Question

СРОЧНО! Вероятность попадания в мишень первого стрелка равна 0,9; второго 0,8; третьего 0,7. Найти вероятность того, что в мишень попадет хотя бы один выстрел из шести, так как каждый стрелок сделал по два выстрела

Answer 1

Объяснение:

если стрелок попадет с первой попытки, то второй раз он не должен попасть. если не попал, то должен попасть. тогда

первый стрелок 0,9×0,1

второй стрелок. 0,8×0,2

третий стрелок. 0,7×0,3

вероятность попадания хотя бы одного из них

0,09+0,16+0,21=0,46

Answer 2

Probability of hitting the target for each shooter

According to the given information, the probability of the first shooter hitting the target is 0.9, the probability of the second shooter hitting the target is 0.8, and the probability of the third shooter hitting the target is 0.7.

Probability of at least one shot hitting the target out of six shots

To find the probability of at least one shot hitting the target out of six shots, we can use the complement rule. The complement rule states that the probability of an event not happening is equal to 1 minus the probability of the event happening.

In this case, we want to find the probability of none of the shots hitting the target. Since each shooter made two shots, we can calculate the probability of a single shooter missing both shots as the product of the probabilities of missing each shot.

For the first shooter, the probability of missing both shots is (1 - 0.9) * (1 - 0.9) = 0.1 * 0.1 = 0.01. For the second shooter, the probability of missing both shots is (1 - 0.8) * (1 - 0.8) = 0.2 * 0.2 = 0.04. For the third shooter, the probability of missing both shots is (1 - 0.7) * (1 - 0.7) = 0.3 * 0.3 = 0.09.

Since the shooters are independent of each other, we can multiply these probabilities together to find the probability of all three shooters missing both shots: 0.01 * 0.04 * 0.09 = 0.000036.

Finally, we can use the complement rule to find the probability of at least one shot hitting the target: 1 - 0.000036 = 0.999964.

Therefore, the probability of at least one shot hitting the target out of six shots is approximately 0.999964.

Calculation Steps:

- Probability of the first shooter missing both shots: (1 - 0.9) * (1 - 0.9) = 0.1 * 0.1 = 0.01- Probability of the second shooter missing both shots: (1 - 0.8) * (1 - 0.8) = 0.2 * 0.2 = 0.04- Probability of the third shooter missing both shots: (1 - 0.7) * (1 - 0.7) = 0.3 * 0.3 = 0.09- Probability of all three shooters missing both shots: 0.01 * 0.04 * 0.09 = 0.000036 - Probability of at least one shot hitting the target: 1 - 0.000036 = 0.999964