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

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At the beginning of a show, a lottery representative put one hundred b [#permalink]

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New post 09 Apr 2015, 03:19
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Question Stats:

39% (02:29) correct 61% (01:47) wrong based on 181 sessions

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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.

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Re: At the beginning of a show, a lottery representative put one hundred b [#permalink]

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New post 09 Apr 2015, 13:07
5
Lucky2783 wrote:
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.


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
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At the beginning of a show, a lottery representative put one hundred b [#permalink]

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New post 07 May 2015, 01:33
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
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Re: At the beginning of a show, a lottery representative put one hundred b [#permalink]

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New post 07 May 2015, 02:07
2
noTh1ng wrote:
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


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.
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At the beginning of a show, a lottery representative put one hundred b [#permalink]

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New post 27 Aug 2015, 06:42
Harley1980 wrote:
Lucky2783 wrote:
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.


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



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
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Re: At the beginning of a show, a lottery representative put one hundred b [#permalink]

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New post 25 Mar 2018, 21:16
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

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Re: At the beginning of a show, a lottery representative put one hundred b   [#permalink] 25 Mar 2018, 21:16
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