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please explain gmat prep question

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by yvonne12 » Sun Apr 01, 2007 3:40 pm
from a bag containing 12 identical 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 best number of yellow balls that must be in the bag?

please explain how I can work out this problem
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Re: please explain gmat prep question

by ajith » Sun Apr 01, 2007 5:21 pm
yvonne12 wrote:from a bag containing 12 identical 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 best number of yellow balls that must be in the bag?

please explain how I can work out this problem
There are 12 balls overall

the maximum number of blue balls = 12*(2/5) = 4.8 = 4 ( let's make it four since it cannot be a fraction

The minimum is of course 0

The number of yellow balls (considering that there is only yellow and blue balls in the bag, maximizing the probability of a yellow ball), varies from 8-12

now from these the best possibility would be 12 yellow balls and 0 blue balls ..


:twisted: It's just a thought, anybody can point out a mistake
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by BTGmoderatorRO » Sun Oct 15, 2017 7:17 am
Number of blue balls=12
number of yellow balls=y
total number of balls in the bag=12+y
probability that one ball randomly removed is blue ball= 12/(12+y)
then, 12/(12+y) < 2/5
60< 24+2y (cross multiplication)
36<2y
y>36/2
y>18

The best number of Y should be the least number. Since Y is greater than 18, the least number will be 19.
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by Brent@GMATPrepNow » Sun Oct 15, 2017 1:01 pm
ORIGINAL QUESTION WITH ANSWER CHOICES
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
If you're not sure where to begin here, TESTING the answer choices will reveal the correct answer in no time.

A) 17
If there are 17 yellow balls, then the TOTAL number of balls = 12 + 17 = 29
So, P(selected ball is blue) = 12/29
12/29 is GREATER THAN 2/5, so we can eliminate A

B) 18
If there are 18 yellow balls, then the TOTAL number of balls = 12 + 18 = 30
So, P(selected ball is blue) = 12/30
12/30 is EQUAL TO 2/5, so we can eliminate B

At this point, we can see that adding 1 more yellow ball (i.e., 19 yellow balls) will make P(selected ball is blue) LESS THAN 2/5
So, the correct answer is C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Sun Nov 24, 2019 5:24 pm
yvonne12 wrote:from a bag containing 12 identical 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 best number of yellow balls that must be in the bag?

please explain how I can work out this problem
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.

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