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From a bag containing 12 identical blue balls, y identical [#permalink]
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Updated on: 22 Sep 2013, 03:54
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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
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Originally posted by Matador on 06 Apr 2006, 15:32.
Last edited by Bunuel on 22 Sep 2013, 03:54, edited 1 time in total.
Renamed the topic, edited the question, added the answer choices and the OA.



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12/x = 2/5
x = 30 total number of balls.
There must be at least 18 yellow balls in the bag.



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Agree with NTLancer.
Since the prob of blue balls is LESS than 2/5, you need at least 19 yellow balls in the bag.



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Agree
The number of yellow balls is atleast 19.
Since yellow balls must be greater than 18...hence least number is 19



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Agree with 19 too.
Since 2/5=12/x, x = 30 so if probability is less than 2/5 that means x is >30, say 31 then 3112=19 yellow balls.



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Re: PS  probability [#permalink]
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09 Apr 2006, 12:38
(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.



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



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Re: From a bag containing 12 identical blue balls, y identical [#permalink]
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25 Nov 2013, 13:52
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> 3012 = 18 Therefore, yellow must be at least 19 Cheers! Kudos Rain J



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Re: From a bag containing 12 identical blue balls, y identical [#permalink]
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22 Nov 2014, 03:39
Can u pls help in explaining the shift in the inequilaties symbol . I mean 12/12+y < 2/5 so cross multiplying 12*5 < (12+y)2 , am i missing out on reversing the sign .... thanks in advance Bunuel wrote: alexgomez2013 wrote: 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 \(P(blue) = \frac{12}{12+y}<\frac{2}{5}\) > \(y>18\) > \(y_{min}=19\). Answer: C. Hope it's clear.



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From a bag containing 12 identical blue balls, y identical [#permalink]
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22 Nov 2014, 06:50
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hanschris5 wrote: Can u pls help in explaining the shift in the inequilaties symbol . I mean 12/12+y < 2/5 so cross multiplying 12*5 < (12+y)2 , am i missing out on reversing the sign .... thanks in advance Bunuel wrote: alexgomez2013 wrote: 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 \(P(blue) = \frac{12}{12+y}<\frac{2}{5}\) > \(y>18\) > \(y_{min}=19\). Answer: C. Hope it's clear. It goes like this: \(\frac{12}{12+y}<\frac{2}{5}\) Cross multiply: \(60< 24+2y\); Subtract 24 from both sides: \(36<2y\); Divide by 2: \(18<y\), which is the same as \(y>18\) (y is greater than 18). Hope it's clear.
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From a bag containing 12 identical blue balls, y identical [#permalink]
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10 Jan 2016, 03:53
Other way of doing is substitute the value.
The question says it probability of blue should be less than 2/5 means it should be less than 40%
Also For a constant numerator and if the denominator is increasing the value of the output will be decreasing.
In our scenario, the numerator is constant which is 12
Based on the given the choice the denominator will be 12/29, 12/30, 12/31, 12/32, 12/33
Since numerator is same and denominator doesn't vary we can conclude as 12/29 > 12/30 > 12/31 > 12/32 > 12/33
Lets choose the number which is easy to solve. Hence choosing 12/30 first which is equal to 2/5
Straight away I can go for "C" from the above answer since 12/29 must be greater than 2/5 and 12/31 should hold the least value of yellow balls which will give the probability as less than 2/5



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Re: From a bag containing 12 identical blue balls, y identical [#permalink]
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15 Dec 2017, 10:39
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 We can create the following inequality: 12/(y+12) < 2/5 60 < 2(y + 12) 30 < y + 12 18 < y Since y is greater than 18, the least number of yellow balls is 19. Answer: C
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Re: From a bag containing 12 identical blue balls, y identical
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