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A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue? 2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue? 3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue? 4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
Originally posted by bmwhype2 on 06 Dec 2007, 14:24.
Last edited by Bunuel on 01 Feb 2019, 04:21, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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21 Jan 2010, 00:35
Bullet wrote: Can any body please explain Question No.3 using probability
Why we need to add twice 7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15
Thanks and appreciated The solution you are posting for the third question is not right. Below is the solution of this question using the "probability". 3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?Probability of at least two marble are blue is the sum of the two probabilities: A. Two marbles are blue and one is red  BBR. This can occur in \(\frac{3!}{2!}=3\) # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. \(3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}\); B. All three marbles are blue  BBB. This can occur only one way, namely BBB. \(\frac{7}{10}*\frac{6}{9}*\frac{5}{8}\) So \(P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}\)
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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29 Aug 2009, 14:50
I would like to make a recap of this problem, as in the prior posts there are 3 different answers for questions 3 and 4. Have 3 red and 7 blue. 1) Pick 2. P of picking at least 1 blue? Combinatorics: P(1b&1r)+P(2b) > (7C1 x 3C1 + 7C2)/10C2 Combinatorics shortcut: 1P(2r) > 1  3C2/10C2 Probability: P(BR)+P(RB)+P(BB) = 7/10*3/9+ 3/10*7/9+ 7/10*6/9 Answer: 14/15 2) Pick 2. P of picking 0 blue? Combinatorics: P(2r) > 3C2/10C2 Probability: P(RR) = 3/10*2/9 Answer: 1/15 3) Pick 3. P of picking at least 2 blue? Combinatorics: P(2b&1r)+P(3b) > (7C2 x 3C1 + 7C3)/10C3 Probability: P(BBR)+P(BRB)+P(RBB)+P(BBB) = 7/10*6/9*3/8+ 7/10*3/9*6/8+ 3/10*7/9*6/8+ 7/10*6/9*5/8 = 7/40+7/40+7/40+7/24 Answer: 49/60 (7/15 and 42/90 are wrong; do the math) 4) Pick 3. P of picking 2 blue? Combinatorics: P(2b&1r) > (7C2 x 3C1)/10C3 Probability: P(BBR)+P(BRB)+P(RBB) = 7/10*6/9*3/8+ 7/10*3/9*6/8+ 3/10*7/9*6/8 Answer: 21/40 (8/15 and 7/40 are wrong; do the math) Hope this may save new visitors time.
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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06 Dec 2007, 14:40
bmwhype2 wrote: 1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
1 [3!/(2!1!)]/[10!/(2!8!)]=14/15
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
[3!/(2!1!)]/[10!/(2!8!)]=1/15
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least two marbles are blue?
[7!/(2!5!)]/[10!/(2!8!)]=7/15
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that two marbles are blue?
[7!/(2!5!)]/[10!/(2!8!)]=7/15
Personally, I really don't know what is the difference between question 3 & 4.



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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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06 Dec 2007, 23:54
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
At least 1 is blue = (1b and 1r) + 2 blue
total = 10c2
prob = (7c1 x 3c1 + 7c2)/10c2 = 42/45 = 14/15
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
prob = 3c2/10c2 = 3/45 = 1/15
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
= (7c2 x 3c1 + 7c3)/10c3
= (63 + 35)/10c3
= 98/120
= 49/60
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
= (7c2 x 3c1 )/10c3
= (63)/10c3
= 63/120
= 21/40



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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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10 Jan 2008, 11:25
bmwhype2 wrote: 1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue? 1: at least one is blue: how bout 1none are blue? 3/10*2/9 > 6/90 11/15 = 14/15 2: 3/10*2/9 = 1/15 3: At least 2 are blue: 11 is blue none are blue 3/10*2/9*1/8 =6/720 3/10*2/9*7/8(3) b/c we have 3!/2! ways to arrange RRB 378/720 > 384/720 192/360 > 96/180> 48/90> 8/15 > 7/15 is our answer 4: 7/10*6/9*3/8 * (3) again b/c 3!/2! 378/720 = 8/15 hrmmm appears I am wrong on 3 and 4



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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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10 Jan 2008, 18:32
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
scenario 1: 1 red, 1 blue scenario 2: 2 blue
total possibility 10C2 (scenario 1 + scenario 2)/total possibilities = (3C1 x 7C1 +7C2)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
scenario 2/total possibilities
3c2/10c2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
3c1*7c2+7c3/10c3
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
7c2*3c1/10c3



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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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04 Oct 2008, 09:23
Thanks bmwhype2 for posting all these questions. And thanks walker for compiling them very well. I am going through your list right now...
My answer :
•1(all red) = 1(3/10*2/9) = 1(1/15) = 14/15 •All red = 3/10*2/9 = 1/15 •3 blue + 3(2 blue, 1red) = RRR+RRB+BRR+RBR = 7/24+7/40+7/40+7/40 = 98/120 = 49/60 •2blue,1 red = 7/10*6/9*3/8 = 7/40 (You can’t look for 3 Blue, because the questions asks for 2 Blue, so the 3rd one has to be red)



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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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06 Feb 2009, 13:08
djveed wrote: I'm not getting it. For A, I would assume the chance of a marble on the first pull is 7/10, and the chance on the second is 6/9. Multiply those two to get the chance that one is blue and you get a 7/15 chance that one of the two marbles pulled are blue. Which is different than what you all have.
And for B, chance of red is 3/10 on first and 2/9 on second, and when multiplied is 1/15 chance. What am I doing wrong? there are lot ways to do this problem.. You are trying to solve the problem using below method.. but missing some steps.. you considered only p(2B) but missed p(1B,1R)+ p(1R,1B) 1. At least one Blue Marble = = p(1B,1R)+ p(1R,1B)+[color=#0000FF]p(2B)[/color] = (7/10)*(3/9)+(3/10)*(7/9)+ (7/10)*(6/9) = 14/45 Another Way. = 7C1*3C1/10C2 + 7C2/10C2 = 21/45 + 21/45 = 14/45 best way = 1 P(2R) = 1 3C2/10C2 = 14/15. Is it clear now..
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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21 Jan 2010, 01:49
Bunuel wrote: Bullet wrote: Can any body please explain Question No.3 using probability
Why we need to add twice 7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15
Thanks and appreciated The solution you are posting for the third question is not right. Below is the solution of this question using the "probability". 3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?Probability of at least two marble are blue is the sum of the two probabilities: A. Two marbles are blue and one is red  BBR. This can occur in \(\frac{3!}{2!}=3\) # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. \(3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}\); B. All three marbles are blue  BBB. This can occur only one way, namely BBB. \(\frac{7}{10}*\frac{6}{9}*\frac{5}{8}\) So \(P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}\) Thanks Bunuel. I posted the solution from the thread as i was confused with the solution. So what you're saying is that we need to multiply the permutations with the probability. Having said that we will do only when we need to choose at least two marbles or any other thing from the whole? Thanks
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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21 Jan 2010, 02:56
Bullet wrote: Thanks Bunuel. I posted the solution from the thread as i was confused with the solution.
So what you're saying is that we need to multiply the permutations with the probability. Having said that we will do only when we need to choose at least two marbles or any other thing from the whole?
Thanks To make it simple: suppose we have the jar of 10 marbles  5 red, 2 blue and 3 green. If five marbles are selected at random, what is the probability that two will be red, one blue and two green? We are looking for all the cases with 2R, 1B and 2G. We can draw these marbles like: RRBGG or GGBRR or RBGGR ... So how many combinations of the drawing of these marbles are there? The answer is as many as there is permutation of the letters GGBRR, which is \(\frac{5!}{2!2!}\). Hence the answer for the above question would be \(\frac{5!}{2!2!}*\frac{5}{10}*\frac{4}{9}*\frac{2}{8}*\frac{3}{7}*\frac{2}{6}\). If the question were: three marbles are selected at random, what is the probability that all three will be red?RRR can occur only in one way: RRR, so the probability would be \(\frac{5}{10}*\frac{4}{9}*\frac{3}{8}\). You can check the Probability and Combination chapters in the Math Book (link below) for more. Also check my posts at: probabilitycoloredballs55253.html#p6375254redchipsand2bluechips85987.html#p644603probabilityqsattention88945.html#p671958pc88431.html?highlight=probability+of+occurring+eventprobability88069.html?highlight=probability+of+occurring+eventcombinationproblemprincentenreview2009bin4q287673.html?highlight=probability+of+occurring+event
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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22 Feb 2012, 13:11
powerka wrote: 3) Pick 3. P of picking at least 2 blue? Combinatorics: P(2b&1r)+P(3b) > (7C2 x 3C1 + 7C3)/10C3 Answer: 49/60 (7/15 and 42/90 are wrong; do the math)
#3. P(at least 2 blues) = P(exactly 2 blues) + P (3 blues) P (exactly 2 blues)= \(7C2/10C3\)=\(21/120\) P (3 blues) = \(7C3/10C3\)=\(35/120\) Therefore, P(at least 2 blues) = 21/120 + 35/120 = 56/120 = 7/15I see that the commented section above has an extra x 3C1 that's probably the cause for my discrepancy. Why do I need to multiply \(7C2\) by 3C1 to get P (exactly 2 blues)? Or, is my answer \(7/15\)correct?



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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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22 Feb 2012, 13:22
fxsunny wrote: powerka wrote: 3) Pick 3. P of picking at least 2 blue? Combinatorics: P(2b&1r)+P(3b) > (7C2 x 3C1 + 7C3)/10C3 Answer: 49/60 (7/15 and 42/90 are wrong; do the math)
#3. P(at least 2 blues) = P(exactly 2 blues) + P (3 blues) P (exactly 2 blues)= \(7C2/10C3\)=\(21/120\) P (3 blues) = \(7C3/10C3\)=\(35/120\) Therefore, P(at least 2 blues) = 21/120 + 35/120 = 56/120 = 7/15I see that the commented section above has an extra x 3C1 that's probably the cause for my discrepancy. Why do I need to multiply \(7C2\) by 3C1 to get P (exactly 2 blues)? Or, is my answer \(7/15\)correct? That's because you are picking 3 marbles and if you pick 2 blue then the third one must be red: BBR. Ways to pick one red marble out of 3 is \(C^1_3\). Complete solution using combinations: \(P(R\geq{2})=P(R=2)+P(R=3)=\frac{C^2_7*C^1_3}{C^3_{10}}+\frac{C^3_7}{C^3_{10}}\). Hope it's clear.
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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07 Mar 2012, 15:59
The below questions are discussed at the below thread, but i have a silly doubt so posting that here: abagof10marblescontains3redmarblesand7blue56728.htmlI understand the solution provided, but is there anything wrong with the simple method as below as i am getting different ans ? A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue? Case 1: Only 2 marbles are blue. Prob: 7/10 * 6/9 * 3/8 Case 2: All 3 marbles are blue. Prob: 7/10 * 6/9 * 5/8 Add 1 and 2 A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue? Only 2 marbles are blue. Prob: 7/10 * 6/9 * 3/8



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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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20 Apr 2014, 14:05
Bunuel wrote: We are looking for all the cases with 2R, 1B and 2G. We can draw these marbles like: RRBGG or GGBRR or RBGGR ... So how many combinations of the drawing of these marbles are there? The answer is as many as there is permutation of the letters GGBRR, which is \(\frac{5!}{2!2!}\).
RRR can occur only in one way: RRR, so the probability would be \frac{5}{10}*\frac{4}{9}*\frac{3}{8}.
Can someone please talk about this a little? I understand that the marbles can be arranged in various ways RBRGG etc. but how do we come up with the \(\frac{5!}{2!2!}\) equation? EDIT: To add to the question above, why can RRR only occur in one way, isn't that 3 separate ways as well? Meaning, R_1, R_2, R3  there are 3 different R's and therefore 3 different ways?



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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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20 Apr 2014, 23:44
russ9 wrote: Bunuel wrote: We are looking for all the cases with 2R, 1B and 2G. We can draw these marbles like: RRBGG or GGBRR or RBGGR ... So how many combinations of the drawing of these marbles are there? The answer is as many as there is permutation of the letters GGBRR, which is \(\frac{5!}{2!2!}\).
RRR can occur only in one way: RRR, so the probability would be \frac{5}{10}*\frac{4}{9}*\frac{3}{8}.
Can someone please talk about this a little? I understand that the marbles can be arranged in various ways RBRGG etc. but how do we come up with the \(\frac{5!}{2!2!}\) equation? EDIT: To add to the question above, why can RRR only occur in one way, isn't that 3 separate ways as well? Meaning, R_1, R_2, R3  there are 3 different R's and therefore 3 different ways? THEORY: Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is: \(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\). For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word. Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice. Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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21 Apr 2014, 00:48
Bunuel wrote: Bullet wrote: Can any body please explain Question No.3 using probability
Why we need to add twice 7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15
Thanks and appreciated The solution you are posting for the third question is not right. Below is the solution of this question using the "probability". 3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?Probability of at least two marble are blue is the sum of the two probabilities: A. Two marbles are blue and one is red  BBR. This can occur in \(\frac{3!}{2!}=3\) # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. \(3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}\); B. All three marbles are blue  BBB. This can occur only one way, namely BBB. \(\frac{7}{10}*\frac{6}{9}*\frac{5}{8}\) So \(P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}\) Bunuel, i just have 1 silly question, here why cant we do: 3!/2!*0.7^2*0.3 , i mean it is not specified in the question whether we replace the ball after selecting it i get confused in such situations, a lot !



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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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21 Apr 2014, 01:05
adymehta29 wrote: Bunuel wrote: Bullet wrote: Can any body please explain Question No.3 using probability
Why we need to add twice 7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15
Thanks and appreciated The solution you are posting for the third question is not right. Below is the solution of this question using the "probability". 3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?Probability of at least two marble are blue is the sum of the two probabilities: A. Two marbles are blue and one is red  BBR. This can occur in \(\frac{3!}{2!}=3\) # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. \(3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}\); B. All three marbles are blue  BBB. This can occur only one way, namely BBB. \(\frac{7}{10}*\frac{6}{9}*\frac{5}{8}\) So \(P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}\) Bunuel, i just have 1 silly question, here why cant we do: 3!/2!*0.7^2*0.3 , i mean it is not specified in the question whether we replace the ball after selecting it i get confused in such situations, a lot ! If it were with replacement it would be specified. Or let me put it this way: proper GMAT question would make it clear whether it's with or without replacement case.
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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29 Nov 2014, 23:43
In a similar question.
In a blue jar there are red, white and green balls. The probability of drawing a red ball is 1/5. The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10. What is the probability of drawing a white ball?
a) 1/5. b) ½. c) 1/3. d) 3/10. e) ¼.
the OA is 1/2..which is just 1/10=1/5*x,so x=1/2.. my question is that returning the ball does not figure into the question in anyway?




Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two
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