(1) Let the probability that the pieces that have been contacted twice before are all white be P 1 and the probability that they are all black be P2.

Then p = p1+p2 = 47× 36+37× 26 = 37;

(2) Let one white chess be pulled out as event A, two white chess be pulled out as event B, three white chess be pulled out as event B, and four white chess be pulled out as event D, then

P(A)=C 14C47=435，P(B)=C24C23C47= 1835，P(C)= c34c 13c 47 = 1235，P(D)=C44C47= 135，

∴eη= 1×435+2× 1835+3× 1235+4× 135= 167.