If M products contain m scraps, take two of them, 5

This question tests the conditional probability (note: [ is a superscript, such as c[1]m means to take 1 from m) solution: record event a as "one is a non-conforming product", b is "the other is also a non-conforming product". Because.

p(a) = p (take out one qualified product, one unqualified product) + p (take out both non-conforming products).

c[2]mm(2m-m-1)/m(m-1)

p(ab)=p (take out both are non-conforming products) = c[2]m c[2]m=m(m-1) m(m-1).

So the probability is as follows:

p(b|a)=p(ab)/p(a)=(m-1)/2m-m-1

With [] is the superscript, and with m is the subscript.

1) C [M-1] (M-1) takes out a scrap product, which means that there are M-1 scrap products in M-1 product, pick one scrap product, 2) 1-C [M-M] M-C [M-1] (M-1) Remove the case where both are scrap products, and subtract both of them are good cases.

3) 1-c [m-m] m minus two are good products.

If there are 2 pieces of waste in 10 products, take 2 of them and find the probability that one of the two pieces is not a waste product, and the other piece is a waste product.

The problem of conditional probability

Let the probability of one piece of scrap be taken out is p(a), and the probability of another piece is also p(ab)p(a)=c2 1 times c8 1 c10 2=17 45p(ab)=(c10 2) 1=1 45

Then both pieces are scrap probabilities p*=p(ab) p(a)=1 17, then the obtained p=1-p*=16 17

The probability of scrap in 2 pieces is: m m%x2

1. There are m defective products in M products, and two of them are taken. On the premise that one of the products is a waste product, the probability that the other is also a waste product is (m-1) (2m-m-1);

2. Probability, also known as "probability", is a reflection of the probability of a random event. Random events are events that may or may not occur under the same conditions;

3. After a lot of trial and error, M n is often getting closer and closer to a certain constant, which is the probability of event a, which is often expressed by p (a).

Solution: Record event A as "one is a non-conforming product" and B is "the other is also a non-conforming product". Because p(a) = p (take out one qualified product, one unqualified product) + p (take out two non-conforming products).

(m,1)*(n-m,1))/(n,2)+(m,2)/(n,2)=(2m(n-m)+m(m-1))/n(n-1)

p(ab)=p (take out both are non-conforming products) = (m,2) (n,2)=m(m-1) n(n-1).

So the probability is as follows:

p(b|a)=p(ab)/p(a)=(m-1)/2n-m-1

Answer: The probability of containing 1 and 2 defective products is c(m-m, 1)xc(m, 1) c(m, 2) and c(m, 2) c(m, 2), and dividing the latter by the sum of the two is the answer, i.e., c(m, 2) [c(m, 2)+c(m-m, 1)xc(m, 1)].

Let a="At least one item is scrap" and b="The other item is scrap".

p(b/a)=p(ab)/p(a)=c(m 2)/c(m 2)=(m-1)/(2m-m-1)

m-1)/(2m-m-1)。

This is a conditional probability problem, let the event that at least one of the defective products is A, and the events where both of them are defective products are B; A1 and A2 are the events when 1 and 2 defective products are just taken. Obviously a=a1+a2

p(b/a)=p(ab)/p(a)=p(b)/p(a)=p(b)/p(a1+a2)

p(b)/[p(a1)+p(a2)]

c(2,m)/c(m,2) /c(m,1)*c(m-m,1)/c(m,2)+c(m,2)/c(m,2)]

m-1)/(2m-m-1)

Regardless of the priority and defective products, N pieces are taken from N pieces, and there are a total of CN [N] methods.

If n pieces contain k defective products, then there must be n-k high-quality products.

There are a total of m pieces of defective products, and there are a total of cm [k] possible to get k pieces.

There are n-m pieces of high-quality products, and there are a total of C(N-M) [N-K] kinds of loose celebrations that may be dismantled and contained when they are taken n-k.

Therefore, the N products taken out contain k defective products, and the combination of n-k high-quality products has a total of cm[k]*c(n-m)[n-k] possibilities.

Divide these conditional methods by the total formulations, and the probability of satisfying the conditions is: cm[k]*c(n-m)[n-k] cn[n].

Conditional probabilities are referred to on the premise that one of them is known to be scrap.

Then this question should be done like this.

p = (probability that both are scrap) (the probability that one of them is a dusty scrap) the probability that both are scrap = c(m, 2) c(n, 2) the probability that one of them is scrap = the probability that both are scrap + 1 is **1 is the probability of scrap.

c(m,2) c(n,2)+c(m,1)c(n-m,1) c(n,2) in town

p=c(m,2)/[c(m,2)+c(m,1)c(n-m,1)]m!/[m!+2!

m-2)!m(n-m)]m(m-1)/[m(m-1)+2m(n-m)](m-1)/(2n-m-1)

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**m defective product n-m

Among the n pieces taken out, **: m pieces, defective products: n-m pieces.

Therefore, there is p=c(m,m)c(n-m,n-m) c(n,n) where c(n,n) means to take n from n.

Hello! The implication of this problem is to find the probability that both of them are scrap (denoted b) under the condition that at least one scrap (denoted a) is, since ab=b, so p(b|a)=p(ab)/p(a)=p(b)/p(a)=/=m(m-1)/。The Economic Mathematics team will help you solve the problem, please adopt it in time. Thank you!