Problem Solving

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 no of yellow balls that must be in the bag?

[u]Options:[/u]
1. 17
2. 18
3. 19
4. 20
5. 21

I followed the below approach-
-> No of outcomes = (12+y)C1= 12+ y
-> No of ways with which blue ball can be selected= 12.
So the probability of selecting a blue ball is -- 12/(12+y) .. is this the correct approach?

baalok88 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 no of yellow balls that must be in the bag?

Options:
1. 17
2. 18
3. 19
4. 20
5. 21

I followed the below approach-
-> No of outcomes = (12+y)C1= 12+ y
-> No of ways with which blue ball can be selected= 12.
So the probability of selecting a blue ball is -- 12/(12+y) .. is this the correct approach?

Sure. Now you can set up your inequality. 12/(12 + y) < 2/5

baalok88 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 no of yellow balls that must be in the bag?

Options:
1. 17
2. 18
3. 19
4. 20
5. 21

I followed the below approach-
-> No of outcomes = (12+y)C1= 12+ y
-> No of ways with which blue ball can be selected= 12.
So the probability of selecting a blue ball is -- 12/(12+y) .. is this the correct approach?

Or just backsolve.

Say we test B. If y = 18, we'll have a total of 30 balls. p(selecting blue) = 12/30 = 2/5. But we know that the probability should be less than 2/5, so the denominator will have to be larger. Answer is C

Hi baalok88,

Another possible route to doing this question:

No. of blue balls = 12
No. of yellow balls = y
Probability of drawing a blue ball < 2/5

Proceed to get to a situation where the probability of drawing a blue ball = 2/5 = 40%
Now, we already know that the number of blue balls = 12.
What would be the total number of balls so that 12 is 40% of that total? To mentally calculate, you could divide it by 4 in your head (to get 10% of the pool) and then multiply it by 10 to arrive at the total number of balls.
Or you can calculate it to say: 2/5x = 12. (x being the total no. of balls). So, x = 12 x 5/2 = 30

Now, 30 is the total number of balls if probability of drawing a blue ball were to be 2/5. But, the probability is less than 2/5, so we need to increase the no. of balls by 1 to equal 31.

Thus, no, of yellow balls = 31 - 12 (no. of blue balls) = 19

Thus, the answer is C.

We know that

12 / (12 + y) < 2/5

and that y ≥ 0. So we multiply both sides:

5*12 < 2 * (12 + y)

60 < 24 + 2y

18 < y

and we're left with y = 19.

Matt@VeritasPrep wrote:We know that

12 / (12 + y) < 2/5

and that y ≥ 0. So we multiply both sides:

5*12 < 2 * (12 + y)

60 < 24 + 2y

18 < y

and we're left with y = 19.

I'm not sure why you said y ≥ 0 and what you did when you multipllied both sides. Can you clarify

hoppycat wrote:[

I'm not sure why you said y ≥ 0 and what you did when you multipllied both sides. Can you clarify

We know y ≥ 0 because we can't have a negative number of yellow marbles. (y = the # of yellows)

When I manipulated both sides, I did the following:

12 / (12 + y) < 2/5

cross multiply:

12 * 5 < 2 * (12 + y)

simplify:

60 < 24 + 2y

subtract:

36 < 2y

divide:

18 < y

Also, the y ≥ 0 isn't terribly important, at least for GMAT purposes. I needed to specify that so I could safely cross-multiply the inequality. (If y ≥ 0, then 12 + y > 0, and I know that when I multiply by (12 + y), I'm multiplying by a positive number.)

Matt@VeritasPrep wrote:

hoppycat wrote:[

I'm not sure why you said y ≥ 0 and what you did when you multipllied both sides. Can you clarify

We know y ≥ 0 because we can't have a negative number of yellow marbles. (y = the # of yellows)

When I manipulated both sides, I did the following:

12 / (12 + y) < 2/5

cross multiply:

12 * 5 < 2 * (12 + y)

simplify:

60 < 24 + 2y

subtract:

36 < 2y

divide:

18 < y

Thanks Matt