Probability of Assigning r Balls in n Urns with m Urns Containing k Balls

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In summary: Your Name]In summary, the conversation discussed a problem of randomly assigning r balls into n urns and finding the probability of exactly m urns containing exactly k balls each. The solution involved using combinations and the stars and bars method to find the total number of ways of distributing the balls into the urns and then calculating the desired probability.
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Yankel
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Hello all,

I am trying to solve this problem:

r balls are randomly assigned into n urns. The assignment is random and the balls are cannot be distinguished. What is the probability that exactly m urns will contain exactly k balls each ?

I know that the probability of each ball to be in each urn is 1/n.
I addition I have (nCm) ways to choose which are the urns to be filled with k balls.
Then I have r−km balls to distribute in the n−m remaining urns.

I do not know how to proceed. Can anyone solve this difficult problem ? :confused:

My current attempt is:

\[\frac{\binom{n}{m}\binom{r-km+n-m-1}{r-km}}{\binom{r+n-1}{r}}\]

It must be wrong...
 
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Hello,

Thank you for bringing this interesting problem to our attention. I will try my best to provide a solution to this problem.

To find the probability that exactly m urns will contain exactly k balls each, we can approach this problem using combinations. Let's break down the problem into smaller parts:

1. Probability of selecting m urns out of n urns: We can do this in \(\binom{n}{m}\) ways.

2. Probability of placing k balls in each of the selected m urns: Since the assignment is random and the balls are indistinguishable, we can think of this as placing k identical balls in m distinct urns. This can be done in \(\binom{r}{k}\) ways.

3. Probability of placing the remaining r-km balls in the remaining n-m urns: We can think of this as placing r-km identical balls in n-m distinct urns. This can be done in \(\binom{r-km+n-m}{r-km}\) ways.

Now, to find the total number of ways of distributing r balls into n urns, we can use the stars and bars method. This method states that if we have r identical balls and n distinct urns, the number of ways to distribute the balls into the urns is \(\binom{r+n-1}{r}\).

Therefore, the probability of exactly m urns containing exactly k balls each can be calculated as:
\[\frac{\binom{n}{m}\binom{r}{k}\binom{r-km+n-m}{r-km}}{\binom{r+n-1}{r}}\]

I hope this helps in solving the problem. Let me know if you have any further questions or if you would like me to explain any step in more detail.
 

1. What is the formula for calculating the probability of assigning r balls in n urns with m urns containing k balls?

The formula for calculating this probability is (mCk * (n-m)Cr) / (nCr), where mCk represents the combination of m urns containing k balls and (n-m)Cr represents the combination of the remaining urns containing the remaining balls.

2. How is this formula derived?

This formula is derived from the concept of combinations, where the total number of ways to choose r balls from n urns is represented as nCr. The probability of choosing k balls from m urns is represented as mCk. The remaining balls and urns are represented by (n-m) and (n-m)Cr, respectively. By dividing the number of ways to choose k balls from m urns and the number of ways to choose the remaining balls and urns by the total number of ways to choose r balls from n urns, we get the probability of assigning r balls in n urns with m urns containing k balls.

3. Can this formula be used for any number of balls and urns?

Yes, this formula can be used for any number of balls and urns as long as the number of balls is equal to or greater than the number of urns, and the number of balls in each urn is equal to or greater than the number of balls being assigned.

4. How does the number of balls and urns affect the probability?

The probability is affected by the number of balls and urns in the sense that as the number of balls and urns increase, the probability of assigning r balls in n urns with m urns containing k balls decreases. This is because as the number of balls and urns increase, there are more possible combinations and the likelihood of choosing a specific combination decreases.

5. Can this formula be applied to real-life situations?

Yes, this formula can be applied to real-life situations such as in genetics, where it can be used to calculate the probability of certain traits being inherited from parents to offspring. It can also be used in probability and statistics to calculate the likelihood of certain events occurring based on a given set of conditions.

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