Or this idea can be changed again ... that is to say, imagine a model like this and turn it into a ball drawing problem ... If an A ball is drawn, then reduce or increase another ball so that the probability of reduction is the same every time. ....

Suppose there are m balls in a box, there are a good ball and M-A bad ball, and the good ball will become a bad ball as soon as it is taken out and put back .. This example will be the same as the title condition, except that P = A/M and P = 1/M can be deduced .. I pushed it to the third time and found the law. ..

The law is .. a (m-1) (n-1)/m n. If you use your p and p, it means: P (1-P) (n- 1) is simpler. ..

The specific approach is:

N= 1:

............. has ...................., but it isn't.

............ Morning/Afternoon ................. (M-A)/M

N=2: ... is ... No, ............. is the ... number. ..

... (a-1)/m ... (m+1-a)/m ... (m-a)/m.

Typesetting is difficult, so I only wrote it twice, and I developed it in the form of a tree, and after every situation, there are two situations. ..

The probability of the first draw is a/m.

The probability of the second draw is a/m * (a-1)/m+(m-a) m * a/m = a (m-1)/m 2.

You can do three by yourself, and you can sum up my formula above ... I'm glad ... ha ... I haven't done math problems for a long time. ..