Probability: 10th ball is blue

Suppose that instead of picking ten balls, you pick all 100 balls and put them into a row in the order you picked them. Every one of the $100!$ possible orders is equally likely, and $20\cdot99!$ have a blue ball in the 10th position. Therefore the probability is exactly $\frac{1}{5}$.

Interesting question.

WLOG, assume that

(1) the balls are either blue (B) or not blue (N), and

(2) there are only two steps:

• (i) Step One: the first 9 balls are chosen in one go ;

• (ii) Step Two: the 10th ball is chosen.

Step One:

The probability of having $i \;(0\le i\le 9)$ blue balls out of $9$ chosen balls is $$\binom {20}i\binom {80}{9-i}\bigg/\binom {100}9$$. This leaves $20-i$ blue balls left, and $91$ balls left in total.

Step Two:

The probability of choosing a blue ball as the 10th ball is $\frac{20-i}{91}$.

In combination, the probability of choosing a blue ball as the 10th ball is $$\sum_{i=0}^9 \binom {20}i\binom {80}{9-i}\frac{20-i}{91}\bigg/\binom{100}9=0.2\qquad\blacksquare$$

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EDIT: Just changed the lower limit of the summation from $1$ to $0$ and the result is $0.2$ (!).