There are $n$ white and $n$ black balls in an urn. We take out $n$ balls. What is the probability that among the $n$ balls taken out, $k$ will be white?
From a set of $2n$ balls, $n$ balls can be drawn in $\binom{2 n}{n}$ ways. Therefore, the number of possible cases is $\binom{2 n}{n}$.

A favorable case is one where there are $k$ white balls and $(n-k)$ black balls. Since $k$ balls can be chosen from $n$ white balls in $\binom{n}{k}$ ways, and $(n-k)$ balls can be chosen from $n$ black balls in $\binom{n}{n-k}=\binom{n}{k}$ ways, the number of favorable cases is: $\binom{n}{k}\binom{n}{k}=\binom{n}{k}^{2}$.

The sought probability: $v=\frac{\binom{n}{k}^{2}}{\binom{2 n}{n}}$.

George Feher (State Fazekas Mihaly r. VII. o. Debrecen). I. Note. Since the number of possible cases includes all cases when $k=0,1,2, \ldots, n$, therefore

$$
\binom{n}{0}^{2}+\binom{n}{1}^{2}+\binom{n}{2}^{2}+\ldots+\binom{n}{n}^{2}=\binom{2 n}{n}
$$

II. Note. In the case where our task would have been to draw $n$ balls one by one, and return the drawn ball, the possible cases would represent the $n$-th order repeated variations of the white and black colors, the number of which is $2^{n}$, while the number of favorable cases would be the number of permutations formed from $n$ elements, where only two types of elements are among them, one in $k$ and the other in $(n-k)$ numbers, that is: $\frac{n!}{k!(n-k)!}=\binom{n}{k}$.

We mention this because in four papers, the result was determined based on this understanding, without, however, highlighting this understanding.[^0]


[^0]:    ${ }^{1}$ At once or one after the other, without returning the drawn ball.