What is the probability that at least two individuals in a group of 13 were born in September? (For simplicity, consider each month equally likely.)
I. Solution: The probability that someone was born in September is $\frac{1}{12}$, and the opposite probability is $\frac{11}{12}$. Therefore, the probability that none of the 13 individuals were born in September is $v_{1}=\left(\frac{11}{12}\right)^{13}$, while the probability that one was born in September and the other 12 in other months is $v_{2}=\left(\frac{13}{1}\right) \frac{1}{12}\left(\frac{11}{12}\right)^{12}$. Thus, the sought probability is

$$
\begin{gathered}
V=1-v_{1}-v_{2}=1-\left(\frac{11}{12}\right)^{13}-\frac{13}{12}\left(\frac{11}{12}\right)^{12}=1-\left(\frac{11}{12}+\frac{13}{12}\right)= \\
=1-2\left(\frac{11}{12}\right)^{12} \approx 0.296
\end{gathered}
$$

Kálmán György (Szolnok, Beloiannisz g. III. o. t.)

II. Solution: The number of all possible cases is $V_{12}^{i, 13}=12^{13}$. In the case of the opposite probability to the sought one, the favorable cases are those where none were born in September, or exactly 1 out of the 13 was born in September. The number of the former is $V_{11}^{i, 13}=11^{13}$, and the number of the latter is $\binom{13}{1} V_{11}^{i, 12}=11^{12}$, and thus the opposite probability to the sought one is

$$
\frac{11^{13}}{12^{13}}+\frac{13 \cdot 11^{12}}{12^{13}}=\left(\frac{11}{12}\right)^{12}\left(\frac{11}{12}+\frac{13}{12}\right)=2 \cdot\left(\frac{11}{12}\right)^{12}
$$

which agrees with the result obtained in the first solution.

Biczó Géza (Bp. II., Rákóczi g, III. o. t.)