Question

В сборочный цех поступают детали с трех конвейеров в отношении 2:2:1. Вероятность поступления бракованной детали с первого конвейера равна 0,1; со второй -0,15; с третьего -0,05. Найти вероятность того, что наудачу проверенная на сборке деталь не будет бракованной.

Answer 1

Событие Р(А)  состоит из двух - Р1 - взять ЛЮБУЮ деталь И -Р2 - взять ГОДНУЮ. 
1) Вероятность взять любую Р1(i) - это доля каждого цеха в выпуске продукции исходя из пропорции в производстве.
р1(1)= р1(2) = 2/5=0,4 и р1(3) = 1/5 = 0,2.
2) Для упрощения (потом будет видно) сосчитаем вероятность взять БРАК, а не годную деталь. Три цеха - три события ИЛИ - для них вероятности СУММИРУЮТСЯ. Для каждого цеха взять БРАК - событие И - И цех И брак- вероятности УМНОЖАЮТСЯ.
Вероятность БРАКОВАННОЙ детали - Q(А) = 0,4* 0,1 + 0,4*0,15 + 0,2* 0,05 = 0,04+0,06+0,01 = 0,11 = 11% - брак.
Вероятность НЕ бракованной -P(A) = 1 - Q(A) = 99% - ГОДНЫХ.
ОТВЕТ: Вероятность НЕ бракованной равна 99%.
Справочно: В таблице приведен расчет и по формуле Байеса из которой видно, что наиболее вероятно это будут детали 1-го или 2-го цехов.

Answer 2

Problem Statement

In a production assembly workshop, parts are received from three conveyor belts in a ratio of 2:2:1. The probability of receiving a defective part from the first conveyor belt is 0.1, from the second conveyor belt is 0.15, and from the third conveyor belt is 0.05. We need to find the probability that a randomly checked part during assembly is not defective.

Solution

To find the probability that a randomly checked part during assembly is not defective, we can use the complement rule. The complement of an event A is the event "not A" or "A does not occur". In this case, the event A is the event that the part is defective.

Let's calculate the probability of the event A for each conveyor belt:

- Probability of receiving a defective part from the first conveyor belt: 0.1 - Probability of receiving a defective part from the second conveyor belt: 0.15 - Probability of receiving a defective part from the third conveyor belt: 0.05

To find the probability of the event "not A" for each conveyor belt, we subtract the probability of A from 1:

- Probability of receiving a non-defective part from the first conveyor belt: 1 - 0.1 = 0.9 - Probability of receiving a non-defective part from the second conveyor belt: 1 - 0.15 = 0.85 - Probability of receiving a non-defective part from the third conveyor belt: 1 - 0.05 = 0.95

Now, let's calculate the overall probability of receiving a non-defective part during assembly. Since the parts are received in a ratio of 2:2:1, we can calculate the weighted average of the probabilities:

- Probability of receiving a non-defective part overall = (2 * 0.9 + 2 * 0.85 + 1 * 0.95) / (2 + 2 + 1)

Simplifying the expression, we get:

- Probability of receiving a non-defective part overall = (1.8 + 1.7 + 0.95) / 5

Calculating the numerator:

- Numerator = 1.8 + 1.7 + 0.95 = 4.45

Calculating the denominator:

- Denominator = 5

Finally, calculating the overall probability:

- Probability of receiving a non-defective part overall = Numerator / Denominator = 4.45 / 5 = 0.89

Therefore, the probability that a randomly checked part during assembly is not defective is 0.89.