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probability

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by smohsin » Mon Oct 27, 2008 6:51 am
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?

Ans: 17, 18, 19, 20, 21

Thanks so much for the help.
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by parallel_chase » Mon Oct 27, 2008 7:09 am
IMO 19.

OA?
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Re: probability

by reacher » Mon Oct 27, 2008 7:11 am
smohsin 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?

Ans: 17, 18, 19, 20, 21

Thanks so much for the help.
The probability of Blue Ball <2/5.

To solve this question, let's say that the probability of drawing the blue ball is 2/5. Then, from the question stem you know that there are 12 blue balls. So, if 2 of numerator denotes 12 blue balls, 5 in the denominator denotes 30 balls in total and thus 18 yellow balls (Total 30 = 12 Blue + 18 Yellow).

Since we want to find the least no of yellow balls so that the probability has to be <2/5, you only have to increase the denominator (in essence, number of yellow balls). So, the least number of yellow balls have to be 18+1= 19 balls.
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by KeyserSoze525 » Mon Oct 27, 2008 7:19 am
You have 12 Blue and Y yellow balls.

The probability of picking a blue ball is:

P(B) = 12/(12+Y) (12 Blue balls out of a total possible 12 + Y balls)

We also know that P(B) < 2/5 so:

12/(12+Y) < 2/5

Cross multiply:

60 < 24 + 2Y

Simplify:

36 < 2Y or 18 < Y

Since Y must be greater than 18, the smallest number it could be is 19.
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