Question

ДОПОМОЖІТЬ! Партія деталей виготовлена трьома робітниками. Перший робітник виготовив 2/5 партії, другий 1/3 партії, а решту- третій. Ймовірність браку для кожного з робітників відповідно дорівнює 1%, 5%, 10%. Яка ймовірність того, що навмання вилучена деталь буде бракованою? Які ймовірності того, що цю деталь виготовив перший, другий третій робітник?

Answer 1

Ответ:

1) вероятность того, что наудачу извлеченная деталь будет бракованной равна 71/1500

2) вероятности того, что эту деталь сделал

первый рабочий   6/71

второй рабочий    25/71

третий рабочий     4/71

Объяснение:

Формула полной вероятности + формула Байерса.

Сначала определим, что изготовленная деталь бракованная.

событие А = {изготовлена бракованная деталь};

Гипотеза Н1 = {эта деталь от первого рабочего};

•        Р(Н1)= 2/5;

• условная вероятность, что бракованную деталь изготовил первый рабочий   $\displaystyle \boldsymbol {P_{H1}(A) = 1\% = 0,01}$

Гипотеза Н2 = {эта деталь от второго рабочего};

•         Р(Н2)=1/3;

• условная вероятность, что бракованную деталь изготовил второй  рабочий   $\displaystyle \boldsymbol {P_{H2}(A) = 5\% = 0,05}$

Гипотеза Н3 = {эта деталь от третьего рабочего};

•        Р(Н3) =     $\displaystyle \bigg(1-\frac{2}{5} -\frac{1}{3} \bigg )=\frac{4}{15}$.

• условная вероятность, что бракованную деталь изготовил третий рабочий   $\displaystyle \boldsymbol {P_{H3}(A) = 10\% = 0,1}$

Тогда по формуле полной вероятности мы вычисляем, что вероятность выбрать бракованную деталь

$\displaystyle P(A) = P(H1)*P_{H1}(A)\;+\;P(H2)*P_{H2}(A)\;+\;P(H3)*P_{H3}(A)\;=\;\\\\\\= \frac{2}{5} *\frac{1}{100} \;+\;\frac{1}{3} *\frac{5}{100} \;+\;\frac{4}{15} *\frac{1}{10} =\frac{71}{1500} \approx 0,0473$

И по формуле Байерса вычисляем, что именно эту деталь сделал

i-тый рабочий

$\displaystyle P_A(Hi) = \frac{P(Hi)*P_{Hi}(A)}{P(A)}$

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Answer 2

Probability of a Defective Part

To find the probability that a randomly selected part is defective, we need to consider the probabilities of each worker producing a defective part and the proportion of parts produced by each worker.

The first worker produced 2/5 of the batch, and the probability of them producing a defective part is 1%. Therefore, the probability of a defective part produced by the first worker is 2/5 * 1% = 2/500.

The second worker produced 1/3 of the batch, and the probability of them producing a defective part is 5%. Thus, the probability of a defective part produced by the second worker is 1/3 * 5% = 5/300.

The remaining parts were produced by the third worker, who has a 10% probability of producing a defective part. Since the third worker produced 1 - (2/5 + 1/3) = 7/15 of the batch, the probability of a defective part produced by the third worker is 7/15 * 10% = 7/150.

To find the overall probability of a defective part, we sum up the probabilities from each worker:

2/500 + 5/300 + 7/150 = 3/500.

Therefore, the probability that a randomly selected part will be defective is 3/500.

Probability of Each Worker Producing the Part

To find the probability that the defective part was produced by each worker, we need to consider the proportion of parts produced by each worker and the overall probability of a defective part.

The first worker produced 2/5 of the batch, and the overall probability of a defective part is 3/500. Therefore, the probability that the defective part was produced by the first worker is (2/5) * (3/500) = 6/2500.

The second worker produced 1/3 of the batch, and the overall probability of a defective part is 3/500. Thus, the probability that the defective part was produced by the second worker is (1/3) * (3/500) = 3/1500.

The remaining parts were produced by the third worker, who has a probability of 7/15 of producing a part. The overall probability of a defective part is 3/500. Therefore, the probability that the defective part was produced by the third worker is (7/15) * (3/500) = 21/7500.

In summary, the probabilities that the defective part was produced by the first, second, and third worker are 6/2500, 3/1500, and 21/7500, respectively.