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

Question

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.

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.

PiyushK · Accepted Answer

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:

:edited the typo error:

Chiranjeevee · Answer

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.

Soothcaster · Answer

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

genuinebot85 · Answer

(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 ?

aniketmunshi · Answer

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
Aniket

blueseas · Answer

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

PiyushK · Answer

yeah that was a typo error

Just edited my solution. thanks

Asifpirlo · Answer

......
sweet problem on conditional probability

Gbv11 · Answer

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

dhawalbp · Answer

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.

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

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