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From a bag containing 12 identical blue balls, y identical [#permalink]
06 Apr 2006, 14:32

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Difficulty:

45% (medium)

Question Stats:

59% (01:45) correct
41% (01:12) wrong based on 138 sessions

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag?

Re: PS - probability [#permalink]
09 Apr 2006, 11:38

(12 blue) / (12 blue + x yellow) = 2/5

12/30 = 2/5

The denominator must be 30. You have 12 already which means you need 18 yellow balls. This makes the equation equal to each other. However, you want less than 2/5 so add one ball to make it 19.

It is 18 P(blue) = 12 / (12+Y) P(blue) < 2/5 solving this: 12 /12+Y = 2/5 y = 18

The answer is 19, not 18. Notice that we have an inequality sign there.

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag? A. 17 B. 18 C. 19 D. 20 E. 21

Re: From a bag containing 12 identical blue balls, y identical [#permalink]
25 Nov 2013, 12:52

Matador wrote:

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag?

A. 17 B. 18 C. 19 D. 20 E. 21

It goes like this.

12 Blue y yellow

12 + y total

12/x = 2/5

So x = 30, x being the total number of ballos Then yellow> 30-12 = 18

Therefore, yellow must be at least 19

Cheers! Kudos Rain J

gmatclubot

Re: From a bag containing 12 identical blue balls, y identical
[#permalink]
25 Nov 2013, 12:52

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