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A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
23 Sep 2013, 07:53
Kudos
Bookmarks
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
This is with replacement case.
\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)
We are multiplying by 2 as there are two possible wining scenarios RW and WR.
Answer: D.
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
This is with replacement case.
\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)
We are multiplying by 2 as there are two possible wining scenarios RW and WR.
Answer: D.
27 Sep 2013, 20:21
Kudos
Bookmarks
imhimanshu
Hello Bunuel,
I m a bit confused about when to consider and when not to consider. I am having a tough time understanding this concept. I was under the impression that in "with replacement" cases, we dont need to consider the cases, however in without replacement cases, scenarios needs to be considered.
In addition,I found this link, where the question is also testing the same concept, but we didn't consider the multiple cases here. Please clarify.
https://gmatclub.com/forum/rich-has-3-gr ... 55253.html
Can you provide a high level conceptual knowledge as in when to consider cases and when not to?
Pls help.
Posted from my mobile device
I m a bit confused about when to consider and when not to consider. I am having a tough time understanding this concept. I was under the impression that in "with replacement" cases, we dont need to consider the cases, however in without replacement cases, scenarios needs to be considered.
In addition,I found this link, where the question is also testing the same concept, but we didn't consider the multiple cases here. Please clarify.
https://gmatclub.com/forum/rich-has-3-gr ... 55253.html
Can you provide a high level conceptual knowledge as in when to consider cases and when not to?
Pls help.
Posted from my mobile device
Responding to a pm:
The status of "replacement" has nothing to do with the "sequence". It only changes the probability.
Say we have 2 red and 3 white balls in a bag. We pull out two one after another with replacement. What is the probability that one is red and the other is white.
Now note that there are 4 ways in which you can pull out two balls from the bag:
1. You pull a Red and then a Red again RR - (2/5)*(2/5) (Note that it is with replacement)
2. You pull a Red and then a White RW - (2/5)*(3/5)
3. You pull a White and then a Red WR - (3/5)*(2/5)
4. You pull a White and then a White WW - (3/5)*(3/5)
Total probability = (2/5)*(2/5) + (2/5)*(3/5) + (3/5)*(2/5) + (3/5)*(3/5) = 1
In how many cases do we have a red and a white ball? In case 2 and case 3.
Probability of picking a red and a white with replacement = (2/5)*(3/5) + (3/5)*(2/5) = (3/5)*(2/5) * 2
Since the probability of picking a red and then a white is same as probability of picking a white and then a red, you simply write down one case and multiply it by 2. You do the same in case of 'without replacement' too. The only thing that changes is the probability.
Without Replacement:
1. You pull a Red and then a Red again RR - (2/5)*(1/4)
2. You pull a Red and then a White RW - (2/5)*(3/4)
3. You pull a White and then a Red WR - (3/5)*(2/4)
4. You pull a White and then a White WW - (3/5)*(2/4)
Probability of picking a red and a white WITHOUT replacement = (2/5)*(3/4) + (3/5)*(2/4) = (3/5)*(2/4) * 2
As for the link you have mentioned, this is exactly what is done there too. Check it out - I will show you how it is done there.
