✔ 最佳答案
a) n classes, k students, and m classes for each student.
assume 1 SPECIFIC class that no one choose,
for the 1st student,
the 1st class chosen is (n-1) / (n)
the 2nd class chosen should be (n-2) / (n-1)
......
the m th class chosen should be (n-m) / (n-m+1)
So, the probability for the 1st student: (n-1)/(n) x (n-2)/(n-1) x (n-3)/(n-2) x ... x (n-m) / (n-m+1) = (n-m) / (n)
the probability for the 2nd student: (n-m) / (n)
......
the probability for the k th student: (n-m) / (n)
suming them up, the probability that 1 SPECIFIC class that no one choose:
(n-m)^k / (n)^k
there exist n classes, so the probability for ONE class that no one choose:
(n) (n-m)^k / (n)^k = (n-m)^k / (n)^(k-1)
b) the expected value of each class:
number of student: k
number of class chosen for each student: m
total number of class chosen: km
number of classes: n
so, the expected value for each class is km / n (evenly distributed)