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
= [tex]\frac{(n-1)^n}{n^n} = (1 - \frac{1}{n})^n[/tex]
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
= [tex]\frac{\binom{n-1}{1}\binom{n}{2}(n-2)!}{n^n}[/tex]
= [tex]\frac{(n-1)n!(n-2)!}{2n^n}[/tex]
= [tex]\frac{n!(n-1)!}{2n^n}[/tex]
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)
= [tex]\frac{n*n!(n-1)!}{2n^n}[/tex]
= [tex]\frac{(n!)^2}{2n^n}[/tex]
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:
=[tex]\frac{ \frac{n!(n-1)!}{2n^n}}{\frac{(1 - n)^n}{n^n}}[/tex]
= [tex]\frac{n!(n-1)!}{2(1-n)^n}[/tex]
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.
