Question

Первый стрелок попадает в цель с вероятностью 0,4, второй с вероятностью 0,8. каждый из стрелков сделал по выстрелу в мишень. определить вероятность попадания в мишень первым стрелком при условии , что мишень была поражена 1 раз.

Answer 1

Вероятности попадания:
$p_1=0,4 \\ p_2=0,8$
Вероятности промахов:
$q_1=0,6 \\ q_2=0,2$

$p=p_1q_1=0,4\cdot 0,2=0,08$

Answer 2

Problem Analysis

We are given that the first shooter hits the target with a probability of 0.4, and the second shooter hits the target with a probability of 0.8. We need to determine the probability of the first shooter hitting the target given that the target was hit once.

Solution

To solve this problem, we can use conditional probability. Let's denote the event that the first shooter hits the target as A and the event that the target was hit once as B. We want to find the probability of A given B, denoted as P(A|B).

We can use Bayes' theorem to calculate P(A|B):

P(A|B) = P(A ∩ B) / P(B)

To calculate P(A ∩ B), we need to find the probability that both the first shooter hits the target and the target was hit once. Since the target was hit once, it means either the first shooter hit the target and the second shooter missed, or the first shooter missed and the second shooter hit.

P(A ∩ B) = P(A) * P(B|A) + P(A') * P(B|A')

Where A' represents the complement of A (i.e., the first shooter misses the target), and P(B|A) and P(B|A') represent the probabilities of the target being hit once given that the first shooter hits or misses, respectively.

Let's calculate these probabilities step by step:

1. P(A) = 0.4 (given) 2. P(B|A) = 0.8 (given that the second shooter hits the target with a probability of 0.8) 3. P(A') = 1 - P(A) = 1 - 0.4 = 0.6 4. P(B|A') = 0.2 (given that the second shooter misses the target with a probability of 0.2)

Now we can substitute these values into the formula:

P(A|B) = (P(A) * P(B|A)) / (P(A) * P(B|A) + P(A') * P(B|A'))

P(A|B) = (0.4 * 0.8) / (0.4 * 0.8 + 0.6 * 0.2)

Calculating this expression gives us the probability of the first shooter hitting the target given that the target was hit once.

Let's calculate it:

P(A|B) = (0.32) / (0.32 + 0.12)

P(A|B) = 0.32 / 0.44

P(A|B) ≈ 0.7273

Therefore, the probability of the first shooter hitting the target given that the target was hit once is approximately 0.7273.

Answer

The probability of the first shooter hitting the target given that the target was hit once is approximately 0.7273.