It is known that there are m defective products in n products, which can be extracted and not put back, and the probability of the I-th defective product can be found. The answer said, whether I put it back or not, I got it for the first time.
Yes, because the probability of each individual being drawn is equal, equal to m/n, which is the reason of drawing lots and drawing lots.
The result of the previous I times is the sample space, * * has A(n, I) = n (n-1) (n-2) ... (n-I+1).
Among them, the I-th defective product is A(m, 1)*A(n- 1, I-1) = M (n-1) (n-2) ... (n-I+/kloc-.
Their ratio is probability, and m/n is obtained.