The day I finally get the second sunny day I leave. What is the probability that I stayed exactly one week? Each day in Iceland, it rains with a probability p=0.8.
Denote X the number of days I stay until I've had 2 sunny days in my holidays.
What is the p.m.f. of X?
The day I finally get the second sunny day I leave.
What is the probability that I stayed exactly one week?
My Attempt
I know I can model this if it was only one sunny day, but I'm not sure how to do it if it were two. For one sunny day, I can model this as a geometric distribution, which is: $X\sim Geom(p)$ where $p=0.8$. So I need to find the probability of staying for $7$ days:
$$\mathbb{P}(X=k)=(1-p)^{k-1}p=(1-0.2)^{7-1}\cdot 0.2$$
But how would I compute it if I wanted to see two sunny days?
 A: The probability that the $k$th day is the second sunny day is the probability that there was exactly one sunny day in the first $k - 1$ days multiplied by the probability that the $k$th day is sunny.  There are $k - 1$ ways for one of the first $k - 1$ days to be sunny.  The probability that $k - 2$ of the first $k - 1$ days are rainy and the other day is sunny is $p^{k - 2}(1 - p)$.  The probability that the $k$th day is sunny is $1 - p$.  Hence, the probability that the $k$th day is the second sunny day is
$$\Pr(X = k) = \binom{k - 1}{1}p^{k - 2}(1 - p)^2 = (k - 1)(0.8)^{k - 2}(0.2)^2$$
Setting $k = 7$ will give you the probability of staying exactly one week.
A: With so many variants of the negative binomial distribution floating around, I would use the binomial distribution coupled with commonsense, as @Sukhoi234 has hinted. This would obviate the need for memorizing a formula, and also help in answering minor deviants.
Only $1$ sunny day in first $6$ followed by $1$ sunny day, so
$\left[\binom61\cdot0.8^5\cdot0.2\right]\times 0.2$
