A problem about C and a formula explained by classical probability If there are 6 tins of a certain beverage in each box, and 2 of them are unqualified, ask the quality supervision personnel to randomly select 2 tins from them, and what is the probability of detecting unqualified products? I can list them, but the teacher gave me two more methods. The algorithms of C and a are not very good. They can only calculate, let alone understand them 1）P=[A(2,2)+C(2,1)C(4,1)A(2,2)]/6X5=3/5 2)[C(2,2)+C(2,1)C(4,1)]/C(6,2)=3/5

A problem about C and a formula explained by classical probability If there are 6 tins of a certain beverage in each box, and 2 of them are unqualified, ask the quality supervision personnel to randomly select 2 tins from them, and what is the probability of detecting unqualified products? I can list them, but the teacher gave me two more methods. The algorithms of C and a are not very good. They can only calculate, let alone understand them 1）P=[A(2,2)+C(2,1)C(4,1)A(2,2)]/6X5=3/5 2)[C(2,2)+C(2,1)C(4,1)]/C(6,2)=3/5

Let me give you the third way. First, find out the qualified probability C (4,2) / C (6,2) = 2 / 5
So the probability of failure is 1-2 / 5 = 3 / 5
There are two possibilities for the first method
1. A (2,2) stands for two pieces, and the quality inspector just picked out two defective pieces
2. C (2,1) C (4,1) a (2,2) stands for only one defective product and the other one is good, so there is another order for C (2,1) C (4,1) to take two products, which is to take the good ones first and then the bad ones, or take the bad ones first and then the good ones, so multiply by a (2,2)
The second kind
C(2,2)+C(2,1)C(4,1)]/C(6,2)=
1. Inside, C (2,2) stands for two pieces, and the quality inspector just picked up two defective pieces
2. C (2,1) C (4,1) stands for only one defective product and the other is good, so C (2,1) C (4,1)
Pay attention to the order, consider the order or not