Question

На предприятии в 1-цеху производится 20 % от всей продукции, во 2-цеху — 30 % и в 3-цеху - 50 %. Вероятности того, что в 1-, 2- и 3-цеху будет произведена годная продукция, соответственно равны 0,95, 0,98 и 0,97. Наудачу выбранная продукция предприятия оказалась негодной. Вероятность того, что эта продукция произведена во 2-цеху, равна

Answer 1

1) 0,2*0,95=0,19 вероятность годный продукции из первого цеха

2) 0,3*0,98=0,294 ....из второго цеха

3) 0,5*0,97=0,485 ... из третьего цеха

4) 0,19+0,294+0,485= 0,969 вероятность вытащить годную деталь любую

5) 1-0,969=0,031 вероятность брака

6) 1-0,98=0,02 брак во втором цеха

7) 0,02/0,031=0,65 искомая вероятность (с округлением)

Answer 2

Calculating the Probability

To calculate the probability that the defective product was produced in the second workshop, we can use the conditional probability formula:

Conditional Probability Formula: The probability of event A given that event B has occurred is given by the formula: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

Where: - \( P(A|B) \) is the probability of event A given that event B has occurred. - \( P(A \cap B) \) is the probability of both events A and B occurring. - \( P(B) \) is the probability of event B occurring.

Applying the Conditional Probability Formula

Given: - The probability of producing a defective product in the first workshop is 5% (0.05). - The probability of producing a defective product in the second workshop is 2% (0.02). - The probability of producing a defective product in the third workshop is 3% (0.03).

We can calculate the probability that the defective product was produced in the second workshop using the conditional probability formula.

\[ P(\text{Defective product from 2nd workshop}) = \frac{P(\text{Defective product}) \cap P(\text{2nd workshop})}{P(\text{Defective product})} \]

\[ P(\text{Defective product from 2nd workshop}) = \frac{0.02 \times 0.30}{0.05 \times 0.20 + 0.02 \times 0.30 + 0.03 \times 0.50} \]

\[ P(\text{Defective product from 2nd workshop}) = \frac{0.006}{0.01 + 0.006 + 0.015} \]

\[ P(\text{Defective product from 2nd workshop}) = \frac{0.006}{0.031} \]

\[ P(\text{Defective product from 2nd workshop}) \approx 0.1935 \]

Therefore, the probability that the defective product was produced in the second workshop is approximately 19.35%.