 2 bags each contains 1 white and 4 black ball; a black ball has been drawn, find the chance that it came from first group.

A. 2/7
B. 6/35
C. 8/25
D. 15/43
E. 3/5

GMAT wont ask such question so if you want you can skip it.

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Originally posted by PiyushK on 05 Aug 2013, 07:18.
Last edited by PiyushK on 21 Oct 2014, 00:05, edited 4 times in total.
Edited the question, renamed the topic.
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Re: There are 5 bags three of which contains 5 white and 2 black  [#permalink]

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New post Updated on: 06 Aug 2013, 03:53
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Please go through my solution and suggest any mistake.

Chances of selection of the first group is 3/5-----------------------------Chances of selection of second group is 2/5
Chances of selecting a black ball from group 1: 2/7----------------------Chances of selecting a black ball from group 2: 4/5

Thus combined probability of section of black ball from group 1:
3/5 x 2/7 = 6/35

Thus combined probability of section of black ball from group 2:
2/5 x 4/5 = 8/25

Out of these chances, chance of occurrence of first case : (6/35) / (6/35 + 8/25) = 15/43

This is a conditional probability case :arrow: :idea:

:edited the typo error:
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Originally posted by PiyushK on 06 Aug 2013, 02:02.
Last edited by PiyushK on 06 Aug 2013, 03:53, edited 1 time in total.
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Re: There are 5 bags three of which contains 5 white and 2 black  [#permalink]

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New post Updated on: 05 Aug 2013, 11:32
The probability of selecting a bag from 1st group is 3C1/5C1 that is 3/5 and from that group the probability of getting a black ball is 2/7. so in my opinion chance of getting black ball from first group is (3*2)/(5*7) = 6/35.

Originally posted by Chiranjeevee on 05 Aug 2013, 09:46.
Last edited by Chiranjeevee on 05 Aug 2013, 11:32, edited 1 time in total.
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Re: There are 5 bags three of which contains 5 white and 2 black  [#permalink]

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New post 05 Aug 2013, 10:28
I don't understand why should we treat every bag separately. Can anyone please explain why can't we take the number of black balls in group one divided by the total number of black balls?

Thanks in advance
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Re: There are 5 bags three of which contains 5 white and 2 black  [#permalink]

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New post 06 Aug 2013, 02:30
PiyushK wrote:
Please go through my solution and suggest any mistake.

Chances of selection of the first group is 3/5-----------------------------Chances of selection of second group is 2/5
Chances of selecting a black ball from group 1: 2/7----------------------Chances of selecting a black ball from group 2: 4/5

Thus combined probability of section of black ball from group 1:
3/5 x 2/7 = 6/35

Thus combined probability of section of black ball from group 2:
2/5 x 4/7 = 8/35

Out of these chances, chance of occurrence of first case : (6/35) / (6/35 + 8/35) = 15/43

This is a conditional probability case :arrow: :idea:



(6/35)/(6/35+8/35) => (6/35)/(14/35) => 6*35/14*35 => 3/7 ( can you explain how it came to 15/43)
Perhaps, I am losing it up somewhere ?
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Re: There are 5 bags three of which contains 5 white and 2 black  [#permalink]

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New post 06 Aug 2013, 02:37
Hi Piyush,

I guess following piece from your reply is wrong


Thus combined probability of section of black ball from group 2:
2/5 x 4/7 = 8/35
it should be 4/5


Thanks
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Re: There are 5 bags three of which contains 5 white and 2 black  [#permalink]

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New post 06 Aug 2013, 02:40
genuinebot85 wrote:
PiyushK wrote:
Please go through my solution and suggest any mistake.

Chances of selection of the first group is 3/5-----------------------------Chances of selection of second group is 2/5
Chances of selecting a black ball from group 1: 2/7----------------------Chances of selecting a black ball from group 2: 4/5

Thus combined probability of section of black ball from group 1:
3/5 x 2/7 = 6/35

Thus combined probability of section of black ball from group 2:
2/5 x 4/7 = 8/35

Out of these chances, chance of occurrence of first case : (6/35) / (6/35 + 8/35) = 15/43

This is a conditional probability case :arrow: :idea:



(6/35)/(6/35+8/35) => (6/35)/(14/35) => 6*35/14*35 => 3/7 ( can you explain how it came to 15/43)
Perhaps, I am losing it up somewhere ?


there is a typo in above solution see highlited area

Thus combined probability of section of black ball from group 1:
3/5 x 2/7 = 6/35

Thus combined probability of section of black ball from group 2:
2/5 x 4/5 =8/25

Out of these chances, chance of occurrence of first case : (6/35) / (6/35 + 8/25) = 15/43

hope this helps
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Re: There are 5 bags three of which contains 5 white and 2 black  [#permalink]

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New post 06 Aug 2013, 03:55
yeah that was a typo error :P
Just edited my solution.
thanks :)
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Re: There are 5 bags three of which contains 5 white and 2 black  [#permalink]

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New post 08 Aug 2013, 21:37
PiyushK wrote:
yeah that was a typo error :P
Just edited my solution.
thanks :)


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sweet problem on conditional probability
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Re: There are 5 bags three of which contains 5 white and 2 black  [#permalink]

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New post 17 Sep 2014, 13:30
PiyushK wrote:
Please go through my solution and suggest any mistake.

Chances of selection of the first group is 3/5-----------------------------Chances of selection of second group is 2/5
Chances of selecting a black ball from group 1: 2/7----------------------Chances of selecting a black ball from group 2: 4/5

Thus combined probability of section of black ball from group 1:
3/5 x 2/7 = 6/35

Thus combined probability of section of black ball from group 2:
2/5 x 4/5 = 8/25

Out of these chances, chance of occurrence of first case : (6/35) / (6/35 + 8/25) = 15/43

This is a conditional probability case :arrow: :idea:

:edited the typo error:



Hi, You are not considering that group contains 3 bags with each of them containing 5W and 2B balls. Similarly, in group 2, each bag contains 1W and 4B balls

So your group 1 probability should be: 3/5 *6/21 = 6/35
group 2 probability should be: 2/5*8/10 = 8 /25

the answer is 6/35 / (6/35 + 8/25) = 15/43. Answer D
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Re: There are 5 bags three of which contains 5 white and 2 black  [#permalink]

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New post 27 Jul 2015, 09:57
PiyushK wrote:
There are 5 bags three of which each contains 5 white and 2 black balls, and remaining 2 bags each contains 1 white and 4 black ball; a black ball has been drawn, find the chance that it came from first group.

A. 2/7
B. 6/35
C. 8/25
D. 15/43
E. 3/5

GMAT wont ask such question so if you want you can skip it.

There are 3 bags having 5W & 2B each, 2bags with 1W & 4B each.
So, step 1 : probability of selecting 1st grp of bags is 3/5.....(1)
step 2: now since you have selected bag from 1st grp, what is the probability of getting a black ball from total 7 balls containing 5W +2B...is (2C1)/(7C1)=2/7...(2)
so, combine probability is 1 & 2 i.e 3/5 * 2/7=6/35.....B.
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Re: There are 5 bags three of which contains 5 white and 2 black   [#permalink] 27 Jul 2015, 09:57
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