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Probability Question- from a bag containing....

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Probability Question- from a bag containing....

by baalok88 » Sun Oct 09, 2016 8:22 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 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?
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by DavidG@VeritasPrep » Sun Oct 09, 2016 8:39 am
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
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by DavidG@VeritasPrep » Sun Oct 09, 2016 8:41 am
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
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by Brent@GMATPrepNow » Sun Oct 09, 2016 9:06 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?
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,
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by MBA Challengers » Tue Oct 11, 2016 3:03 am
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.
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by Matt@VeritasPrep » Fri Oct 14, 2016 1:54 am
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.
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by hoppycat » Sun Jun 04, 2017 8:27 am
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
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by [email protected] » Sun Jun 04, 2017 10:04 am
Hi hoppycat,

The prompt tells us that there are Y identical yellow balls, so Y cannot be a negative value (Y is either 0 or a positive integer).

Mathematically, we can multiply both sides of an inequality by the same positive value and keep the 'direction' of the inequality exactly the way it is. By extension, we can multiply both sides of the inequality by 5 and by (12 + Y); since we know that (12+Y) will be POSITIVE, we don't have to worry about whether we have to 'flip' the inequality or not.

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by Jay@ManhattanReview » Sun Jun 04, 2017 9:08 pm
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?
We have

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

=> 12/(12+y) < 2/5

=> 6/(12+y) < 1/5

=> 30 < 12+y; we can do so since y is a positive integer

=> y > 18

The least integer value of y is 19.

The correct answer: C

Hope this helps!

Relevant book: Manhattan Review GMAT Combinatorics and Probability Guide

-Jay
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by Matt@VeritasPrep » Mon Jun 05, 2017 11:32 pm
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
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by Matt@VeritasPrep » Mon Jun 05, 2017 11:33 pm
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.)
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by hoppycat » Wed Jun 14, 2017 7:19 am
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
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by Scott@TargetTestPrep » Thu Jun 15, 2017 3:31 pm
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
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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