# consecutive cases

• Oct 25th 2010, 08:08 PM
chutiya
mean and variance
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!
• Oct 25th 2010, 10:35 PM
Sambit
consider it as a geometric distribution with probability of success (that is, probability of finding a disease) = $p^3$
• Oct 26th 2010, 06:43 AM
chutiya
I still am not sure if this is really a geometric distribution:

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

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