Solution: Let A, B, C and D be the first, second, third and fourth questions respectively, and use Mi(i= 1, 2, 3, 4) to indicate that the answer to the first question of A is correct.

If Ni (I = 1, 2,3,4) indicates that the answer to question I is wrong, then Mi and Ni (I = 1, 2,3,4) are opposite events.

P(m 1)= 3/4 P(M2)= 1/2 P(M3)= 1/3 P(M4)= 1/4；

Then p (n1) =1/4p (N2) =1/2p (n3) = 2/3p (n4) = 3/4;

(i) Remember "Students can enter the next round" as event Q,

Then q = m1m2m3+n1m2m3m4+m1n2m3m4+m1m2n3 m4+n1m2n3 m4.

Because the result of each question is independent,

∴p(q)=p(m 1m2m3+n 1m2m3m4+m 1n2m3m4+m 1m2n3m4+n 1m2n3m4)

= P(m 1 m2 m3)+P(n 1 m2 m3 M4)+P(m 1 N2 m3 M4)+P(m 1 m2 n3m 4)+P(n 1 m2 n3m 4)= 1/4

(2) According to the meaning of the question, the possible values of the random variable ξ are 2, 3 and 4.

Because the answer to each question is relatively independent,

∫P(ξ= 2)= P(n 1 N2)= 1/8

P(ξ= 3)= P(m 1m2m 3)+P(n 1n2n 3)= 3/8

p(ξ=4)= 1﹣p(ξ=2)﹣p(ξ=3)= 1/2

∴eξ=2* 1/8+3*3/8+4* 1/2=27/8