Question

3 стрелка выстрелили одновременно. Вероятность попадания для них всех одинакова и равно 0,6. Найти вероятность того,что а) никто не попалб) только один стрелок попалв) только два стрелка попалиг) все стрелки попалид) хотя бы один стрелок промахнулсяпомогитеее пожалуйстааа!!!

Answer 1

Ответ:

Если в-ть попадания 0.6, то промаха 0.4.

а) Никто не попал, т.е. все промахнулись: 0.4*0.4*0.4.

б) Только 1 попал:

> пусть попал 1й, другие 2 не попали: 0.6*0.4*0.4

> пусть попал 2й: 0.4*0.6*0.4

> пусть попал 3й: 0.4*0.4*0.6.

Так как тут либо-либо, складываем это все (или умножаем на 3, числа везде идентичны): 0.6*0.4*0.4*3.

в) Только 2 попали. По формуле сочетаний найдем сколькими способами можно выбрать 2 стрелков из 3: С(2, 3) = 3!/(2!*1!)=3. Будет 3 случая, все по результату одинаковые, как и в прошлый раз. Разберем один из них:

> 1й и 2й попали: 0.6*0.6*0.4.

Ну и домножим на 3, аналогично предыдущему пункту: 0.6*0.6*0.4*3.

г) Все попали: 0.6*0.6*0.6

д) Хотя бы 1 промахнулся. Тут можно разбором случаев, как раньше, но это нудно. Проще найти противоположное событие: никто не промахнулся. А это мы знаем из пункта г). Поэтому просто вычтем эту в-ть из 1: 1 - 0.6^3.

Answer 2

To solve this problem, we can use the binomial probability formula. Let's denote the event of hitting the target as "success" and the event of missing the target as "failure". The probability of success for each archer is 0.6, and the probability of failure is 0.4.

a) Probability that no one hit the target:

To find the probability that none of the archers hit the target, we need to calculate the probability of failure for all three archers. Since the events are independent, we can multiply the probabilities together.

The probability that a single archer misses the target is 0.4. Therefore, the probability that all three archers miss the target is:

P(none hit) = (0.4) * (0.4) * (0.4) = 0.064

b) Probability that only one archer hit the target:

To find the probability that only one archer hit the target, we need to consider three cases: the first archer hits while the other two miss, the second archer hits while the other two miss, and the third archer hits while the other two miss. Again, since the events are independent, we can calculate the probabilities separately and then add them together.

The probability that the first archer hits while the other two miss is:

P(first hits) = (0.6) * (0.4) * (0.4) = 0.096

Similarly, the probabilities for the second and third archers hitting while the other two miss are also 0.096.

Therefore, the probability that only one archer hit the target is:

P(only one hit) = P(first hits) + P(second hits) + P(third hits) = 0.096 + 0.096 + 0.096 = 0.288

c) Probability that only two archers hit the target:

To find the probability that only two archers hit the target, we need to consider three cases: the first two archers hit while the third one misses, the first and third archers hit while the second one misses, and the second and third archers hit while the first one misses. Again, we calculate the probabilities separately and then add them together.

The probability that the first two archers hit while the third one misses is:

P(first two hit) = (0.6) * (0.6) * (0.4) = 0.144

Similarly, the probabilities for the other two cases are also 0.144.

Therefore, the probability that only two archers hit the target is:

P(only two hit) = P(first two hit) + P(first and third hit) + P(second and third hit) = 0.144 + 0.144 + 0.144 = 0.432

d) Probability that all three archers hit the target:

To find the probability that all three archers hit the target, we simply multiply the probability of success for each archer together.

The probability that a single archer hits the target is 0.6. Therefore, the probability that all three archers hit the target is:

P(all hit) = (0.6) * (0.6) * (0.6) = 0.216

e) Probability that at least one archer missed the target:

To find the probability that at least one archer missed the target, we can subtract the probability that all three archers hit the target from 1.

P(at least one miss) = 1 - P(all hit) = 1 - 0.216 = 0.784

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