So the probability that student A receives the activity information is 1-( 1-kn)2=2kn? k2n2
(II) When k=n, m can only take n, with P(X=m)=P(X=n)= 1.
When k < n, the integer m satisfies k≤m≤t, where t is the smaller of 2k and m. Because the total number of basic events contained in "Miss Li and Teacher Zhang send activity information to K independently and randomly" is (Ckn)2, when X=m, the number of students who receive information sent by two teachers at the same time is 2k-m, and the number of students who only receive information forwarded by Miss Li or Teacher Zhang is m-.mkcm. kn？ k=CknCm？ kkCm？ kn？ k
P(X=M)=CknC2k？ mkCm？ kn？ k(Ckn)2=C2k？ mkCm？ kn？ kCkn？
When k ≤ m < t, p (x = m) < p (x = m+ 1)? (m-k+ 1)2≤(n-m)(2k-m)？ m≤2k-(k+ 1)2n+2
If k ≤ 2k-(k+ 1) 2n+2 < t holds, then when (k+ 1)2 is divisible by n+2,
K ≤ 2k-(k+1) 2n+2 < 2k+1-(k+1) 2n+2 < t, so P(X=M) is in m = 2k-(k+1) 2n.
When (k+ 1)2 is not divisible by n+2, P(X=M) reaches the maximum value at m=2k-[(k+ 1)2n+2] (note: [x] means the largest integer not exceeding x).
It is proved that k ≤ 2k-(k+1) 2n+2 < t.
Because 1 ≤ k < n, 2k-(k+ 1)2n+2-k=kn? k2？ 1n+2≥k(k+ 1)？ k2？ 1n+2=k？ 1n+2≥0
And 2k-(k+ 1)2n+2-n=? (n？ K+ 1) 2n+2 < 0, so 2k-(k+ 1) 2n+2 < n, obviously 2k-(k+ 1) 2n+2 < 2k.
Therefore, k ≤ 2k-(k+1) 2n+2 < t.