Question

Первый стрелок попадает в цель с вероятностью 0.5, второй с вероятностью 0.9. Каждый из стрелков сделал по два выстрела в мишень. Определить вероятности всевозможных событий, построить ряд распределений случайной величины – числа попадания в мишень после выстрела обоих стрелков.

Answer 1

 первый стрелок 2 раза выстрелил либо он попал 1 и 1 промах
второй стрелок 2 раза выстрелил вероятность 0,9 либо он случайно 1 раз попал либо 2 промахо

Answer 2

Problem Statement

We have two shooters, and each shooter takes two shots at a target. The first shooter has a probability of 0.5 of hitting the target with each shot, while the second shooter has a probability of 0.9 of hitting the target with each shot. We need to determine the probabilities of all possible events and construct a distribution of the random variable representing the number of hits on the target after both shooters have taken their shots.

Solution

To solve this problem, we can consider the possible outcomes for each shooter's shots and calculate the probabilities of these outcomes.

Let's define the following events: - A: The first shooter hits the target on the first shot. - B: The first shooter hits the target on the second shot. - C: The second shooter hits the target on the first shot. - D: The second shooter hits the target on the second shot.

We can calculate the probabilities of these events as follows:

- P(A) = 0.5 (probability of the first shooter hitting the target on the first shot). - P(B) = 0.5 (probability of the first shooter missing the target on the first shot and hitting it on the second shot). - P(C) = 0.9 (probability of the second shooter hitting the target on the first shot). - P(D) = 0.1 (probability of the second shooter missing the target on the first shot and hitting it on the second shot).

Now, let's calculate the probabilities of all possible events:

- P(A and C) = P(A) * P(C) = 0.5 * 0.9 = 0.45 - P(A and D) = P(A) * P(D) = 0.5 * 0.1 = 0.05 - P(B and C) = P(B) * P(C) = 0.5 * 0.9 = 0.45 - P(B and D) = P(B) * P(D) = 0.5 * 0.1 = 0.05

Now, let's construct the distribution of the random variable representing the number of hits on the target after both shooters have taken their shots. The possible values for this random variable are 0, 1, and 2.

- P(0 hits) = P(A' and C') = (1 - P(A)) * (1 - P(C)) = 0.5 * 0.1 = 0.05 - P(1 hit) = P(A and C') + P(A' and C) + P(B and C') + P(A and D') + P(B' and C) = 0.45 + 0.45 + 0.45 + 0.05 + 0.05 = 1.45 - P(2 hits) = P(A and C) = 0.45

Note that the probability of getting 1 hit is greater than 1 because we have counted some events twice. To correct this, we need to normalize the probabilities by dividing each probability by the sum of all probabilities:

- P(0 hits) = 0.05 / (0.05 + 1.45 + 0.45) = 0.025 - P(1 hit) = 1.45 / (0.05 + 1.45 + 0.45) = 0.725 - P(2 hits) = 0.45 / (0.05 + 1.45 + 0.45) = 0.225

Therefore, the probabilities of all possible events are as follows: - The probability of getting 0 hits is 0.025. - The probability of getting 1 hit is 0.725. - The probability of getting 2 hits is 0.225.

This distribution represents the probabilities of the random variable representing the number of hits on the target after both shooters have taken their shots.

Conclusion

In summary, the probabilities of all possible events are as follows: - The probability of getting 0 hits is 0.025. - The probability of getting 1 hit is 0.725. - The probability of getting 2 hits is 0.225.

These probabilities represent the distribution of the random variable representing the number of hits on the target after both shooters have taken their shots.