Question

Suppose n balls are distributed in n boxes in such a way that each ball chooses a box indepen- dently of each other. (a) What is the probability that Box #1 is empty? (b) What is the probability that only Box #1 is empty? (c) What is the probability that only one box is empty? (d) Given that Box #1 is empty, what is the probability that only one box is empty? (e) Given that only one box is empty, what is the probability that Box #1 is empty?

Answer 1

Answer:

Check the explanation

Step-by-step explanation:

a) Probability that Box #1 is empty is computed here as:

= Number of ways to distribute each of the the n balls in remaining (n - 1) boxes / Total ways to distribute n balls into n boxes

= $\frac{(n-1)^n}{n^n} = (1 - \frac{1}{n})^n$

This is the required probability here.

b) Probability that only 1 box is empty is computed here as:

= Number of ways to choose one of the remaining (n - 1) boxes such that it will have 2 balls * Number of ways to select 2 balls of that box* permutation of remaining (n - 2) balls into (n - 2) boxes / Total ways to distribute n balls into n boxes

= $\frac{\binom{n-1}{1}\binom{n}{2}(n-2)!}{n^n}$

= $\frac{(n-1)n!(n-2)!}{2n^n}$

= $\frac{n!(n-1)!}{2n^n}$

This is the required probability here.

c) Probability that only 1 box is empty is computed here as:

= Number of ways to select a box which would be empty * Probability that only that box would be empty ( from previous part)

= $\frac{n*n!(n-1)!}{2n^n}$

= $\frac{(n!)^2}{2n^n}$

This is the required probability here.

d) Given that box #1 is empty, probability that only 1 box is empty is computed here as:

= Probability that only box 1 is empty / Probability that box 1 is empty

We will use the answers from parts a) and b) here to get:

=$\frac{ \frac{n!(n-1)!}{2n^n}}{\frac{(1 - n)^n}{n^n}}$

= $\frac{n!(n-1)!}{2(1-n)^n}$

This is the required probability here.

e) Given that only 1 box is empty, probability that box #1 is empty is computed here as:

= 1/n as each box is equally likelt to be empty

Therefore 1/n is the required probability here.