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# probability math(combinations (nCk)) help needed

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probability math(combinations (nCk)) help needed [#permalink]

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18 Jan 2010, 15:39
I think i have some understanding of the term nCk. It's the number of ways you can select a k-set from a n-set, if i am correct?

One of the two math problems im having difficulty with right now is:
"Find the probability of getting 3 green marbles and 2 red marbles, if five marbles are picked from a bag containing: 6 green marbles and 4 red ones"

I have the answer right in front of me, (6C4 * 4C2)/10C5.
The thing i don't understand is why 6C4 and 4C2 are used.

It might be fairly simple, but for some reason my brain cant comprehend it.

The second math problem is "A jar holds 4 red, 3 green and 6 white marbles. How many different ways can you pick 6 marbles so that you have at least one of each color" I can solve this using the long ways
"4C4 * 3C1 * 6C1 + 4C3*3C2*6C2 + 4C2*3C3*6C1 + 4C2*3C2 * 6C2 + 4C2 * 3C1 * 6C3 + 4C1 * 3C3 * 6C2 + 4C1*3C2*6C3 + 4C1 * 3C1 * 6C4 = 1416" and the one with "13C6 - all the instances where one color is absent"
But is there another shorter way to solve this?

Btw. sorry if my mathematical English sucks, I'm from Iceland, going to the final exam in probability math tomorrow, and I was scourging the internet to find explanations for the problems I'm currently not getting, and then i found you, seemingly a utopia of math discussion
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Re: probability math(combinations (nCk)) help needed [#permalink]

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18 Jan 2010, 16:29
Check these posts:
- Probability
- Combinatorics

succybuzz wrote:
"Find the probability of getting 3 green marbles and 2 red marbles, if five marbles are picked from a bag containing: 6 green marbles and 4 red ones"

I have the answer right in front of me, (6C4 * 4C2)/10C5.
The thing i don't understand is why 6C4 and 4C2 are used.

Maybe (6C3 * 4C2)/10C5

6C3 - the number of options of getting 3 green marbles out of 6.
4C2 - the number of options of getting 2 red marbles out of 4.
10C5 - the total number of options of getting 5 marbles out of 10.
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Re: probability math(combinations (nCk)) help needed [#permalink]

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17 Mar 2010, 13:05
I did it in another way:
6/10*5/9*4/8*4/7*3/6 = 1/42

Is it correct?
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Re: probability math(combinations (nCk)) help needed [#permalink]

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17 Mar 2010, 13:32
fruit wrote:
I did it in another way:
6/10*5/9*4/8*4/7*3/6 = 1/42

Is it correct?

First of all 6/10*5/9*4/8*4/7*3/6 equals to 1/21.

But the above is still incorrect. We need 3 green and 2 red marbles. But GGGRR can happen in $$\frac{5!}{3!2!}=10$$ ways: GGGRR, GRRGG, RRGGG, ... (Basically the number of permutations of 5 letters out of which there are 3 identical G-s and 2 identical R-s).

So we have to multiply $$\frac{1}{21}$$ by $$\frac{5!}{3!2!}=10$$. Answer: $$\frac{10}{21}$$.
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Re: probability math(combinations (nCk)) help needed [#permalink]

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12 Apr 2010, 08:22
Find the probability of getting 3 green marbles, 2 yellow and 4 red marbles, if nine marbles are picked from a bag containing: 5 green marbles, 4 yellow and 5 red.

Answer:
150/1001 ?
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Re: probability math(combinations (nCk)) help needed [#permalink]

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12 Apr 2010, 09:04
fruit wrote:
Find the probability of getting 3 green marbles, 2 yellow and 4 red marbles, if nine marbles are picked from a bag containing: 5 green marbles, 4 yellow and 5 red.

Answer:
150/1001 ?

$$\frac{C^3_5*C^2_4*C^4_5}{C^9_{14}}=\frac{150}{1001}$$

OR: $$(\frac{9!}{3!2!4!})*(\frac{5}{14}*\frac{4}{13}*\frac{3}{12})*(\frac{4}{11}*\frac{3}{10})*(\frac{5}{9}*\frac{4}{8}*\frac{3}{7}*\frac{2}{6})=\frac{150}{1001}$$
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Re: probability math(combinations (nCk)) help needed   [#permalink] 12 Apr 2010, 09:04
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# probability math(combinations (nCk)) help needed

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