Thread: consecutive cases

Hi I have this question, which has completely stumped me..Does anyone has any suggestions on how to do it?

Can anyone give any hint?

Thanks!

consider it as a geometric distribution with probability of success (that is, probability of finding a disease) =$\displaystyle p^3$

I still am not sure if this is really a geometric distribution:

so, am i supposed to use the pmf $\displaystyle p^3(1-p)^x$??

that would mean we are calculating for $\displaystyle p^3 , \mbox{and} , q^3$. thats either 3 success or 3 failures.. but we do have $\displaystyle ppq $ (2 successes and 1 failure) and $\displaystyle pqq$(1 success and 2 faliures) as well. Dont we need to count those? Deriving the pmf using these scenarios is really confusing me.