From a set of nine cards numbered 1 through 9, we draw three numbers - without replacement - one after another: \(a, b, c\). Using the drawn numbers, we form the equation

$$
a x^{2}+b x+c=0
$$

What is the probability that:
a) in 10 trials, at least once the two roots are equal?
b) in 5 trials, exactly once the product of the two roots is a decimal number with a finite number of decimal places that is less than 1?
a) In each experiment, the number of possible values for $a, b$, and $c$ is $V_{9}^{3}=9 \cdot 8 \cdot 7$.

The two roots are equal if the discriminant is 0, i.e., $b^{2}=4 a c$. Therefore, $b$ must be an even number, and since $a, b, c$ are distinct numbers, we can quickly determine by trial that only $b=6, a=1,9$ and $c=9,1$ satisfy the condition.

Thus, the number of favorable cases is 2, and the probability that the two roots are equal in one experiment is

$$
v_{a}=\frac{2}{9 \cdot 8 \cdot 7}=\frac{1}{252}
$$

The probability that the two roots are never equal in 10 experiments is

$$
\left(1-v_{0}\right)^{10}=\left(\frac{251}{252}\right)^{10}
$$

and thus the desired probability is

$$
V_{a}=1-\left(\frac{251}{252}\right)^{10} \approx 0.039
$$

b) It is known that the product of the two roots is $x_{1} x_{2}=\frac{c}{a}$.

Numbers that can be expressed as finite decimal fractions are those fractions whose denominators do not contain prime factors other than 2 and 5.

In this case, therefore, only the values $a=2,4,5,8$ are considered, and (due to the possibility of simplification) the value $a=6$, provided that $c=3$ (where $c<a$).

Since the problem states that $c<a$, the favorable cases are:

$$
\begin{array}{ll}
a=2 & c=1 \\
a=4 & c=1,2,3 \\
a=5 & c=1,2,3,4 \\
a=8 & c=1,2, \ldots, 7, \\
a=6 & c=3
\end{array}
$$

Thus, the number of favorable cases is 16 (independent of $b$). The number of possible cases is $V_{9}^{2}=9 \cdot 8$, and thus the probability that the conditions of the problem are met in one experiment is

$$
v_{b}=\frac{16}{9 \cdot 8}=\frac{2}{9}
$$

and that they are met exactly once in 5 experiments is

$$
V_{b}=\binom{5}{1} v_{b}\left(1-v_{b}\right)^{4}=5 \cdot \frac{2}{9} \cdot\left(\frac{7}{9}\right)^{4}=\frac{10 \cdot 7^{4}}{9^{5}} \approx 0.407
$$

Zsombok Zoltán (Bp., IV., Konyves Kalman g. II. o. t.)