Question

Вероятность брака детали в партии из 200 деталей равна 0,01. Найти вероятность того, что в партии не более двух бракованных деталей.

Answer 1

Ответ:

0.13534667490752794

Пошаговое объяснение:

Вероятность того, что одна деталь без брака = 1 - 0.01 = 0.99

Вероятность того, что все детали без брака = 0.99 ^ 200

Вероятность того, что одна деталь с браком, а остальные без = 0.01 * 0.99 ^ 199

Вероятность того, что две детали с браком, а остальные без = 0.01 * 0.01 * 0.99 ^ 198

Чтобы получить ответ надо сложить все эти числа. Получится примерно 0.13534667490752794

Answer 2

Probability of Defective Parts in a Batch

To find the probability of having no more than two defective parts in a batch of 200 parts, we need to use the given information that the probability of a part being defective is 0.01.

Let's break down the problem step by step:

1. Find the probability of a part being non-defective: - The probability of a part being defective is given as 0.01. - Therefore, the probability of a part being non-defective is 1 - 0.01 = 0.99.

2. Calculate the probability of having no defective parts in the batch: - Since each part is independent, we can use the binomial probability formula. - The formula for the probability of having exactly k successes in n independent Bernoulli trials is: P(X = k) = (n choose k) * p^k * (1 - p)^(n - k) - n is the number of trials (200 parts in this case) - k is the number of successes (0 defective parts in this case) - p is the probability of success (0.99 in this case)

- Plugging in the values, we get: P(X = 0) = (200 choose 0) * 0.99^0 * (1 - 0.99)^(200 - 0)

3. Calculate the probability of having one defective part in the batch: - Using the same formula as above, we calculate: P(X = 1) = (200 choose 1) * 0.99^1 * (1 - 0.99)^(200 - 1)

4. Calculate the probability of having two defective parts in the batch: - Again, using the same formula: P(X = 2) = (200 choose 2) * 0.99^2 * (1 - 0.99)^(200 - 2)

5. Calculate the probability of having no more than two defective parts: - To find the probability of having no more than two defective parts, we sum the probabilities calculated in steps 2, 3, and 4: P(X <= 2) = P(X = 0) + P(X = 1) + P(X = 2)

Let's calculate the probabilities:

- P(X = 0) = (200 choose 0) * 0.99^0 * (1 - 0.99)^(200 - 0) = 0.1353 - P(X = 1) = (200 choose 1) * 0.99^1 * (1 - 0.99)^(200 - 1) = 0.2707 - P(X = 2) = (200 choose 2) * 0.99^2 * (1 - 0.99)^(200 - 2) = 0.2707

Now, let's calculate the probability of having no more than two defective parts:

- P(X <= 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1353 + 0.2707 + 0.2707 = 0.6767

Therefore, the probability of having no more than two defective parts in a batch of 200 parts is 0.6767.