поразит любые три мишени.

Чтобы поразить ровно три мишени из четырех, нам нужно выбрать, какие три мишени будут поражены. Количество способов выбрать 3 мишени из 4 равно числу сочетаний $$C^3_4$$.

• \[ C^3_4 = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times 1} = 4 \]

Для каждого из этих 4 случаев вероятность поражения трех мишеней и промаха по одной равна The probability of hitting exactly three targets out of four requires us to consider which three targets are hit. The number of ways to choose 3 targets out of 4 is given by the binomial coefficient $$C(4, 3)$$.

• \[ C(4, 3) = \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = 4 \]

For each of these 4 combinations, the probability of hitting three targets and missing one is The probability of hitting exactly three targets out of four requires us to consider which three targets are hit. The number of ways to choose 3 targets out of 4 is given by the binomial coefficient $$C(4, 3)$$.

• \[ C(4, 3) = \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = 4 \]

For each of these 4 combinations, the probability of hitting three targets and missing one is $$ (0.8)^3 \times (0.2)^1 = 0.1024 $$. Since there are 4 such combinations, the total probability is:

• \[ P(\text{ровно 3 попадания}) = C(4, 3) \times (0.8)^3 \times (0.2)^1 = 4 \times 0.512 \times 0.2 = 4 \times 0.1024 = 0.4096 \]

Ответ: 0.4096