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A bag of 10 marbles contains 3 red marbles and 7 blue

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Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]  07 Mar 2012, 23:02
Owsum Bunuel ... Thanks !
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Re: Combinatorics - at least, none .... [#permalink]  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}.

You can check the Probability and Combination chapters in the Math Book (link below) for more.
Also check my posts at:
probability-colored-balls-55253.html#p637525
4-red-chips-and-2-blue-chips-85987.html#p644603
probability-qs-attention-88945.html#p671958
p-c-88431.html?highlight=probability+of+occurring+event
probability-88069.html?highlight=probability+of+occurring+event
combination-problem-princenten-review-2009-bin-4-q2-87673.html?highlight=probability+of+occurring+event

That was an awesome explanation. Thank you!
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]  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?

Just wondering
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J
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]  09 Feb 2014, 00:32
Expert's post
jlgdr 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?

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.
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Re: Combinatorics - at least, none .... [#permalink]  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: Combinatorics - at least, none .... [#permalink]  20 Apr 2014, 23:44
Expert's post
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: Combinatorics - at least, none .... [#permalink]  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: Combinatorics - at least, none .... [#permalink]  21 Apr 2014, 01:05
Expert's post
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 [#permalink]  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] 21 Apr 2014, 17:21
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