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

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.

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

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

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

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

\(\left( 2 \right)\,\,\,\left( {{\rm{blue}}\,\,{\rm{:}}\,\,{\rm{yellow}}} \right)\,\,\,{\rm{removed}}\,\,{\rm{ratio}}\,\,\,\, = \,\,\,\,\left( {{\rm{blue}}\,\,{\rm{:}}\,\,{\rm{yellow}}} \right)\,\,\,{\rm{original}}\,\,{\rm{ratio}}\,\,\,\,\)

\(\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {b,y} \right)\,\,{\rm{original}} = \left( {20,80} \right)\,\,\,\,\,\left[ {{{20} \over {100}} = 20\% } \right]\,\,\,\,\,\,\,AND\,\,\,\,\,\left( {b,y} \right)\,\,{\rm{removed}} = \left( {10,40} \right)\,\,\,\,\,\,\left[ {{{10} \over {50}} = 20\% } \right]\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,b = 20\,\,{\rm{viable}}\,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {b,y} \right)\,\,{\rm{original}} = \left( {40,60} \right)\,\,\,\,\,\left[ {{{40} \over {100}} = 40\% } \right]\,\,\,\,\,\,\,AND\,\,\,\,\,\left( {b,y} \right)\,\,{\rm{removed}} = \left( {20,30} \right)\,\,\,\,\,\,\left[ {{{20} \over {50}} = 40\% } \right]\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,b = 40\,\,{\rm{viable}}\,\,\,\, \hfill \cr} \right.\)

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


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

Regards,
Fabio.

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

\(b+y=100\)
\(y/100>b/100…y-b>0\)

(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.

\(y-4/100<b+4/100…100(y-b)<800…y-b<8…(y-b>0)\)
\(0<y-b<8…(b=100-y)…0<y-(100-y)<8…0<2y-100<8…50<y<54\)
\(y=(51,52,53),b=(49,48,47)\)

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

\(b/100=b-x/50…50b=100b-100x…-50b=-100x…b=2x=even.integer\)

(1&2) sufic

\(b=even=(49,48,47)=48\)

Ans (C)

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.

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.

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.




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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.



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