Question

В ящике находится 7 деталей первого сорта, 5 второго и 3 - третьего. Из ящика последовательно вынимают три детали. Найти вероятность того, что первая вынутая деталь окажется первого сорта, вторая второго и третья - третьего.

Answer 1

Вероятность A = вытянуть в первый раз деталь первого сорта = 7 / (7 + 5 + 3) = 7/15
Вероятность B|A = вытянуть во второй раз деталь второго сорта (при условии, что A выполнено) = 5 / (6 + 5 + 3) = 5/14
Вероятность C|AB вытянуть в третий раз деталь третьего сорта (при условии выполнения A и B)  = 3 / (6 + 4 + 3) = 3/13

P(ABC) = P(A)P(B|A)P(C|AB) = 1/26

Answer 2

Problem Analysis

We are given a box containing 7 parts of the first type, 5 parts of the second type, and 3 parts of the third type. We need to find the probability that the first part drawn is of the first type, the second part drawn is of the second type, and the third part drawn is of the third type.

Solution

To find the probability, we need to calculate the probability of each event happening and then multiply them together.

Let's calculate the probability of each event:

1. Probability of drawing the first part of the first type: There are a total of 7 parts of the first type out of a total of 7 + 5 + 3 = 15 parts in the box. Therefore, the probability of drawing the first part of the first type is 7/15.

2. Probability of drawing the second part of the second type: After drawing the first part, there are now 14 parts left in the box, out of which 5 are of the second type. Therefore, the probability of drawing the second part of the second type is 5/14.

3. Probability of drawing the third part of the third type: After drawing the first and second parts, there are now 13 parts left in the box, out of which 3 are of the third type. Therefore, the probability of drawing the third part of the third type is 3/13.

Now, let's multiply the probabilities together to find the overall probability:

P = (7/15) * (5/14) * (3/13)

Calculation

Let's calculate the probability:

P = (7/15) * (5/14) * (3/13) = 0.02857142857142857

Answer

The probability that the first part drawn is of the first type, the second part drawn is of the second type, and the third part drawn is of the third type is approximately 0.0286.

Please note that the above calculation assumes that the parts are drawn without replacement, meaning that once a part is drawn, it is not put back into the box before drawing the next part.