Probability question- picking marbles from a bag

  • Thread starter Apollonian
  • Start date
  • Tags
    Probability
In summary: 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
  • #1
Apollonian
9
0
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.
 
Physics news on Phys.org
  • #2
0, you are picking 5 marbles and you want 6: 3 red and 3 blue. It is impossible.
 
  • #3
Apollonian said:
Pick 5 marbles ... probability of picking 3 red and 3 blue?

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
My mistake, it was meant to be 3 red and 2 blue. Sorry for the misunderstanding.
 
  • #5
Total number of marbles: 18
The probability of picking 3 red: [itex]\dfrac{7\cdot 6\cdot 5}{18\cdot 17\cdot 16}[/itex]
Do you see how I got that? If you do, just do the same for the blues.
 
  • #6
Millennial said:
Total number of marbles: 18
The probability of picking 3 red: [itex]\dfrac{7\cdot 6\cdot 5}{18\cdot 17\cdot 16}[/itex]
Do you see how I got that? If you do, just do the same for the blues.

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
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
operationsres said:
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.

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
operationsres said:
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.

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
 

1. How do you calculate the probability of picking a specific marble from a bag?

The probability of picking a specific marble from a bag can be calculated by dividing the total number of desired outcomes (picking the specific marble) by the total number of possible outcomes (picking any marble from the bag).

2. What is the difference between theoretical probability and experimental probability?

Theoretical probability is based on mathematical calculations and assumes that all outcomes are equally likely. Experimental probability is based on actual experiments or observations and may vary from the theoretical probability due to chance or other factors.

3. How does the number of marbles in the bag affect the probability of picking a specific marble?

The number of marbles in the bag does not affect the probability of picking a specific marble, as long as the number of desired outcomes and total number of possible outcomes remain the same. However, as the number of marbles in the bag increases, the probability of picking a specific marble decreases.

4. Can the probability of picking a specific marble change if marbles are removed or added to the bag?

Yes, the probability of picking a specific marble can change if marbles are removed or added to the bag. This is because the total number of possible outcomes and desired outcomes may change, thus altering the probability calculation.

5. How can probability be used in real life situations?

Probability is used in real life situations to make predictions and informed decisions. For example, in gambling, understanding probability can help determine the likelihood of winning a game. In insurance, probability can be used to calculate the likelihood of an event occurring and the associated risk. In science, probability is used to analyze experimental data and make conclusions about the likelihood of certain outcomes.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
11
Views
2K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
619
  • Precalculus Mathematics Homework Help
Replies
2
Views
871
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
8
Views
1K
  • Precalculus Mathematics Homework Help
Replies
8
Views
6K
  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
  • Precalculus Mathematics Homework Help
Replies
1
Views
857
  • Precalculus Mathematics Homework Help
Replies
11
Views
3K
Back
Top