# Homework Help: Probability question- picking marbles from a bag

1. Aug 8, 2012

### Apollonian

Pick 5 marbles from 7 red, 5 blue and 6 green. What is the probability of picking 3 red and 3 blue?
I'm hopeless with probability questions so any help would be fantastic.

2. Aug 8, 2012

### Millennial

0, you are picking 5 marbles and you want 6: 3 red and 3 blue. It is impossible.

3. Aug 8, 2012

### AGNuke

Incorrect question. Corrections can be
1. Pick 6 marbles.
2. 3 red OR 3 blue

And probability is just the game of guessing all possibilities and desired possibilities. Also, knowledge of Permutation and Combination and Set Theory (They helped me so far) can help in Probability.

4. Aug 8, 2012

### Apollonian

My mistake, it was meant to be 3 red and 2 blue. Sorry for the misunderstanding.

5. Aug 8, 2012

### Millennial

Total number of marbles: 18
The probability of picking 3 red: $\dfrac{7\cdot 6\cdot 5}{18\cdot 17\cdot 16}$
Do you see how I got that? If you do, just do the same for the blues.

6. Aug 8, 2012

### Ray Vickson

Of course, that would give the probability of RRRBB in that order; the OP would still need to find the probabilities of getting 3 red and 2 blue in other orders, such as RBBRR, BRBRR, etc.

RGV

7. Aug 8, 2012

### operationsres

For 5 draws, the sample space cardinality is |Ω|=18!/(5!13!)=8568 (this is the number of possible combinations of 5 marbles from 18 in total).

You then have to find the cardinality of the constrained Ω, where each state of the world is a drawing of 3 red balls and 2 blue balls. You then divide this number by 8568 and this will be your solution.

8. Aug 9, 2012

### operationsres

Ok I've been working on this for a few hours and concluded that this methodology is incorrect.

I don't think the sample space can be calculated by using the simple application of a permutation or combination formula. Order doesn't matter in the sense that {b1,b2,g1,g2,g3} is the same as {b2,b1,g1,g2,g3} but it does matter in the sense that {b1,g1,g2,g3,b2} is not the same as {b1,b2,g1,g2,g3}, which makes the problem quite challenging.

you also can't simply divide the cardinalities of both sample spaces because the probability of selecting an element from one set of members of a certain color is different to doing the same with another.

9. Aug 9, 2012

### Ray Vickson

The binomial coefficient C(n,m) = n!/[m!*(n-m)!] plays *two* roles: (1) it is the number of distinct subsets of size m contained in a set of size n; and (2) it is the number of different strings of type ABABB...A that are of length n and have m 'A's and (n-m) 'B's.

To see the latter, let F(n,m) be that number of strings. Clearly, we have F(n,0) = F(n,n) =1. We also have F(n,1) = F(n,n-1) = n, because if we have just one 'A' it can occupy any of the n places. Now suppose 1 < m < n-1. If the first letter is 'A' we have to put a string of m-1 'A's and (n-m) 'B's after it, and the number of such strings is F(n-1,m-1). If the first letter is 'B' we have F(n-1,m) strings remaining. Altogether, we get F(n,m) = F(n-1,m) + F(n-1,m-1), which is just Pascal's triangle. Together with the boundary conditions, we get (by induction) that F(n,m) = C(n,m).

RGV