According to the meaning of the question, all the values of x are 2 3 4 5 6.

(There is an explanation below. When solving this problem, we should think that no matter how many times we draw, there must be only two black balls in the ball, and the last one must be a black ball. For example, x=5, then the fifth ball must be a black ball, and the first four balls are randomly arranged. With a clear idea, you can start to solve the problem. Solving problems requires knowledge of permutation and combination. I think you should be a student in Grade Two or Grade Three.

P(x=2)=C(2 1)/A(62)=2/30

p(x = 3)= C(2 1)C(4 1)A(22)/A(63)= 16/ 120

p(x = 4)= C(2 1)C(42)A(33)/A(64)= 72/360

p(x = 5)= C(2 1)C(43)A(44)/A(65)= 192/720

p(x = 6)= C(2 1)C(44)A(55)/A(66)= 240/720

Yes: (distribution list. Draw it yourself ..)

∴ex = 2 * 2/30+3 * 16/ 120+4 * 72/360+5 * 192/720+6 * 240/720 = 14/3。