Solution: (1) There are n white balls in the bag, which means that n/7 = 3/7. ∴ n = 3。

That is, there are three white balls in the bag.

Remember that the event of "taking the ball twice was terminated" was A, so the first time A didn't get the white ball, and the second time B got the white ball.

So P(A)=(4/7)*(3/6)=2/7.

(2) Because A takes the ball first, A only takes the ball at 1, 3 and 5 times.

Note that the event "A gets the white ball" is B, and "The ball taken out for the first time is a white ball" is Ai, (I = 1, 2, …5).

Then P(B)=P(A 1+A3+A5).

=P(A 1)+P(A3)+P(A5)

=3/7+(4/7)*(3/6)*(3/5)+(4/7)*(3/6)*(2/5)*( 1/4)=22/35

=22/35