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](https://tex.z-dn.net/?f=%5Cfrac%7B%28n-1%29%5En%7D%7Bn%5En%7D%20%3D%20%281%20-%20%5Cfrac%7B1%7D%7Bn%7D%29%5En)
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}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cbinom%7Bn-1%7D%7B1%7D%5Cbinom%7Bn%7D%7B2%7D%28n-2%29%21%7D%7Bn%5En%7D)
= ![\frac{(n-1)n!(n-2)!}{2n^n}](https://tex.z-dn.net/?f=%5Cfrac%7B%28n-1%29n%21%28n-2%29%21%7D%7B2n%5En%7D)
= ![\frac{n!(n-1)!}{2n^n}](https://tex.z-dn.net/?f=%5Cfrac%7Bn%21%28n-1%29%21%7D%7B2n%5En%7D)
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}](https://tex.z-dn.net/?f=%5Cfrac%7Bn%2An%21%28n-1%29%21%7D%7B2n%5En%7D)
= ![\frac{(n!)^2}{2n^n}](https://tex.z-dn.net/?f=%5Cfrac%7B%28n%21%29%5E2%7D%7B2n%5En%7D)
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}}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5Cfrac%7Bn%21%28n-1%29%21%7D%7B2n%5En%7D%7D%7B%5Cfrac%7B%281%20-%20n%29%5En%7D%7Bn%5En%7D%7D)
= ![\frac{n!(n-1)!}{2(1-n)^n}](https://tex.z-dn.net/?f=%5Cfrac%7Bn%21%28n-1%29%21%7D%7B2%281-n%29%5En%7D)
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.