# k extractions of colored balls from numbered boxes

Suppose there are $n+1$ boxes numbered from $0$ to $n$. The $i$-th box contains $i$ white balls and $n−i$ black balls.

A box is chosen randomly and a ball is selected from the box, the ball is returned to the box afterwards. This procedure is repeated $k$ times.

1. What is the probability of the $k$ balls extracted being white?
2. A box is chosen and then $k$ balls are drawn with replacement from the selected box. Calculate the probability of the $k$ balls being white.

1) If $A_j=\{\text{j-th extracted ball is white}\}$, I want to calculate $$P(A_1 \cap\dots \cap A_k)$$

If I could show that these events are independent, then $$P(A_1 \cap\dots \cap A_k)=P(A_1) \cdots P(A_k)$$

For each $j$, the probability of the $j-th$ ball extracted being white clearly depends on the box we've chosen. If I call $V_i=\{\text{the i-th box is chosen}\}$, then$$P(A_j)=\sum_{i=0}^nP(A_j|V_i)P(V_i)$$$$=\sum_{i=0}^n\dfrac{1}{n+1}\dfrac{i}{n}$$$$=\dfrac{1}{2}$$

So we would have that the probability of the original event is $$P(A_1 \cap \dots \cap A_k)=\dfrac{1}{2^k}$$

2) If $A=\{\text{the k balls are white}\}$, we have $$P(A)=\sum_{i=0}^n P(A|V_i)P(V_i)$$$$=\sum_{i=0}^n\dfrac{1}{n+1}(\dfrac{i}{n})^k$$$$=\dfrac{1}{(n+1)n^k}(\sum_{i=0}^n i^k)$$

Is this correct? Is there an easy way to show the events are independent (providing that they actually are independent, something which I am not quite sure of)? Thanks in advance for the help.

The events are independent and you have a 1/2 probability of choosing a white ball each time due to the symmetry. So the answer is $\frac{1}{2^k}$ for the first question. Second answer is $[\sum_{i=0}^n(\frac{i}{n})^k]/(n+1)$