Two shooters aim at the same target. The first one hits the target with a probability of 0.8, the second one with a probability of 0.6. The first shooter fires twice, the second one fires three times. What is the probability that the target is

a) hit by at least one of them,

b) hit once by the first and twice by the second,

c) hit at least twice by at least one of the shooters,

d) hit by only one bullet?
The solution to the task will utilize the following relationships:

I. If events $A$ and $B$ are independent, then the probability of their joint occurrence is equal to the product of their individual probabilities:

$$
P(A B)=P(A) \cdot P(B)
$$

II. If events $A$ and $B$ are mutually exclusive, then the probability that at least one of them occurs is equal to the sum of the two probabilities:

$$
P(A+B)=P(A)+P(B)
$$

III. If the probability of an event occurring is $P$, then the probability of the event not occurring is $(1-P)$.

In solving the task, we will also assume that each shot represents independent events (whether from the same or different archers).

a) The event in question does not occur if none of the five shots hit. The first archer has a 0.2 probability of missing the target according to III, so if they shoot twice, the probability of missing both times is $0.2 \cdot 0.2$ according to I. The second archer has a 0.4 probability of missing the target, so the probability of missing all three shots is $0.4 \cdot 0.4 \cdot 0.4$ according to I. The probability that none of the five shots hit is $0.2^{2} \cdot 0.4^{3}$ according to I. Finally, the probability of the event in question is $1-0.2^{2} \cdot 0.4^{3}=0.99744$ according to III.

b) Similarly to the previous part, the probability that the first archer's first shot hits and the second misses is $0.8 \cdot 0.2$. The same probability applies if the first shot misses and the second hits. Thus, according to II, the probability that the first archer hits exactly once is $2 \cdot 0.8 \cdot 0.2$.

By similar reasoning, the probability that the second archer hits exactly twice out of three shots is $3 \cdot 0.6 \cdot 0.6 \cdot 0.4$.

Thus, the probability that both of these events occur simultaneously (i.e., in a single trial) is, according to I,

$$
(2 \cdot 0.8 \cdot 0.2) \cdot\left(3 \cdot 0.6^{2} \cdot 0.4\right)=0.13824
$$

c) First, let's determine the probabilities $p, q$ of the following events:

1. The first archer hits at least twice,
2. The second archer hits at least twice.

The probability $p$ of the first event is, similarly to the previous parts: $p=0.8^{2}$.

The second event occurs if the second archer hits either exactly twice or exactly three times, so

$$
q=3 \cdot 0.6^{2} \cdot 0.4+0.6^{3}
$$

The event in question can occur if either of the two events listed here occurs. Let's break this joint occurrence into three independent parts:

1. The first event occurs, the second does not;
2. The second event occurs, the first does not;
3. Both the first and the second events occur.

Thus, using the relationships listed, the probability of the event in question is

$$
p(1-q)+q(1-p)+p q=p+q-p q=0.87328
$$

d) The probability that the first archer hits the target once and the second archer does not hit it at all is $2 \cdot 0.8 \cdot 0.2 \cdot 0.4^{3}$. The probability that the second archer hits once and the first archer does not hit at all is

$$
0.2^{2} \cdot 3 \cdot 0.4^{2} \cdot 0.6
$$

Thus, the probability that one of these two events occurs, according to II, is the sum of these: 0.032.

With this, we have completed the solution.

Remarks. 1. Some interpreted part c) of the task to mean that at least one target shot hits the target exactly twice. In this case, the probability $q$ would be replaced by $3 \cdot 0.6^{2} \cdot 0.4$, and the final result would be

$$
p+q-p q=0.79552
$$

2. The original result of part c) can also be derived from $1-(1-p)(1-q)$. The interpretation is left to the reader.