There are two boxes of similar parts, the first box has 50 pieces, of which 10 are first class

1) If it is the first box, the probability is: (1 2) * (10 50) * (9 49) + (1 2) *) 40 50) * (10 49) = 1 10

If it is the second box, the probability is: (1 2) * (18 30) * (17 29) + (1 2) * (12 30) * (18 29) = 3 10

So the probability is: 4 10

2) If it is the first box, the probability is: (1 2) * (9 49) = q1

If it is the second box, the probability is: (1 2) * (17 29) = q2

The probability is q1+q2

3) If it is the first box, the probability is: (1 2) * (40 50) * (39 49) = q1

If it is the second box, the probability is: (1 2) * (12 30) * (11 29) = q2

The probability is q1+q2

The specific result of the landlord, you can calculate it yourself!

The probability of selecting each box is 1 2, and there are two types of cases: the first box is selected, and there are 49 left after the first time the first class is taken, of which the first class is 9, so the probability of the second time taking out the first class is 9 49;Select the second box, and there are 29 left after the first time you take out the first class, of which 17 are the first class, so the probability of the second time you take out the first class is 17 29So the probability is 1 2 * 9 49 + 1 2 * 17 29 = 547 1421

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Event A: The first part taken out is first-class, and Time B: The second part taken out is first-class.

The probability of a first-class product in the first box is 1 5

The probability of a first-class product in the second box is 3 5

The probability that the part taken out first is a first-class product p(a)=1 5*use the conditional probability to find p(b|a) = p(ab) p(a) let the probability of the first box being selected for the first time be p(c)=

The probability of picking the second box for the first time is p(d)=

p(ab)=p(c)p(ab|c)+p(d)p(ab|d)=p(ab|c) understood as the probability of selecting the first box for the first time and taking out the first-class products for the first and second time].

So p(b|a)=p(ab)/p(a)=

Summary. Okay, according to the correspondence, a small box has as many parts as a large box, so there are about 100 identical parts in a large box.

There are about 100 parts in a small box and about () the same parts in a large box.

Okay, according to the correspondence, a small box has as many parts as a large box, so there are about 100 identical parts in a large box.

Is it wrong in the textbook?

No, there is no error in this question.

The problem solving process is shown below:

The quantification of the probability of an event occurring introduces "probability". The total number of independent repetitions n, the frequency of event a, the frequency of event a(a)= n, is there a stable value for the frequency fn(a) of a? If there is, the stable value p of frequency n is said to be the probability of event a occurring, denoted as p(a)=p (statistical definition of probability).

p(a) is objective, while fn(a) is empirical. In statistics, the value of fn(a) when n is very large is sometimes used as an approximation of probability.

There are two boxes of parts of the same kind, the first box contains 50, of which 10 are first-class, the second box is 30, of which 18 are first-class, now pick a box from any, and then change the box from a part, the probability of getting a first-class product is.

The calculation process is as follows:

It can be calculated according to the title:

The probability of getting the first box, and the probability of getting the first class in the first box.

The probability of getting the second box, and the probability of getting the first class in the second box.

Total probability: So the probability of getting a first-class product is.

The probability of getting the first box, and the probability of getting the first class in the first box.

The probability of getting the second box, the probability of getting the first class in the second box, the total probability.

The probability of the first drawing a ball from the first box is 1 2, the probability of drawing a first class is 1 5, and the multiplication of the two is 1 10The probability of drawing a ball from the second box is 1 2, the probability of drawing a first class is 18 30, multiplying the two is 9 30, and adding the probabilities of the two cases gives 1 10 + 9 30 = 6 15

The probabilities are 2 5 and 690 1421, respectively. The specific steps for answering the question are as follows:

Conditional probability, using the Bayesian formula.

Equal to 690 1421