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cloaked_vessel
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nero44
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bewakoof
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only way you can get the sum of 6, if 1st ball is 1 and 2nd ball is 5 or 1st ball is 5 and second ball is 1

so P(1 , 5) = 2/5* 2/4 =1/5

and P (5, 1) = 2/5*2/4 = 1/5

so P(1 , 5) or P(5, 1) = 1/5 + 1/5 = 2/5

so answer is 2/5
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nero44
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bewakoof
nero44
Balls:
(1) (1) (5) (5) (15)

Total possibility to draw balls: 5c2 = 5x4/2 = 10
Possibility drawing two balls, which sum up to 6 are: (1) and (5)
(5) and (1) = 2 possibilities

probability is 2/10 = 1/5

(1)(1)(5)(5)(15)
A B C D E

So, someone can choose AC AD BC or BD. hence the total probability of drawing two balls with sum 6 = 4/10 = 2/5


True! D a m n , making a lot of minor mistakes in quant these days. Maybe I'm doing too much verbal :?
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marth750
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nakib77
only way you can get the sum of 6, if 1st ball is 1 and 2nd ball is 5 or 1st ball is 5 and second ball is 1

so P(1 , 5) = 2/5* 2/4 =1/5

and P (5, 1) = 2/5*2/4 = 1/5

so P(1 , 5) or P(5, 1) = 1/5 + 1/5 = 2/5

so answer is 2/5


I like this method. thanks nakib77
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GMATT73
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I also get 4/10 or 2/5. However, this is only possible if the balls are not replaced after removal (dependant probability).



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