M32-04

Question

A bag contains only blue and red balls. If the balls are picked without replacement, what is the probability that the second ball we pick will be blue, given that the first ball we picked was red?

(1) Initially, there was one more red ball than blue balls in the bag.
 
(2) Initially, there were 7 red balls in the bag.

Bunuel · Accepted Answer

Official Solution:

A bag contains only blue and red balls. If the balls are picked without replacement, what is the probability that the second ball we pick will be blue, given that the first ball we picked was red? 
 
Let  b  be the number of blue balls and  r  the number of red balls. After picking a red ball, there are  b  blue balls and  (r-1)  red balls left, with  (b+r-1)  balls in total. The probability of picking a blue ball next is  \frac{b}{b+r-1} .
 
(1) Initially, there was one more red ball than blue balls in the bag. 
 
This implies that  r=b+1 . Therefore, the probability is  \frac{b}{b+r-1}=\frac{b}{b+(b+1)-1}=\frac{b}{2b}=\frac{1}{2} . Sufficient.
 
(2) Initially, there were 7 red balls in the bag. 
 
This information is insufficient, as we do not know the number of blue balls in the bag.

Answer: A

nealz · Answer

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.

gmatravi · Answer

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?

AdiTank · 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  = 1 , as there is no other possible outcome other than what already came

Using your example,
Let blue balls be 3 and red balls be 4 :

1/1 * 3/6 = 1/2

Hence, it's correct.

Canteenbottle · Answer

I don't agree with the explanation. The probability of first red, and then blue should be   *  . How could you guarantee you always get red ball first?

Bunuel · Answer

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.

jjsozal · Answer

I think this is a  high-quality  question and I agree with explanation.

psaikrishna90 · Answer

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.

leonardBe · Answer

I think this is a  high-quality  question and I agree with explanation.

danklin · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

Gemmie · Answer

R: Number of Red balls
B: Number of Blue balls

Question:  \frac{B}{B+R-1} = ?

Statement 1: R = B + 1

=> R - 1 = B

=>  \frac{B}{B+R-1} = \frac{B}{B+B} = \frac{B}{2B} = \frac{1}{2}

Therefore, statement 1 is sufficient.

Statement 2: R = 7

\frac{B}{B+R-1} = \frac{B}{B+6}

B is not given

Therefore, statement 2 is not sufficient.

==> Only statement 1 is sufficient to determine the probability.

TC97 · Answer

I think this is a  high-quality  question and I agree with explanation.

JuniqueLid · Answer

I think an improvement can be made to the question by explicitly stating that there are at least 1 of each colored ball in the bag. Otherwise, in the event there was initially 1 red ball and no blue ball, the probability formula constructed with (b+r-1) as the denomenator would not be meaningful - it can still happen in this case that, having already picked a red ball, the probability of picking out a blue ball is 0.

Bunuel · Answer

"A bag contains only blue and red balls" already implies that there is at least one of each color. So, the question is already correct, and no improvement is needed.

kanwar08 · Answer

KarishmaB   Bunuel

I am facing a big challenge when addressing such problem. To me this looks entirely similar to the Conditional Probability Ques, which say given A occured what's the probablity of B.

Typically answer in those scenarios if P(A) * P(B) or P(A intersection B)/P(A)

Now, how do I know it's not a CP question? Can you please clarify this doubt?

KarishmaB · Answer

It is a CP question. We are given that the first ball was red. So we know that red balls have reduced by 1. We can calculate from that step onwards. 
Or you can use P(B given A) = P(A and B)/P(A) here and the answer would be the same just that using the formula is a round about way of arriving at the answer.

ext{P('second ball blue' given 'first ball red')} = \frac{	ext{P('first ball red' and 'second ball blue')}}{	ext{P('first ball red')}}   = \frac{R/(R+B) * B/(R- 1 + B)}{R/(R+B)} = \frac{B}{(R- 1 + B)}  and you get the same answer

kanwar08 · Answer

Thanks KarishmaB for clarifying this. On this line I solve few more questions and got one doubt, is this a hard formula we need to use while address problems or we can somehow use logic to solve such ques? https://gmatclub.com/forum/download/file.php?id=84829 In this question, this is also a CP question but solved via a different means. Can you please help me with a general strategy or a concept note to approach such questions? GMAT-Club-Forum-v6zdo3ow.png

M32-04

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