What is the probability of drawing 8 cards at once from a deck of Hungarian cards, that among them a) 4 are lower cards?

b) at least 5 are red cards?
The number of possible cases is $C_{32}^{8}=\binom{32}{8}$. Favorable are those 8-card combinations where there are 4 low cards. Clearly, there are as many such groups as 4-card combinations can be formed from the other 28 cards, i.e., $C_{28}^{4}=\binom{28}{4}$, and thus the sought probability is

$$
\begin{aligned}
v_{1}=\frac{\binom{28}{4}}{\binom{32}{8}} & =\frac{28 \cdot 27 \cdot 26 \cdot 25}{1 \cdot 2 \cdot 3 \cdot 4} \cdot \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{32 \cdot 31 \cdots \cdots \cdot 25}= \\
& =\frac{5 \cdot 6 \cdot 7 \cdot 8}{32 \cdot 31 \cdot 30 \cdot 29}=\frac{7}{4 \cdot 31 \cdot 29}=\frac{7}{3596}(\approx 0.002)
\end{aligned}
$$

b) The number of possible cases remains unchanged $\binom{32}{8}$. Favorable are the groups containing 8, 7, 6, or 5 red cards. Let's calculate the number of such groups in turn.

8 red cards are contained in only one 8-card combination. 7 red card groups can be formed by selecting 7 out of the 8 red cards in $\binom{8}{7}$ ways and pairing each such 7-card combination of red cards with each of the 24 non-red cards. Thus, the total number of such groups is $\binom{8}{7} \cdot 24=8 \cdot 24=192$.

Similarly, the number of groups containing 6 red cards is $\binom{8}{6}\binom{24}{2}=\frac{8 \cdot 7}{1 \cdot 2}, \frac{24 \cdot 23}{1 \cdot 2}=28 \cdot 12 \cdot 23=7728$, and the number of groups containing 5 red cards is $\binom{8}{5}\binom{24}{3}=113344$.

Thus, the sought probability is

$$
v_{2}=\frac{1+192+7728+113344}{\binom{32}{8}}=\frac{121265}{10518300}=\frac{24253}{2103660}(\approx 0.012)
$$

As we can see, both probabilities are quite small, but the latter is about 6 times larger than the former.

Tolnai Tibor (Szombathely, Nagy Lajos g. II.o.t.)