I want to know the probability that, when rolling a die 7 times in a row,

a) we roll at least one 4?

b) we roll exactly one 4?

c) we roll at most one 4?
It is again advisable to calculate the opposite probability. The probability of not rolling a 4 on a die is $\frac{5}{6}$; that of not rolling a 4 in 7 throws is $\left(\frac{5}{6}\right)^{7}$, and thus the sought probability is

$$
v_{1}-1-\left(\frac{5}{6}\right)^{7} \approx 0.721
$$

b) The probability that, for example, we roll a 4 on the first throw and do not roll a 4 in the subsequent throws is $\frac{1}{6} \cdot\left(\frac{5}{6}\right)^{6}$. This probability is the same if we say the second, third, ..., seventh throw instead of the first, so according to the addition theorem, the sought probability is

$$
v_{2}=7 \cdot \frac{1}{6} \cdot\left(\frac{5}{6}\right)^{6}=\frac{7 \cdot 5^{6}}{6^{7}} \approx 0.391
$$

c) Rolling a 4 at most once in 7 throws means either not rolling a 4 at all, or rolling it exactly once.

The probability of the former is $\left(\frac{5}{6}\right)^{7}$ based on part a), and that of the latter is $\frac{7 \cdot 5^{6}}{6^{7}}$ based on part b). Since these two events exclude each other, the sought probability, using the addition theorem, is

$$
v_{3}=\left(\frac{5}{6}\right)^{7}+\frac{7 \cdot 5^{6}}{6^{7}} \approx 0.279+0.391=0.670
$$

Lábos Elemér (Sátoraljaújhely, Kossuth g. II. o. t.)