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# M32-04

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Math Expert
Joined: 02 Sep 2009
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13 Jun 2016, 06:40
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75% (hard)

Question Stats:

40% (01:31) correct 60% (01:14) wrong based on 60 sessions

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A bag contains only blue and red balls. What is the probability that the second ball we pick will be blue, if the first ball we picked was red?

(1) Initially, the number of red balls was one more than the number of blue balls.

(2) Initially, the number of red balls was 7.

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13 Jun 2016, 06:40
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Official Solution:

A bag contains only blue and red balls. What is the probability that the second ball we pick will be blue, if the first ball we picked was red?

Say, initially there were $$b$$ blue balls and $$r$$ red balls. The probability that the second ball we pick will be blue, if the first ball we picked was red will therefore be $$\frac{b}{b+r-1}$$

(1) Initially, the number of red balls was one more than the number of blue balls. This implies that $$r=b+1$$, hence the probability is $$\frac{b}{b+r-1}=\frac{b}{b+(b+1)-1}=\frac{b}{2b}=\frac{1}{2}$$. Sufficient.

(2) Initially, the number of red balls was 7. Clearly insufficient.

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22 Aug 2016, 21:34
This probability defined for picking blue given that first red doesnt seem right to me as its a conditional probability and should not be straight up b/(r+b-1). I might be wrong but would be good to confirm this.
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24 Aug 2016, 04:57
sharmaanurag wrote:
This probability defined for picking blue given that first red doesnt seem right to me as its a conditional probability and should not be straight up b/(r+b-1). I might be wrong but would be good to confirm this.

No, it's right. When you pick 1 red you are left with b blue balls and r red balls. The probability that the second ball we pick will be blue, if the first ball we picked was red will therefore be $$\frac{b}{b+(r-1)}$$
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06 Sep 2016, 11:23
Bunuel,

I think I am missing something here. I tried with an example:

Lets blue balls be 3 and red balls be 4 :

4/7 * 3/6 = 2/7

but if

B=5 and R=6 then it will be 3/11

How you are getting 1/2 as the answer?
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07 Sep 2016, 08:12
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gmatravi wrote:
Bunuel,

I think I am missing something here. I tried with an example:

Lets blue balls be 3 and red balls be 4 :

4/7 * 3/6 = 2/7

but if

B=5 and R=6 then it will be 3/11

How you are getting 1/2 as the answer?

What you've calculated here is conditional probability.

3/11 would've been correct if we were asked to calculate the probability of getting the FIRST ball red and THEN picking the second ball blue.

But here we're given that the FIRST has already been picked and it was red. That event has already happened, so we don't need to worry about that.

In mathematical terms,

P[first ball red] = 1 , as there is no other possible outcome other than what already came

Let blue balls be 3 and red balls be 4 :

1/1 * 3/6 = 1/2

Hence, it's correct.
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11 Sep 2016, 05:00
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We can solve this question while reading the question ...
Given one red ball was picked.

condition 1 > red ball was one more than the blue ball.
as we have picked one red ball now so red balls equals blue ball.

So the probability of picking the blue ball is half.
Hence condition 1 is sufficient.

condition 2 > not sufficient.
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22 Mar 2017, 04:42
Hi Bunuel. I don't agree with the explanation,

Statement (1): What's up if there are zero blue balls and only one red ball in the bag? The probability of getting a blue ball would be zero (because there are no blue balls in the bag!) - I agree that for any other case, the P is 1/2.

So we need statement (2) to solve.

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22 Mar 2017, 05:08
jcriedel wrote:
Hi Bunuel. I don't agree with the explanation,

Statement (1): What's up if there are zero blue balls and only one red ball in the bag? The probability of getting a blue ball would be zero (because there are no blue balls in the bag!) - I agree that for any other case, the P is 1/2.

So we need statement (2) to solve.

Since we are told that "A bag contains only blue and red balls", then I think it's safe to assume that the number of blue balls is more than 0. If it were otherwise then we could say that the bag contains 0 number of any other color balls.
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14 Mar 2018, 06:07
I don't agree with the explanation. The probability of first red, and then blue should be [r][/b+r]*[b][/b+r-1]. How could you guarantee you always get red ball first?
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14 Mar 2018, 06:14
1
Canteenbottle wrote:
I don't agree with the explanation. The probability of first red, and then blue should be [r][/b+r]*[b][/b+r-1]. How could you guarantee you always get red ball first?

This is called conditional probability. The question asks to find the probability that the second ball we pick will be blue, PROVIDED the first ball we picked is red.
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10 Sep 2018, 16:54
I think this is a high-quality question and I agree with explanation.
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21 Nov 2018, 10:07
I think this is a high-quality question and I agree with explanation. This is a nice trap question. We need to understand the difference between getting first one red and first is red.
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31 Jan 2019, 19:54
Hi all!

I have one doubt I would like to discuss. If statement number (1) would have been the following:

(1) Initially, the number of red balls was TWO more than the number of blue balls

This statement would not have been sufficient, right? As the probability of getting a blue ball after getting a red one would depend on the total number of balls in the bag. Only when the statement said that there was one more red ball, it is guaranteed that the probability of getting a blue ball after getting a red one will be 1/2.

Thanks!
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31 Jan 2019, 20:25
1
matmen wrote:
Hi all!

I have one doubt I would like to discuss. If statement number (1) would have been the following:

(1) Initially, the number of red balls was TWO more than the number of blue balls

This statement would not have been sufficient, right? As the probability of getting a blue ball after getting a red one would depend on the total number of balls in the bag. Only when the statement said that there was one more red ball, it is guaranteed that the probability of getting a blue ball after getting a red one will be 1/2.

Thanks!

the number of red balls was TWO more than the number of blue balls
=> r = b + 2
Probability = b/(b+r-1) = b/(b+b+2-1) = b/(2b+1) => Insufficient.
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11 Jun 2019, 03:44
what will be the case say blue balls are 0 and the red ball is 1..then what will be the scenario? after all, we don't know how many blue balls are there
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11 Jun 2019, 03:46
what will be the case say blue balls are 0 and the red ball is 1..then what will be the scenario? after all, we don't know how many blue balls are there
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11 Jun 2019, 03:48
gaurav2m wrote:
what will be the case say blue balls are 0 and the red ball is 1..then what will be the scenario? after all, we don't know how many blue balls are there

Sorry, cannot understand what you are implying there...

As for possibility of blue (or red) balls being 0, I addressed this above:

Since we are told that "A bag contains only blue and red balls", then I think it's safe to assume that the number of blue balls is more than 0. If it were otherwise then we could say that the bag contains 0 number of any other color balls.
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Re: M32-04   [#permalink] 11 Jun 2019, 03:48
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# M32-04

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