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5 blue marbles, 3 red marbles and 4 purple marbles are [#permalink]

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16 Oct 2006, 10:42

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5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn without replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?

5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn WITHOUT replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?

Given: 5B+3R+4P=12 marbles. Question: what is the probability of NOT having scenario BBPP.

Calculate the probability of an opposite event and subtract this value from 1.

Opposite event is that out of 4 marbles we will have 2 blue and 2 purple marbles (scenario BBPP): \(P(opposite \ event)=\frac{4!}{2!2!}*\frac{5}{12}*\frac{4}{11}*\frac{4}{10}*\frac{3}{9}=\frac{4}{33}\), we are multiplying by \(\frac{4!}{2!2!}\) as scenario BBPP can occur in several ways: BBPP, PPBB, BPBP, ... So scenario BBPP can occur in \(\frac{4!}{2!2!}\) # of ways (# of permuations of 4 letters BBPP out of which 2 B's and 2 P's are identical);

Or with combinatorics \(P(opposte \ event)=\frac{C^2_5*C^2_4}{C^4_{12}}=\frac{4}{33}\).

So \(P=1-\frac{4}{33}=\frac{29}{33}\).

5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn WITH replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?

Again Calculate the probability of an opposite event and subtract this value from 1.

Opposite event is that out of 4 marbles we will have 2 blue and 2 purple marbles: \(P(opposite \ event)=\frac{4!}{2!2!}*\frac{5}{12}*\frac{5}{12}*\frac{4}{12}*\frac{4}{12}=\frac{25}{216}\), the same reason to multiply by \(\frac{4!}{2!2!}\);

Yes excellent. With or without replacement gives same answer. Can you give me an example where The results are different?

Also while in the topic of probability can you answer the following please, my guesses are mentioned

Distribute 10 different rocks among 15 people ? 10^15 Distribute 10 identical rocks among 15 ? Not sure Distribute 15 different rocks among 10 ? 15^10 Distribute 15 identical rocks among 10? 15+9 C 9 thanks

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Distribute 10 different rocks among 15 people? 10^15 Distribute 10 identical rocks among 15? (10+15-1)C(15-1)=24C14 Distribute 15 different rocks among 10? 15^10 Distribute 15 identical rocks among 10? (15+10-1)C(10-1)=24C9

Direct formulas for identical objects: The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r-1C_{r-1}\).

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n-1C_{r-1}\).

I doubt you'll see division of identical object involving such a big numbers on GMAT or division of different objects when each can get at least n objects (the question you've asked before) involving such a big numbers. With smaller numbers you can solve with different approaches depending on problem. So I would not worry about this questions too much. Better to concentrate on other areas.
_________________

Excellent Bunuel. Now if you can clarify one more point. Same cases as 1 and 2 above, but in this case the question is - we pull balls 1by1 , what is the probability of seeing a blue one in the 9th draw (i) with replacement - i think this is easy 5/12 (ii) without replacement - i believe the answer here is also 5/12. My question how?

A jar contains 5 blue balls and 7 black balls. One by one, every ball is selected at random without replacement. What is the probability that the 9th ball selected is blue?

The probability of drawing blue ball is 5/12 and it will not change for ANY successive drawing: second, third, fourth...

--> WHY??

If I draw 1 blue ball only 4 will be left (no replacement...) - next prob. of drawing blue should be 4/11 ??

I think you misinterpreted the question (or I made it ambiguous). The question is: what is the probability that the 9th ball will be blue and it's meant that we don't know the results of first 8 draws (or drawing has not started yet and we are just curious what is the probability that 9th ball will be blue). If I pick one by one and every time I know the result then of course the probability of drawing of a blue ball will change depending on results of previous drawings.

Consider the following: these 12 balls were put in a line randomly. Now, what is the probability that the 9th ball will be blue? ANY ball in a line has equal chances to be blue and ANY ball has equal chances to be black. Why would the probability of say 4th and 7th balls of being blue be different? For any ball in a line p=5/12 to be blue and for any ball p=7/12 to be black: for any ball in a line the probability of being blue plus the probability of being black must be 1, as there are only black and blue balls in a line. This example is basically the same as our original question, so answer for original question is also 5/12.

Follow the links in my previous post for similar problems.

Its without replacement, and above u have done the equivalent of with replacement. E.g whats the probability of picking 2 blue marbles out of a box containing 5 blue, 2 red, 3 black without replacement?

blue blue purple purple: 5/12*4/11*4/10*3/9
purple purple blue blue: 4/12*3/11*5/10*4/10
purple blue purple blue: 4/12*5/11*3/10*5/9
blue purple blue purple: 5/12*4/11*4/10*3/9

1-sum of all those values... must be a shortcut though

Re: PS : Probability : GMAT made me lose my marbles [#permalink]

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11 Aug 2010, 06:51

I'm thinking that with replacement you would have a total number of possible draws of 12^4 (could choose any of the 12 with any single draw). I think the numerator would look like 5*5*4*4 (5 ways to choose the first blue marble times 5 ways to choose the second blue marble times 4 ways to choose the first purple marble, etc.). Hence we would ultimately have

1 - 25*16/12^4 = 98% chance of not having this draw

Howver, after reading this post probability-colored-balls-55253.html#p637525 I'm confused. Can anyone chime in an clear up the correct way to compute the probability with replacement?

Re: PS : Probability : GMAT made me lose my marbles [#permalink]

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11 Aug 2010, 08:37

Excellent Bunuel. Now if you can clarify one more point. Same cases as 1 and 2 above, but in this case the question is - we pull balls 1by1 , what is the probability of seeing a blue one in the 9th draw (i) with replacement - i think this is easy 5/12 (ii) without replacement - i believe the answer here is also 5/12. My question how?
_________________

Excellent Bunuel. Now if you can clarify one more point. Same cases as 1 and 2 above, but in this case the question is - we pull balls 1by1 , what is the probability of seeing a blue one in the 9th draw (i) with replacement - i think this is easy 5/12 (ii) without replacement - i believe the answer here is also 5/12. My question how?

A jar contains 5 blue balls and 7 black balls. One by one, every ball is selected at random without replacement. What is the probability that the 9th ball selected is blue?

The probability of drawing blue ball is 5/12 and it will not change for ANY successive drawing: second, third, fourth...

Obviously for case with replacement the answer would be the same - every time we are drawing from 12 balls out of which 5 are blue, so every time probability of drawing blue ball is 5/12.

Re: PS : Probability : GMAT made me lose my marbles [#permalink]

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11 Aug 2010, 13:28

Yes excellent. With or without replacement gives same answer. Can you give me an example where The results are different?

Also while in the topic of probability can you answer the following please, my guesses are mentioned

Distribute 10 different rocks among 15 people ? 10^15 Distribute 10 identical rocks among 15 ? Not sure Distribute 15 different rocks among 10 ? 15^10 Distribute 15 identical rocks among 10? 15+9 C 9 thanks

Re: PS : Probability : GMAT made me lose my marbles [#permalink]

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12 Aug 2010, 03:40

Bunuel wrote:

mainhoon wrote:

Excellent Bunuel. Now if you can clarify one more point. Same cases as 1 and 2 above, but in this case the question is - we pull balls 1by1 , what is the probability of seeing a blue one in the 9th draw (i) with replacement - i think this is easy 5/12 (ii) without replacement - i believe the answer here is also 5/12. My question how?

A jar contains 5 blue balls and 7 black balls. One by one, every ball is selected at random without replacement. What is the probability that the 9th ball selected is blue?

The probability of drawing blue ball is 5/12 and it will not change for ANY successive drawing: second, third, fourth...

--> WHY??

If I draw 1 blue ball only 4 will be left (no replacement...) - next prob. of drawing blue should be 4/11 ??

Re: PS : Probability : GMAT made me lose my marbles [#permalink]

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13 Aug 2010, 04:44

Thanks Bunuel. I just chose the numbers at. Random, I see GMAT use smaller numbers, and then some start existing options, but the formula is more powerful. Also was curious to know how identical vs different changed the answers.

I wouldn't have known the replacement vs nonreplacement giving the same probability, thanks for the explanation to the prioranswer.

Re: PS : Probability : GMAT made me lose my marbles [#permalink]

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04 Jun 2012, 07:19

Bunuel wrote:

5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn WITHOUT replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?

Given: 5B+3R+4P=12 marbles. Question: what is the probability of NOT having scenario BBPP.

Calculate the probability of an opposite event and subtract this value from 1.

Opposite event is that out of 4 marbles we will have 2 blue and 2 purple marbles (scenario BBPP): \(P(opposite \ event)=\frac{4!}{2!2!}*\frac{5}{12}*\frac{4}{11}*\frac{4}{10}*\frac{3}{9}=\frac{4}{33}\), we are multiplying by \(\frac{4!}{2!2!}\) as scenario BBPP can occur in several ways: BBPP, PPBB, BPBP, ... So scenario BBPP can occur in \(\frac{4!}{2!2!}\) # of ways (# of permuations of 4 letters BBPP out of which 2 B's and 2 P's are identical);

Or with combinatorics \(P(opposte \ event)=\frac{C^2_5*C^2_4}{C^4_{12}}=\frac{4}{33}\).

So \(P=1-\frac{4}{33}=\frac{29}{33}\).

5 blue marbles, 3 red marbles and 4 purple marbles are placed in a bag. If 4 marbles are drawn WITH replacement, what is the probability that the result will not be 2 blue and 2 purple marbles?

Again Calculate the probability of an opposite event and subtract this value from 1.

Opposite event is that out of 4 marbles we will have 2 blue and 2 purple marbles: \(P(opposite \ event)=\frac{4!}{2!2!}*\frac{5}{12}*\frac{5}{12}*\frac{4}{12}*\frac{4}{12}=\frac{25}{216}\), the same reason to multiply by \(\frac{4!}{2!2!}\);

Is there a way to solve WITH replacement cases using combinatorics? I notice that all WITH replacement cases have only been solved using only one method.

Re: 5 blue marbles, 3 red marbles and 4 purple marbles are [#permalink]

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17 Feb 2014, 14:14

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