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A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]
06 Dec 2007, 14:24

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Difficulty:

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Question Stats:

43% (03:05) correct
56% (01:20) wrong based on 23 sessions

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?

Re: Combinatorics - at least, none .... [#permalink]
06 Dec 2007, 14:40

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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.

Re: Combinatorics - at least, none .... [#permalink]
06 Dec 2007, 23:54

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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?

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?

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?

Re: Combinatorics - at least, none .... [#permalink]
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 1-none are blue? 3/10*2/9 --> 6/90 1-1/15 = 14/15 2: 3/10*2/9 = 1/15 3: At least 2 are blue: 1-1 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

Re: Combinatorics - at least, none .... [#permalink]
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?

I also got question 3 and 4 wrong on the first try.

For 3, I did (7C2 + 7C3) / 10C3 = 7/15 b/c number of ways of picking 2 blue (7C2) plus number of ways of picking three blue (7C3) divided by the total number of ways of picking a group of three out of ten possible (10C3).

However, it seems as though 7C2 is not equal to the number of ways of picking 2 blue, because you are ignoring all of the ways to pick one Red marble (which must happen if you pick two blue). this can be represented as...

7C2 X 3C1

So, if you use the above as the number of ways to pick two blue marbles, you get exactly what GMAT TIGER got.

= (7c2 x 3c1 + 7c3)/10c3 = (63 + 35)/10c3 = 98/120 = 49/60 which is correct.

does this make sense? I am going through Walker's excellent list of comb.and prob. problems, and am really trying to understand the theory.

Re: Combinatorics - at least, none .... [#permalink]
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)

Re: Combinatorics - at least, none .... [#permalink]
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

Re: Combinatorics - at least, none .... [#permalink]
06 Feb 2009, 14:30

since I need practice with these types of questions, I attempted nevertheless.

1. probability that no blue is picked= 3C2/10C2 = 3x2/(10x9) = 1/15 Hence, atleast one blue is picked = 14/15 2. clearly 1/15 3. atleast two blue=> 2 out of 3 blue or all 3 blue. Thus, (7c2x3c1+7c3)/10c3 = (7x6x3+7x6x5)/10x9x8 = 42/90 4. two are blue=> 7c2x3c1/10c3 = 7x6x3/(10x9x8) = 7/40.

dominion 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?

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?

Re: Combinatorics - at least, none .... [#permalink]
29 Aug 2009, 14:50

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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: 1-P(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.
_________________

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?

Soln: (7C2 + 7C1*3C1)/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?

Soln: 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?

Soln: (7C2*3C1 + 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?

Re: Combinatorics - at least, none .... [#permalink]
21 Jan 2010, 00:35

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Expert's post

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} _________________

Re: Combinatorics - at least, none .... [#permalink]
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?

Re: Combinatorics - at least, none .... [#permalink]
21 Jan 2010, 02:56

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Expert's post

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}.

Re: Combinatorics - at least, none .... [#permalink]
21 Jan 2010, 04:31

Bunuel wrote:

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}.

I 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 7/15correct?

I 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 7/15correct?

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}}.

I 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

Re: 3 red marbles and 7 blue marbles [#permalink]
07 Mar 2012, 16:10

Expert's post

rohitgoel15 wrote:

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

The point is that BBR case can occur in 3 different ways: BBR, BRB, and RBB. So you should multiply 7/10 * 6/9 * 3/8 by 3.