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A bag contains 10 marbles - 3 red, 2 green and 5 blue

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Senior Manager
Joined: 15 Aug 2004
Posts: 327
A bag contains 10 marbles - 3 red, 2 green and 5 blue [#permalink]

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27 Jul 2006, 01:25
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A bag contains 10 marbles - 3 red, 2 green and 5 blue marbles. What is the probability for all these conditions.

1. Choosing 2 red marbles.
2. Choosing 1 red, 1 green and 3 blue marbles.
3. Choosing 2 marbles such that the first one is green and the second is red.
4. Choosing 3 marbles such that the first one is green, the second is red and the third is blue.

Just trying to understand how to deal with this when order matters and when it doesn't.
SVP
Joined: 30 Mar 2006
Posts: 1728
Re: Probability Question - Marbles [#permalink]

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27 Jul 2006, 01:52
sumitsarkar82 wrote:
A bag contains 10 marbles - 3 red, 2 green and 5 blue marbles. What is the probability for all these conditions.

1. Choosing 2 red marbles.
2. Choosing 1 red, 1 green and 3 blue marbles.
3. Choosing 2 marbles such that the first one is green and the second is red.
4. Choosing 3 marbles such that the first one is green, the second is red and the third is blue.

Just trying to understand how to deal with this when order matters and when it doesn't.

1) Ways of choosing 2 marbles = 10C2 = 45
Ways of choosing two red marbles = 3C2 = 3
Probability = 3/45 = 1/15

2) Ways of Choosing 5 marbles = 10C5
ways of choosing 1 red 1 green and 3 blue = 3C1*2C1*5C3

Probabilty = 3C1*2C1*5C3/10C5 = 5/21

3) Probabiliy that first marble is green = 2/10
Probability that second marble is red = 3/9
Total = 2/10 * 3/9 = 1/15

4) Probabiliy that first marble is green = 2/10
Probability that second marble is red = 3/9
Probability that third marble is blue = 5/8
Total = 2/10 * 3/9 * 5/8 = 1/24
Senior Manager
Joined: 15 Aug 2004
Posts: 327
Re: Probability Question - Marbles [#permalink]

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27 Jul 2006, 02:03
jaynayak wrote:
sumitsarkar82 wrote:
A bag contains 10 marbles - 3 red, 2 green and 5 blue marbles. What is the probability for all these conditions.

1. Choosing 2 red marbles.
2. Choosing 1 red, 1 green and 3 blue marbles.
3. Choosing 2 marbles such that the first one is green and the second is red.
4. Choosing 3 marbles such that the first one is green, the second is red and the third is blue.

Just trying to understand how to deal with this when order matters and when it doesn't.

1) Ways of choosing 2 marbles = 10C2 = 45
Ways of choosing two red marbles = 3C2 = 3
Probability = 3/45 = 1/15

2) Ways of Choosing 5 marbles = 10C5
ways of choosing 1 red 1 green and 3 blue = 3C1*2C1*5C3

Probabilty = 3C1*2C1*5C3/10C5 = 5/21

3) Probabiliy that first marble is green = 2/10
Probability that second marble is red = 3/9
Total = 2/10 * 3/9 = 1/15

4) Probabiliy that first marble is green = 2/10
Probability that second marble is red = 3/9
Probability that third marble is blue = 5/8
Total = 2/10 * 3/9 * 5/8 = 1/24

Jaynayak,

I would like to know when to use this method - x/y * z/k....

and when to use the combination method... I am a bit confused with this...

I learned that x/y * z/k ... should only be used when there is a similar group but you used them in Part 3 & 4...
Manager
Joined: 14 Mar 2006
Posts: 208

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27 Jul 2006, 08:34
same here, I also get confused about when to use which method. I am getting same answer as jay for 1,3, and 4 but not 4. A detailed explanation would be much appreciated.

Is it wrong to use this method for 2?

3/10*2/9*5/8*4/7*3/6=360/30240

Thanks
CEO
Joined: 20 Nov 2005
Posts: 2894
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

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27 Jul 2006, 08:58
Just my 2 cents:

When order matters then use x/y , a/b method.

When order doesn't matter then use combinations.

I think this is a very subjective thing. You will learn by practice.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Intern
Joined: 17 Jul 2006
Posts: 26
Re: Probability Question - Marbles [#permalink]

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27 Jul 2006, 09:26
sumitsarkar82 wrote:
A bag contains 10 marbles - 3 red, 2 green and 5 blue marbles. What is the probability for all these conditions.

1. Choosing 2 red marbles.
2. Choosing 1 red, 1 green and 3 blue marbles.
3. Choosing 2 marbles such that the first one is green and the second is red.
4. Choosing 3 marbles such that the first one is green, the second is red and the third is blue.

Just trying to understand how to deal with this when order matters and when it doesn't.

Note the difference between the following:

When the question says 1 red and 1 green marble, the order is not important. This includes the two possibilities RG and GR, meaning the two chioces where first one is Red, second one is Green and first one is Green, second one is Red.

The problems where order is not important can actually be solved in one of the two ways (Combinations or Multiplications) as explained below:

for example, take question 1 above, what is the probability of getting 2 Red marbles:

problem 1:

Method I:

Total # of outcomes = Total # of different ways 2 marbles can be chosen from 10 marbles = 10C2 = (10*9)/(1*2) = 45

Total # of ways 2 Red marbles can be chosen from 3 red marbles = 3C2 = 3

The probabilty of getting 2 red marbles = 3/45

Method II:

To get 2 red marbles, the marbles that are pulled out from the bag should come out in the order RR (any other order is not permitted).

Hence, the problem boils down to two events where

event 1 - picking 1 red marble out of 10 marbles in which 3 are red
event 2 - picking 1 red marble out of 9 marbles in which 2 are red (note that event two will have only a total of 9 marbles; out of which ONLY 2 are red having already pulled a red one out in event 1).

So, the probability of event 1 = 3/10
probability of event 2 = 2/9

probability of the combined event (1 & 2) = (3/10) * (2/9) = 3/45

Problem 2:

Method I:

choosing 1 red, 1 green and 3 blue out of 10 marbles - 3 red, 2 green and 5 blue marbles

So, again the order is not important.

total # of outcomes = 10C5 = (10*9*8*7*6)/(1*2*3*4*5) = 252

# of selections of 1 red out 3 red marbles = 3C1 = 3
# of selections of 1 green out of 2 green = 2C1 = 2
# of selections of 3 blue out of 5 blue = 5C3 = (5*4)/(1*2) = 10

So, the combined probability = (3*2*10)/(252) = 5/21

Method II:

Note that 1 red, 1 green and 3 blue marbles can come out in any order.

For example, selecting RGBBB, GRBBB, BBRGB, BBRBG, ... etc will all satisfy the condition that you will have 1 red, 1 green, 3 blue marbles at the end.

So, if you want to apply this method to solve the problem, you would have to add the individual probabilities p(RGBBB) + p(GRBBB)+p(BBRGB)+...(all these combinations that would result in 1 red, 1 green and 3blue marbles).

Even though you can take this approach to solve the problem, you notice that the problem has become cumbersome to solve this way.

This is given to show that the problem can be solved either way.

problem 3:

Choosing 2 marbles such that the first one is green and the second is red.

Method I:

There are two events - first one which should yield green and second one which should yield red. You are looking for GR.

Note that getting a Red and then a Green does not satisfy this condition.

Selecting a red and a green includes both these choices "GR" and "RG".

So, the probability of a "GR" is - 2C1/10C1 * 2C1/9C1 = (2*2)/(10*9) = 2/45.

Method II:

In this case, the second method virtually becomes same as method I.

So, the probability = 2/10 * 2/9 = 2/45.

Problem 4

Choosing 3 marbles such that the first one is green, the second is red and the third is blue.

Method I: probability = 2C1/10C1 * 2C1/9C1 * 5C1/8C1 = 2*2*5/(10*9*8) = 1/36

Method II: virtually same = 2/10* 2/9 * 5/8 = 1/36

-mathguru
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Manager
Joined: 14 Mar 2006
Posts: 208

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27 Jul 2006, 12:49
Thanks guys. Makes much more sense now. Seems like the combination method is usually more easier and conveniant to use.
Manager
Joined: 26 Jun 2006
Posts: 152

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27 Jul 2006, 21:07
Just a note. You should clearly state (or so that it can be inferred) if the selection of marbles is WITH or WITHOUT replacement. The calculations are totally different for these 2 cases
27 Jul 2006, 21:07
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