Question

На автозаводе три конвейерных линии, причем на первой из них собирается 35% всех изделий, на второй 25%, на третьей - 45%. Вероятность брака для изделий, собранных на первой линии, равна 0,2; на второй -0,1, на трктьей -0,15. Приобретенный покупателем автомобиль не имеет брака. Найти вероятность того, что он собран на первой линии.

Answer 1

Находим по формуле Байеса
Р=0,35*0,8/(0,35*0,8+0,25*0,9+0,45*0,85)=0,315

Answer 2

Problem Analysis

We are given that there are three conveyor lines at an automobile factory, and the percentage of products assembled on each line. We are also given the probability of defects for products assembled on each line. We need to find the probability that a car purchased by a customer was assembled on the first line.

Solution

To find the probability that a car was assembled on the first line, we need to use Bayes' theorem. Bayes' theorem states that the probability of an event A given event B can be calculated as:

P(A|B) = (P(B|A) * P(A)) / P(B)

In this case, event A is the car being assembled on the first line, and event B is the car not having any defects.

Let's calculate the required probabilities step by step:

1. Calculate the probability of the car being assembled on the first line, given that it has no defects: - P(A) = 35% (given in the question) = 0.35 - P(B|A) = probability of no defects for products assembled on the first line = 1 - 0.2 (given in the question) = 0.8 - P(B) = probability of no defects for any product = (P(B|A) * P(A)) + (P(B|A') * P(A')) + (P(B|A'') * P(A'')) = (0.8 * 0.35) + (0.9 * 0.25) + (0.85 * 0.45) = 0.28 + 0.225 + 0.3825 = 0.8875 - P(A|B) = (P(B|A) * P(A)) / P(B) = (0.8 * 0.35) / 0.8875 = 0.28 / 0.8875 ≈ 0.3155

Answer

The probability that the car purchased by the customer was assembled on the first line is approximately 0.3155.

Note: The probabilities used in this calculation are based on the information provided in the question and may not represent the actual probabilities in a real-world scenario.