With two cannons, which we always fire simultaneously, we shoot at a stationary target. The probability of a short shot, hit, and long shot for one cannon is $0.3, 0.2, 0.5$; for the other, it is $0.6, 0.1, 0.3$. What is the probability that
a) out of 20 pairs of shots, we hit the target with both cannons at least once?
b) out of 3 pairs of shots, we hit the target with at least one cannon every time?
c) out of 2 pairs of shots, there are at most 3 short shots?

d) How many pairs of shots must be fired so that the probability of "at least one double hit" is at least 0.9?
a) The probability of hitting with a pair of shots is $0.1 \cdot 0.2=0.02$, and the probability of not hitting with a pair of shots is $1-0.02=0.98$.

Therefore, the probability of not hitting with 20 pairs of shots is $0.98^{20}$, and thus the sought probability is

$$
v_{a}=1-0.98^{20} \approx 1-0.667=0.333
$$

b) The probability of not hitting with a pair of shots is $0.9 \cdot 0.8=0.72$, and thus the probability of hitting at least once is 0.28.

Therefore, the sought probability is

$$
v=0.28^{3} \approx 0.022
$$

c) The probability that all 4 shots in 2 pairs of shots are short is $(0.3 \cdot 0.6)^{2}=0.18^{2}$, and thus the sought probability is

$$
v=1-0.18^{2} \approx 1-0.032=0.968
$$

d) In part b), we saw that the probability of not hitting with one pair of shots is 0.72, so the probability of not hitting with $x$ pairs of shots is $0.72^{x}$. Let's determine $x$ such that the probability of not hitting is 0.1. Therefore,

$$
0.72^{x}=0.1
$$

from which

$$
x=\frac{\lg 0.1}{\lg 0.72}=\frac{-1}{0.8573-1}=\frac{1}{0.1427}=7.007
$$

Thus, after 8 pairs of shots, the probability of "hitting at least once" is already greater than 0.9.

Roboz Ágnes (Bp., VI., Varga Katalin g. III. o. t.)