Question

При стрельбе по мишени вероятность промаха равна 0,2. Вычислить вероятность не более двух попаданий при четырех выстрелы.

Answer 1

Вроде как-то так. 0,2 - вероятность промаха => 0,8 - вероятность попадания. А дальше просто разбираем случаи.

Answer 2

Calculating the Probability of Not More Than Two Hits in Four Shots

When the probability of missing the target is 0.2, the probability of hitting the target is 1 - 0.2 = 0.8. This means that the probability of hitting the target in a single shot is 0.8.

To calculate the probability of not more than two hits in four shots, we can use the binomial probability formula:

P(X=k) = (n choose k) * (p^k) * ((1-p)^(n-k))

Where: - n = total number of trials (in this case, 4 shots) - k = number of successful trials (in this case, 0, 1, or 2 hits) - p = probability of success on an individual trial (in this case, 0.8)

Let's calculate the probabilities for 0, 1, and 2 hits in four shots using the given probability of hitting the target.

Calculations

- Probability of 0 hits (k=0): - P(X=0) = (4 choose 0) * (0.8^0) * ((1-0.8)^(4-0)) - P(X=0) = 1 * 1 * (0.2^4) - P(X=0) = 1 * 1 * 0.0016 - P(X=0) = 0.0016

- Probability of 1 hit (k=1): - P(X=1) = (4 choose 1) * (0.8^1) * ((1-0.8)^(4-1)) - P(X=1) = 4 * 0.8 * 0.2^3 - P(X=1) = 0.512

- Probability of 2 hits (k=2): - P(X=2) = (4 choose 2) * (0.8^2) * ((1-0.8)^(4-2)) - P(X=2) = 6 * 0.64 * 0.04 - P(X=2) = 0.1536

Final Result

The probability of not more than two hits in four shots is the sum of the probabilities for 0, 1, and 2 hits: P(X<=2) = P(X=0) + P(X=1) + P(X=2) P(X<=2) = 0.0016 + 0.512 + 0.1536 P(X<=2) = 0.6672

Therefore, the probability of not more than two hits in four shots is 0.6672.