Binomial Distribution confusion

#1
Hi, I'm currently confused on a question and don't quite understand how to solve it.

The question is Suppose 75% of all households with telephone service have voicemail. You are conducting a survey about voicemail usage and you randomly call 10 households.

What is the probability that the first 7 households have voicemail and the last 3 do not?


So, how exactly would I do both the first 7 households have voicemail and the last 3 don't? I've tried doing P(X<=7) but not sure how to incorporate the "3 do not" part.
 

rogojel

TS Contributor
#2
hi,
it is not a binomial distribution stricly. You need to calculate P(A*B) where A is getting 7 voicemails out of 7 and B is getting 3 without a voicemail out of 3.

Regards