『From an urn containing $n$ balls, a certain number of balls are drawn. What is the probability that the number of balls drawn is even?[^0]


[^0]:    ${ }^{1} \mathrm{~A}$ solution can also be submitted by university students.
It is known that the mathematical probability is given by the ratio of the number of favorable cases to the total number of possible cases. Since we can draw $1,2,3, \ldots n-1, n$ balls, the number of possible cases $l$ is equal to the sum of the 1-combinations, 2-combinations, 3-combinations, $\ldots n$-combinations that can be formed from $n$ elements; thus,

$$
l=\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\ldots+\binom{n}{n-1}+\binom{n}{n}
$$

To determine this sum, let

$$
(a+b)^{n}=\binom{n}{0} a^{n}+\binom{n}{1} a^{n-1} b+\binom{n}{2} a^{n-2} b+\ldots+\binom{n}{n-1} a b^{n-1}+\binom{n}{n} b^{n}
$$

be the equality where $a=b=1$. Then

$$
2^{n}=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\ldots+\binom{n}{n-1}+\binom{n}{n}
$$

thus

$$
l=2^{n}-1
$$

Next, let's determine the number of favorable cases. Since 2, 4, 6, $\ldots$ balls can be drawn in $\binom{n}{2},\binom{n}{4},\binom{n}{6}, \ldots$ ways, the number of favorable cases $k$ will be:

$$
k=\binom{n}{2}+\binom{n}{4}+\binom{n}{6}+\ldots
$$

First, let $n$ be an even number. Since

$$
(1-1)^{n}=0=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\ldots+\binom{n}{n-2}-\binom{n}{n-1}+\binom{n}{n}
$$

then

$$
\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\binom{n}{6}+\ldots+\binom{n}{n}=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\ldots+\binom{n}{n-1}=\frac{2^{n}}{2}
$$

thus

$$
k=2^{n-1}-\binom{n}{0}=2^{n-1}-1
$$

Similarly, if $n$ is an odd number:

$$
k^{\prime}=\binom{n}{2}+\binom{n}{4}+\ldots+\binom{n}{n-1}=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\ldots+\binom{n}{n}-\binom{n}{0}=\frac{2^{n}}{2}-1=2^{n-1}-1
$$

We see that the number of favorable cases is $2^{n-1}-1$ in both cases.

Thus, the sought probability is:

$$
v=\frac{2^{n-1}-1}{2^{n}-1}=\frac{1}{2}-\frac{1}{2} \cdot \frac{1}{2^{n}-1}
$$

The reverse probability, i.e., the probability that the number of drawn balls is odd:

$$
v=\frac{1}{2}+\frac{1}{2} \cdot \frac{1}{2^{n}-1}
$$

Our results show that the probability of drawing an odd number of balls is greater. Both probabilities approach $\frac{1}{2}$ more closely as $n$ increases.

(Kürti Imre, Eger.)

The problem was also solved by: Bartók I. and Schmidl I., university students, as well as Ádámffy E., Dömény I., Fodor H., Fuchs I., Haar A., Krampera Gy., Messer P., Pichler S., Rássy P., Schwarz Gy.