Actually, the probability that he fails in the first n trials is 0.9 * 0.99 * 0.999 * ... * (1-(0.1^n)) --> 0.89 as n goes to oo. Thus it is very likely that he can not pass at all and this probability is around 0.89, very high.
You misread the question. The success rate at the n'th round is p^n. So, the probability of no success up to step n is: (1-p)(1-p^2)...(1-p^n). To understand whether this converges to 0, we take log. We have: sum(log(1-p^n)) As n->infty, this is O(sum(p^n)). This converges to a finite number for all choices of p, except if p=1, which is a trivial case. That means, the original sequence does not converge to zero. Even if p=.99999, there is still a non-zero probability of never hitting a success. -iCare-
99999, there is still a non-zero probability of never hitting a
【在 I***e 的大作中提到】 : You misread the question. : The success rate at the n'th round is p^n. : So, the probability of no success up to step n is: : (1-p)(1-p^2)...(1-p^n). : To understand whether this converges to 0, we take log. We have: : sum(log(1-p^n)) : As n->infty, this is O(sum(p^n)). This converges to a finite number : for all choices of p, except if p=1, which is a trivial case. : That means, the original sequence does not converge to zero. Even if p=.99999, there is still a non-zero probability of never hitting a : success.