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?

  • 1
    $\begingroup$ This is a version of the negative binomial distribution (take care: there are at least eight different versions) $\endgroup$
    – Henry
    Apr 23, 2021 at 7:58
  • 1
    $\begingroup$ 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 $\endgroup$
    – user898773
    Apr 23, 2021 at 8:18

2 Answers 2


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.


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

$\left[\binom61\cdot0.8^5\cdot0.2\right]\times 0.2$


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