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

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01 Apr 2013, 08:51

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 \(\frac{5}{10}*\frac{4}{9}*\frac{3}{8}\).

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

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08 Feb 2014, 13:21

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?

In number 4 we need to specify that probability of EXACTLY two marbles are blue. Otherwise it would be the same as 3 no?

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?

In number 4 we need to specify that probability of EXACTLY two marbles are blue. Otherwise it would be the same as 3 no?

Just wondering Cheers J

No. If it meant at least two, then it would be written that way.
_________________

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

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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?

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 \(\frac{9!}{4!3!2!}\).
_________________

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

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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 !

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.
_________________

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

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21 Apr 2014, 17:21

Thanks Bunuel, very clear!

That leads me to my second question -- sorry it's a little long winded.

When we solve this using probability, let's take question 3 for example(10 Marbles, 3R, 7B, if 3 are selected what is the prob that at least 2 will be blue). Following that, if I use probability, it is clear that the solution is P(BBR)+P(BRB)+P(RBB)+P(BBB).

On the other hand, if we use combinatorics, according to the solutions posted above, why aren't permutations taken into account. What I mean by that is, the correct solution is (7C2 x 3C1 + 7C3)/10C3 which implies that it's BBR OR BBB. Why are we ignoring BRB and RBB -- shouldn't it be ((7C2 x 3C1)3) + 7C3)/10C3?

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

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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?

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?

The probability of drawing a red ball is 1/5: P = (red)/(total) = 1/5; The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10: (red)/(total)*(white)/(total) = 1/10.

Since (red)/(total) = 1/5, then 1/5*(white)/(total) = 1/10 --> (white)/(total) = 1/2.

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

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12 Mar 2015, 08:04

nevergiveup wrote:

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.

Having trouble with this. For the first question the way that I am attempting to reason through it is 10!/2!8! . 2! represents the 2 marbles selected and 8! is for the other 8 NOT selected. I then have probability first one is not blue ,3/10, times this and answer subtracted from 1. So 1- [10!/2!/8!(3/10)]

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.

Having trouble with this. For the first question the way that I am attempting to reason through it is 10!/2!8! . 2! represents the 2 marbles selected and 8! is for the other 8 NOT selected. I then have probability first one is not blue ,3/10, times this and answer subtracted from 1. So 1- [10!/2!/8!(3/10)]

hi GMAT01, you do not require the coloured portion above... wherever the questions of this form are there , the best is to take probability none is there and it seems you begun that way but went wrong mid way.. prob of both not being blue means both are red.. prob of first red=3/10 and second red=2/9..... prob of both red 3/10*2/9=6/90.. so prob of atleast one blue=1-6/90=84/90... hope its clear
_________________

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

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12 Mar 2015, 09:34

Chetan2u thank you for clarifying - I get it now. I was just confusing myself trying to approach this through combinatorics. I just made the problem more complex for me to understand instead of sticking to a simple method.

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 \(7/15\)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.

Hello, sir. Could you you please expain why we should assume that 7 blue marbles are not identical? As I understand, when we deal with ditinct objects, say 7 distinct people A, B, C, D, E, F, and want to choose 2 of them, we find it by 7C2= 7!/2!5!. But how can be 7 blue marbles be distinct? shouldn't their 2-place combination equal to 1?

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

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24 Oct 2016, 01:59

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A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

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14 Nov 2016, 09:50

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? For above question, can you show me how to solve by 1- p(all blue) as I am getting different answer by doing 1- (7/10*6/9) = 8/15

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