# 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?

• This is a version of the negative binomial distribution (take care: there are at least eight different versions) Apr 23, 2021 at 7:58
• Can't we find the probability , by considering that , upto the sixth day , there have been 5 rainy days/1 sunny day which can be easily calculated by binomial distribution formula ??@Henry
– user898773
Apr 23, 2021 at 8:18

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.
Only $$1$$ sunny day in first $$6$$ followed by $$1$$ sunny day, so
$$\left[\binom61\cdot0.8^5\cdot0.2\right]\times 0.2$$