At the beginning of a show, a lottery representative put one hundred b

Question

At the beginning of a show, a lottery representative put one hundred balls in a bag. Any ball is equally likely to be selected. 
Each ball is either blue or yellow. If someone picks one ball at random, it probably will be a yellow one.
How many blue balls has the representative put in the bag?

(1) If four yellow balls had been repainted blue, it would have been more likely to select a blue ball rather than a yellow one.

(2) After the representative removes fifty balls, the probability of selecting a blue ball is the same as it was initially.

Harley1980 · Accepted Answer

From task we know that probability of blue more than probability of yellow P(Y) > P(B)
Total 100 balls

1) From this statement we can make infer that there is 3 possible variants:
Y=53, B=47 
probability of yellow equal to  \frac{53}{100}  and probability of blue equal to  \frac{47}{100}

ANd if we repaint 4 yellow balls to blue it'll be  P(Y) = \frac{49}{100}  and  P(B) = \frac{51}{100}

Next variants: Y=52, B=48 and Y=51, B=49
Insufficient

2) This statement says that we remove 50 balls and received the same probability (so we removed equal quantity of borh colors). 
So it should be symmetric remove. And we can have symmetric remove only if case in which we have even number of balls.
But it can be any variant that include even numbers Y=60 B=40; Y=72 B=28 etc
Insufficient

1+2) From first statement we have 3 variants: two with odd numbers and one with even numbers. And from second statement we know that we should have even numbers
So answer is Y=52, B=48:  P(Y)=\frac{52}{100}  and  P(B) = \frac{48}{100} 
If we repaint 4 balls we received such probabilities: 
 P(Y)=\frac{48}{100}  and  P(B) = \frac{52}{100}

And if we remove 25 balls of each color we received  P(Y) = \frac{26}{50}  and  P(B)=\frac{24}{50}

Sufficient.

Answer is C

hellosanthosh2k2 · Answer

Good question, solved it this way.

Given : Y > B, => B <= 49

Statement 1:

4 yellow balls are repainted blue.

B+4 > Y-4
 B+8 > Y
Also, Y > B => B+8 > Y > B
Implies that B = {47, 48, 49}

Still insufficient

Statement 2:

Let B1 number of blue balls after 50 balls removal.

B/100 = B1/50
B = 2B1.
Implies B is even, but still can't find B

1+2, B = 48 is only possible even - C

Posted from my mobile device

noTh1ng · Answer

Hi  Harley1980

can you please explain what a symmetric remove is and why we can infer that the numbers then have to be even? Thanks

PrepTap · Answer

Hi,

The second statement says 50 balls (= 1/2 of 100 balls) are removed and the probability of selecting a blue ball remains the same.
So we can infer that the ratio of blue balls to yellow balls is unchanged.
This can only happen if you remove 1/2 of the blue balls and 1/2 of the yellow balls as 50 balls (= 1/2 of 100 balls) are removed.
Now, if the number of blue balls or yellow balls is odd, can we divide it equally? No, eg: 15 -> 15/2 = 7.5 (as 0.5 ball has no meaning, the two groups will contain 7 and 8 balls)
Hence, only when we have even number of balls we can divide it equally. So it can be inferred that we have even number of blue balls and even number of yellow balls.

Hope this helps. Do let us know if you have more questions.

GauravSolanky · Answer

Awesome.The key here was to pick set with even no. balls, as set with odd number of balls won't result in realistic numbers as far as balls are concerned. +1 for this solution. Thanks, Gaurav :-D

fskilnik · Answer

Beautiful problem! (Be careful to avoid coming to wrong conclusions!)

100\,\,{m{balls}}\,\,\,\left\{ \matrix{
 \,b\,\,{m{blue}}\,\,\,\,\,,\,\,\,\,\,\,1 \le b \le 49\,\,\,\left( * ight) \hfill \cr 
 \,100 - b\,\,{m{yellow}} \hfill \cr} ight.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,? = b

\left( 1 ight)\,\,\,\,{{b + 4} \over {100}} > {1 \over 2}\,\,\,\,\,\,\, \Rightarrow \,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,b > 46\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,\left\{ \matrix{
 \,{m{Take}}\,\,b = 47 \hfill \cr 
 \,{m{Take}}\,\,b = 48 \hfill \cr} ight.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m{INSUFF}}.

\left( 2 ight)\,\,\,\left( {{m{blue}}\,\,{m{:}}\,\,{m{yellow}}} ight)\,\,\,{m{removed}}\,\,{m{ratio}}\,\,\,\, = \,\,\,\,\left( {{m{blue}}\,\,{m{:}}\,\,{m{yellow}}} ight)\,\,\,{m{original}}\,\,{m{ratio}}\,\,\,\,

\left\{ \matrix{
 \,{m{Take}}\,\,\left( {b,y} ight)\,\,{m{original}} = \left( {20,80} ight)\,\,\,\,\,\left \,\,\,\,\,\,\,AND\,\,\,\,\,\left( {b,y} ight)\,\,{m{removed}} = \left( {10,40} ight)\,\,\,\,\,\,\left \,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,b = 20\,\,{m{viable}}\,\, \hfill \cr 
 \,{m{Take}}\,\,\left( {b,y} ight)\,\,{m{original}} = \left( {40,60} ight)\,\,\,\,\,\left \,\,\,\,\,\,\,AND\,\,\,\,\,\left( {b,y} ight)\,\,{m{removed}} = \left( {20,30} ight)\,\,\,\,\,\,\left \,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,b = 40\,\,{m{viable}}\,\,\,\, \hfill \cr} ight.

\left( {1 + 2} ight)\,\,\,\,\,\left\{ \matrix{
 \,b = 47\,\,\,\, \Rightarrow \,\,\,y = 53\,\,\,\, \Rightarrow \,\,\,{{47} \over {100}} 
e {x \over {50}}\,\,\,\,,\,\,\,\,x\,\,{\mathop{m int}} \hfill \cr 
 \,b = 48\,\,\,\, \Rightarrow \,\,\,y = 52\,\,\,\, \Rightarrow \,\,\,{{48} \over {100}} = {x \over {50}}\,\,\,\left( { = {{2x} \over {100}}} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 24\,\,\left( {{\mathop{m int}} } ight)\,\,\,,\,\,\,{m{viable}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m{only}}\,\,\,{m{survivor!}} \hfill \cr 
 \,b = 49\,\,\,\, \Rightarrow \,\,\,y = 51\,\,\,\, \Rightarrow \,\,\,{{49} \over {100}} 
e {x \over {50}}\,\,\,\left( { = {{2x} \over {100}}} ight)\,\,\,,\,\,\,\,x\,\,{\mathop{m int}} \hfill \cr} ight.

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

exc4libur · Answer

b+y=100 y/100>b/100…y-b>0 (1) If four yellow balls had been repainted blue, it would have been more likely to select a blue ball rather than a yellow one. y-4/1000) 0

antartican · Answer

Is it not important to put that there is at the lease one Yellow and at the least one Blue ball. 
What if there is no Blue ball ?

Or may be that is concluded after adding stmt 1 and 2.

CrackverbalGMAT · Answer

It’s given, the probability of selecting a yellow ball is more than the probability of selecting blue ball
 P(Y) > P(B)
A total of 100 balls are there.

Statement 1.
If four yellow balls are repainted as blue, P(B) > P(Y)
Case 1:
Y = 51 balls , B= 49 balls.
P(Y) = 51/100 and P(B) = 49/100
P(Y) > P(B)
After repainting four yellow balls as blue, y= 47 and B = 51
P(Y) = 47/100 and P(B) = 51/100
P(B) > P(Y)

Case 2:
Y = 52 balls, B= 48 balls.
P(Y) = 52/100 and P(B) = 48/100
P(Y) > P(B)
After repainting four yellow balls as blue, y= 48 and B = 52
P(Y) = 48/100 and P(B) = 52/100
P(B) > P(Y)
 
Case 3:
Y = 53, B = 47
After repainting 4 yellow as blue, y = 49, B =51
 
These are the only 3 possible cases where after repainting, P(B) > P(Y)
(Check by yourself what happens when Y=54, B= 46)
Our question is to find no of blue balls in the bag. As per the statement 1, it could be 49,48 or 47.
We will not get a definite answer for B, hence insufficient.

Statement 2:

It says, after we remove 50 balls, the probability of a getting a blue ball remains same as before.
B= no of blue balls initially
B’ = no of blue balls after removing 50 balls

B/100 = B’/50
That means B’ = B/2.

We can conclude that no of blue balls should be even and we will not be able to get a definite value for B
Statement 2 is not sufficient.

Combining Statement 1 and 2.
From statement 1: B could be 49,48 or 47
From statement 2: B is even
Combining both, we can say that value of B = 48.
Option C is the answer.

Fdambro294 · Answer

Regarding statement 1 alone and statement 2 alone, neither being sufficient: it is explained rather well by everyone above.

Statement 1 allows 3 possible cases given the original constraints (Yellow > Blue):

Case 1: Y = 51 and B = 49

Case 2: Y = 52 and B = 48

Case 3: Y = 53 and B = 47

Together:

given that we have only the 3 cases from statement 1:

S2 says: after the representative removes 50 balls, the probability of selecting a Blue ball is the SAME AS IT WAS Initially.

As it is stated above, it is only possible to make the Probability the Same if we “remove the 50 balls symmetrically” from the Yellow and Blue side.

For anyone reading, this does NOT mean we remove 25 from each side. If we remove
25 blue balls and 25 yellow balls, we end up with different probabilities than those we had initially.

Initial probability for each ball in case 2, the only case that can work:

P(Blue) = 48/100 = .48

P(Yellow) = 52/100 = .52

We could have the probability of selecting a Blue stay the same by removing (1/2) of the balls from each side. This is what is meant by symmetrically removing the “same” number of balls from each side. We are decreasing each of the balls by (1/2) and the total removed must be 50.

Blue: remove 24 from initial 48——- now have 24 blue

Yellow: remove 26 from initial 52—— now have 26 Yellow

24 + 26 = 50 balls removed.

P(Blue) = 24 / (24 + 26)
 
= 24/50 = 48/100

= .48 ——> same as the initial probability of randomly selecting a Blue

Together the statements are therefore sufficient as the other 2 cases have numbers in which we can not “symmetrically” remove 50 balls and have the probabilities stay the same.

C together sufficient.

Posted from my mobile device

Fdambro294 · Answer

Great! 👍

Thank you.

The only part that should be edited about the 1st post is the part that says:

“25 balls need to be removed from each side”

But without even worrying about that, you can get to the answer following number properties.

Posted from my mobile device

bumpbot · Answer

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