Question

We are to distribute 10 colored balls into five containers. Do the following:

1. Compute the probability of having at least one ball in each bucket if the balls are all different and the buckets are all labeled differently.

2. Compute the probability of having at least one ball in each bucket if the balls are all different and the buckets are all gray.

3. Compute the probability of having at least one ball in each bucket if the balls are gray and the buckets are all labeled differently.

4. Which one gives the most likelihood?

Answer 1

1. We have 10 different balls and 5 buckets.
   Using the definition of permutation, the balls can be arranged in ¹⁰P₅ ways = 30240 ways.------ (i)
   Each of the bucket can be chosen in ⁵C₁ ways=5 ways. ----(ii)
   Therefore, all possible cases in this setup is ¹⁰P₅ * ⁵C₁ = 151200.
   Now, each ball can be placed in 5 ways (under unrestricted setup).
   Therefore, 10 balls can be placed in 5¹⁰ ways = 9765625.
   Therefore, the required probability is : ¹⁰P₅ * ⁵C₁ / 5¹⁰ = 0.01548.
   
   
2. Now, since all the buckets are same, (ii) will be neglected.
   Therefore, all possible cases in this setup is ¹⁰P₅= 30240.
   Therefore, the required probability is : ¹⁰P₅ / 5¹⁰ = 0.003096576.
   
   
3. Now, since all the balls are of same type, (i) is divided by 5!
   Therefore, all possible permutations in this case is ¹⁰P₅ / 5! =252.
   Therefore, the required probability is : (¹⁰P₅ / 5!)/5¹⁰=0.000025804.
   
   
4. The first one gives the most likelihood, since, the probability is higher than the other cases.
   
   
   

I hope these answers will help you. Thank you. :)