According to the call center statistics of a company, on average, only 1 in 80 calls made by a teleseller can he/she approach a potential client

a) Find the probability that a teleseller fails to approach any potential client in 1000 calls made

b) What is the least number of calls that a teleseller has to make in order to give a probability greater than 0.9 of approaching at least one potential client?

I've worked out a), which comes down to 0.0000034443. However, I'm stuck on b) as I'm not sure how to start. Originally I thought setting 1-P(X=0) would help me get the answer, but then the 0.9 probability threw me off. Could someone give me some advice?