Question

Два стрелка стреляют по одной мишени. Вероятность попадания у первого равна 0.8, у второго 0.4, найдите вероятность что в мишени будет две пробоины.

Answer 1

Ответ:

0.32

Пошаговое объяснение:

для этого надо умножить вероятность одного события на вероятность другого:
0.8 * 0.4 = 0.32

Answer 2

Вероятность того, что будет ровно 2 попадания, равна:

P2=p1⋅p2=0.8⋅0.4=0.32.

Answer 3

Problem Analysis

We are given two shooters who are shooting at the same target. The probability of the first shooter hitting the target is 0.8, and the probability of the second shooter hitting the target is 0.4. We need to find the probability that there will be two hits on the target.

Solution

To find the probability that there will be two hits on the target, we need to consider all the possible outcomes where both shooters hit the target.

Let's denote the event that the first shooter hits the target as A, and the event that the second shooter hits the target as B.

The probability that both shooters hit the target can be calculated using the formula:

P(A and B) = P(A) * P(B)

Substituting the given probabilities, we have:

P(A and B) = 0.8 * 0.4 = 0.32

Therefore, the probability that both shooters hit the target is 0.32.

Now, we need to find the probability that there will be two hits on the target, which means that both shooters hit the target and there are no misses.

Let's denote the event that there are two hits on the target as C.

The probability that there will be two hits on the target can be calculated using the formula:

P(C) = P(A and B) * P(no misses)

To calculate the probability of no misses, we need to consider all the possible outcomes where there are no misses. This can be calculated as:

P(no misses) = P(A and B) + P(A' and B')

Where A' represents the event that the first shooter misses the target, and B' represents the event that the second shooter misses the target.

The probability of the first shooter missing the target is 1 - 0.8 = 0.2, and the probability of the second shooter missing the target is 1 - 0.4 = 0.6.

Substituting the given probabilities, we have:

P(no misses) = 0.32 + (0.2 * 0.6) = 0.32 + 0.12 = 0.44

Now, we can calculate the probability that there will be two hits on the target:

P(C) = P(A and B) * P(no misses) = 0.32 * 0.44 = 0.1408

Therefore, the probability that there will be two hits on the target is approximately 0.1408.

Answer

The probability that there will be two holes in the target is approximately 0.1408.