Binomial Distribution Question

#1
Hey guys, I have a bit of trouble understanding this binomial distribution question

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?
 
#2
The probability of not reaching a potential client after one call is (79/80). The probability of not reaching a potential client after k calls is (79/80)^k. Therefore, the probability of reaching at least one potential client after k calls is 1–(79/80)^k, which must be greater than 0.9.

Solve the resulting inequality for k.
 
#3
Hey, thanks very much for your help. I tried working around the question based on your advice. Could you check if it's on the right track?

1-(79/80)^k > 0.9

-(79/80)^k > -0.1

(79/80)^k < 0.1

ln(79/80^k) < ln(0.1)

k*ln(79/80) < ln(0.1)

k < 183.0531013

I feel like I'm missing something here, any help would be greatly appreciated. Thanks once again.
 

Dason

Ambassador to the humans
#4
Is your issue that you end up with k < 183.05 instead of k > 183.05? Check your very last step. Hint: evaluate ln(79/80) and check the sign.
 
#5
Oh I see, I didn't realize the negative sign for ln(79/80). Thanks a bunch :).
By the way, does "k" in this context refer to "n-x" as per the binomial formula for probability nCx*p^x*(1-p)^n-x?
 
#6
My pleasure. Yup, Dason's lesson is manifest: Always check the sign of a divisor or multiplier when manipulating an inequality.

No, k is the number of successive calls made, which is the same as n. Your x would be the number of successful calls in k (or unsuccessful calls, depending on which way around you frame the binomial expansion).