Question

Стрелок три раза стреляет по мишеням. Вероятность попадания в мишень при одном выстреле равна 0,7. Найдите вероятность того, что стрелок первые два раза попал в мишени, а последний раз промахнулся.

Answer 1

Ответ:

$0.7 \times 0.7 \times 0.3 = 0.147$

14,7%

Answer 2

Problem Analysis

We are given that a shooter fires three shots at targets, and the probability of hitting the target with a single shot is 0.7. We need to find the probability that the shooter hits the target on the first two shots and misses on the third shot.

Solution

To find the probability of hitting the target on the first two shots and missing on the third shot, we can multiply the probabilities of each event together.

Let's denote the event of hitting the target as "H" and the event of missing the target as "M". The probability of hitting the target on a single shot is given as 0.7, so the probability of missing the target is 1 - 0.7 = 0.3.

Therefore, the probability of hitting the target on the first two shots and missing on the third shot can be calculated as follows:

P(H, H, M) = P(H) * P(H) * P(M)

Substituting the given probabilities:

P(H, H, M) = 0.7 * 0.7 * 0.3

Calculating the result:

P(H, H, M) = 0.147

Therefore, the probability that the shooter hits the target on the first two shots and misses on the third shot is 0.147.

Answer

The probability that the shooter hits the target on the first two shots and misses on the third shot is 0.147.