I think the question is not exactly correct. we can select 1 blue ball from 12 identical blue ball in 1 way and 1 ball from 12 identical blue and y identical y in 2 ways(the ball will be either blue or yellow),so probability should be 1/2

but if I ignore the term identical,the probability of selecting one blue ball is 12/(12+y)

this is less than 2/5
so 12/(12+y)<2/5
or 60<(24+2y)
or 36<2y
or y>18..since y is integer ,minimum value of y is 19 which is given as ans.
Can some one please confirm about the identical part?

selango 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 must be the least number of yellow balls that must be in the bag?

A) 17

B)18

C)19

D) 20

E)21

OA 19

I went this way!!

since the to have a probability of 2/5 ,total number of balls should be a multiple of 5. So the yellow balls+ blue balls = multiple of 5.

Going thru options I see only 18 will fit the bill (18+12=30, multiple of 5).

And it is mentioned p is less than 2/5, so in that case , Total number needs to greater than 30,

Since the minimum no is asked for, I picked 31 and that mounts to 19 yellow balls

my approach:

set up ratio
y= yellow
x=blue

x/(x/(x+y))<(y/(x+y))

solve for y

12/(2/5)<y/(3/5)

18<y

It's important to note that there are no other types of balls (besides blue and yellow) because it means that the probability that a blue is picked is b/(b+y) or 12/(12+y). Since this probability is less than 2/5, we can simply setup the inequality 12/(12+y) < (2/5) and isolate y. If you do the math properly, you will find that y > 18. As a result the least number of yellow possible is 19. The answer is C

This is GMATPrep question 1305. You can practice similar questions if you have access to the Solutions Engine drill generator by selecting topic="Combinatorics" and difficulty="600-700"

Good luck,
-Patrick