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A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]
26 Jul 2004, 09:33
6
This post was BOOKMARKED
00:00
A
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Difficulty:
65% (hard)
Question Stats:
46% (02:06) correct
54% (00:59) wrong based on 363 sessions
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?
agree with Dookie....here the events are independent....
can you post a detail explanation ? thanks
This is how I came up with 2/27. the probability of getting one red is 3/9 (nine is the total number of balls). The probability of getting a white ball is 2/9 (9 again because the ball is put back after each draw) so 3/9*2/9 + 6/81 = 2/27
according to Dookie (who is right) if they are asking for the balls to draw one of the the other, which they are (successive draws) you have to multiply 2/27 by 2 = 4/27.
First of all we have
Probability of drawing a Red ball is 3/9
Probability of drawing a White ball is 3/9
There are two ways in which the balls can be drawn
Case 1: Red ball in the first draw and white in the second draw
Hence the combined Probability is 3/9*2/9=6/81
Case 2: White ball in the first draw and red in the second draw
Hence the combined Probability is 2/9*3/9=6/81
both these cases satisfy our requirement
Hence either of them will do i.e OR
Hence the final probability comes to be
Case 1 OR Case 2 = 6/81 + 6/81 (OR means addition)
Hence the Ans is 12/81=4/27
Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]
23 Sep 2013, 06:53
Expert's post
2
This post was BOOKMARKED
imhimanshu wrote:
Bumping this thread as I am looking for experts to provide an explanation and confirm the OA.
Thanks
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.
Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]
23 Sep 2013, 19:22
1
This post received KUDOS
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.
Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]
27 Sep 2013, 19:21
2
This post received KUDOS
Expert's post
imhimanshu wrote:
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.
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. _________________
A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]
22 Sep 2014, 05:33
VeritasPrepKarishma wrote:
imhimanshu wrote:
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.
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.
Thank You Karishma. I got a key concept here
But i am more comfortable by Combination method
1C3*1C2/ (1C9 * 1C9) = 6/81 = 2/27
Now we can get this in two ways (as described by u) 2* 2/27 = 4/27 _________________
Consider +1 Kudos Please
gmatclubot
A bag contains 3 red, 4 black and 2 white balls. What is the
[#permalink]
22 Sep 2014, 05:33
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