Use binomial expansion for (p+(1-p))^10=p^10+10p^9(1-p)+45p^8(1-p)^2+120p^7(1-p)^3+... where each term represents an exact probability. In this expansion p can be either 0.1 or 0.9 because the coefficients of the expansion are symmetrical. If p=0.9 then the first term is the probability that none will withdraw, the second that one will withdraw, and so on. So we need 120p^7(1-p)^3 where p=0.9.
This is 120*0.9^7*0.1^3=0.0574 or 5.74% approx.