consecutive cases

**chutiya** · Oct 25th 2010, 08:08 PM

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!

**Sambit** · Oct 25th 2010, 10:35 PM

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

**chutiya** · Oct 26th 2010, 06:43 AM

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.

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