a) The condition for the roots to be rational is that $b^{2}-4 a c$ must be a perfect square or zero.

Considering that there are no two equal numbers among $a, b, c$, we count the favorable cases by freely choosing $b$ from 1 to 9, and finding the corresponding $a \cdot c$ products for each $b$. (Since $a$ and $c$ are interchangeable, each $a \cdot c$ pair must be counted twice.)

For $b=1$ and $b=2$, there are no suitable $a \cdot c$.

| $b=3$ | in the case of | $1 \cdot 2$ | fits | (2 cases), |
| :--- | :---: | :---: | :---: | :---: |
| $b=4$ | $\ll$ | $1 \cdot 3$ | " | (2 cases), |
| $b=5$ | $\ll$ | $1 \cdot 4,1 \cdot 6,2 \cdot 3$ | 《 | (6 cases), |
| $b=6$ | $\ll$ | $1 \cdot 5,1 \cdot 8,1 \cdot 9,2 \cdot 4$ | $\ll$ | (8 cases), |
| $b=7$ | $\ll$ | $1 \cdot 6,2 \cdot 3,2 \cdot 5,2 \cdot 6,3 \cdot 4$ | $\ll$ | (10 cases), |
| $b=8$ | $\ll$ | $1 \cdot 7,2 \cdot 6,3 \cdot 4,3 \cdot 5$ | 《 | (8 cases), |
| $b=9$ | $\ll$ | $1 \cdot 8,2 \cdot 4,2 \cdot 7,3 \cdot 6,4 \cdot 5$ | 《 | (10 cases). |

Thus, the number of favorable cases is $2+2+6+8+10+8+10=46$.

The number of possible cases is the number of permutations of three elements from nine without repetition

$$
V_{9}^{3}=9 \cdot 8 \cdot 7
$$

Therefore, the probability that the roots are rational in one trial is

$$
v_{a}=\frac{46}{9 \cdot 8 \cdot 7}=\frac{23}{252}
$$

Thus, the probability that the roots are never rational in 10 trials is

$$
\left(1-v_{a}\right)^{10}=\left(\frac{229}{252}\right)^{10}
$$

The probability that the roots are rational once and not rational nine times in 10 trials is

$$
\binom{10}{1} v_{a}\left(1-v_{a}\right)^{9}=10 \cdot \frac{23}{252}\left(\frac{229}{252}\right)^{9}
$$

Therefore, the sought probability is

$$
\begin{gathered}
V_{a}=1-\left(\frac{229}{252}\right)^{10}-\frac{230}{252}\left(\frac{229}{252}\right)^{9}=1-\left(\frac{229}{252}\right)^{9}\left(\frac{229}{252}+\frac{230}{252}\right)= \\
=1-\left(\frac{229}{252}\right)^{9} \cdot \frac{459}{252}=1-\left(\frac{229}{252}\right)^{9} \cdot \frac{51}{28} \approx 0.2308 \\
\text { b) } \frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{2}+x_{1}}{x_{1} x_{2}}=\left(-\frac{b}{a}\right): \frac{c}{a}=-\frac{b}{c}
\end{gathered}
$$

Thus, the favorable values are independent of $a$:

$$
\begin{array}{lll}
c=1, & b=2,3, \ldots \ldots, 9 & (8 \text { cases }) \\
c=2, \quad b=4,6,8 & (3 \text { cases }) \\
c=3, \quad b=6,9 & (2 \text { cases }) \\
c=4, \quad b=8 & \text { (1 case) }
\end{array}
$$

Thus, the number of favorable cases (ignoring $a$) is 14, and the number of possible cases is $V_{9}^{2}=9 \cdot 8=72$, so the sought probability in one trial is

$$
v_{b}=\frac{14}{72}=\frac{7}{36}
$$

An event can occur at most twice if it does not occur at all, or if it occurs exactly once, or exactly twice.

In this case, the sought probability is

$$
\begin{gathered}
V_{b}=\left(1-v_{b}\right)^{20}+\binom{20}{1}\left(1-v_{b}\right)^{19} v_{b}+\binom{20}{2}\left(1-v_{b}\right)^{18} v_{b}^{2}=\left(\frac{29}{36}\right)^{20}+ \\
+20\left(\frac{29}{36}\right)^{19} \frac{7}{36}+190\left(\frac{29}{36}\right)^{18}\left(\frac{7}{36}\right)^{2}=\left(\frac{29}{36}\right)^{18}\left(\frac{29^{2}}{36^{2}}+\frac{29}{36} \cdot \frac{140}{36}+\frac{49 \cdot 190}{36^{2}}\right)= \\
=\left(\frac{29}{36}\right)^{18} \frac{14211}{36^{2}}=\frac{29^{18} 14211}{36^{20}} \approx 0.2238
\end{gathered}
$$

Pál Bártfai (Bp. I., Petőfi g. III. o. t.)
From nine cards numbered 1 to 9, we sequentially draw - without replacement - three numbers: $a, b$, and $c$. If we form the equation

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

using the drawn numbers, what is the probability that
a) in 10 trials, the roots are rational at least twice?
b) in 20 trials, the sum of the reciprocals of the roots is an integer at most twice?