Question

Стрелок стреляет по мишени один раз. В случае промаха стрелок делает второй выстрел по той же мишени. Вероятность попасть в мишень при одном выстреле равна 0,6. Найдите вероятность того, что мишень будет поражена (одним из выстрелов). из тестовых ЕГЭ по математике (профильный уровень!). Можно с объяснением? 25 баллов!

Answer 1

При 100 опытах в 60 попадание первый раз.
Потом из 40 оставшихся попадание в 40*0,6=24 случаях
Итого из 100 опытов попали в 84. Вероятность 0,84.
С использованием теоремы. Вероятность не попасть 2 раза 0,6*0,6. Противоположное событие - попасть хотя бы один раз.
1-0,6*0,6=0,84

Answer 2

Problem Analysis

We are given that a shooter takes two shots at a target. If the first shot misses, the shooter takes a second shot at the same target. The probability of hitting the target with a single shot is 0.6. We need to find the probability that the target is hit with either the first or the second shot.

Solution

To find the probability of hitting the target with either the first or the second shot, we can use the concept of complementary probability. The complementary probability of an event A is the probability of the event not occurring, denoted as P(~A).

Let's calculate the probability of missing the target with a single shot, denoted as P(miss). Since the probability of hitting the target with a single shot is 0.6, the probability of missing the target with a single shot is 1 - 0.6 = 0.4.

Now, let's calculate the probability of missing the target with both shots, denoted as P(miss, miss). Since the second shot is taken only if the first shot misses, the probability of missing both shots is the product of the probabilities of missing each shot. Therefore, P(miss, miss) = P(miss) * P(miss) = 0.4 * 0.4 = 0.16.

The probability of hitting the target with either the first or the second shot is equal to 1 minus the probability of missing both shots. Therefore, P(hit) = 1 - P(miss, miss) = 1 - 0.16 = 0.84.

The probability that the target will be hit with either the first or the second shot is 0.84.

Calculation Steps

1. Probability of missing the target with a single shot: P(miss) = 1 - P(hit) = 1 - 0.6 = 0.4. 2. Probability of missing the target with both shots: P(miss, miss) = P(miss) * P(miss) = 0.4 * 0.4 = 0.16. 3. Probability of hitting the target with either the first or the second shot: P(hit) = 1 - P(miss, miss) = 1 - 0.16 = 0.84.

Therefore, the probability that the target will be hit with either the first or the second shot is 0.84.

I hope this helps! Let me know if you have any further questions.