From the n prize volumes of 1, 2, 3, ..., n, all the methods for extracting k prize volumes of 1≤k≤n are as follows:
(n- 1)(n-2)(n-3)..(n-k+ 1)
From the probability formula of classical probability
P=(n? 1)(n? 2)(n? 3)..(n? k+ 1)n(n? 1)(n? 2)(n? 3)..(n? k+ 1)= 1n。
So the answer is1n.