Average time between successive occurrences of earthquakes? In any given year, the probability of an earthquake greater than Magnitude $6$ occurring in the Garhwal Himalayas is $0.04$. The average time between successive occurrences of such earthquakes is ____ years.

My attempt:
Somewhere, answer is given $25$ years.

This means $4$ out of $100$ years will face an earthquake greater than Magnitude 6.So average time between successive earthquakes will be $\frac{100}{4}=25$ years.

I'm not getting how I apply $\frac{1}{0.04}=25$ with given probability and average time.

Can you explain in formal way? Please.

 A: Let $X$ be the number of years to the next earthquake. 
Then $X$ is a geometric random variable and the
probability that the next earthquake happens in $k$ years is
$P(X = k) = (1-p)^k \cdot p$
where $p = 0.04$.
The expected value of $X$ is
$E(X) = 1/p = 1/0.04 = 25 \:\text{years.}$
And this can also be interpreted as if you waited a long period
of time the average time between earthquakes would be $25$ years. 
(or you can think of it that the distribution of the length of the gaps between earthquakes will follow a geometric distribution)
A: Suppose the probability that an earthquake occurs is p in one year. We assume that the probability of an earthquake is independent from the number of eqrthquake in the past.  Therefore the probability that you do not occour an earthquake in n years is $(1-p)^n$. Now can consider smaller intervals than one year like half a year. We assume that the occurence of an earthquake in half a year is $\frac{p}{2}$. And the number of half-year intervals is $ n\cdot 2$. At least n years you do not occur an earthquake. We denote the random variable as $X$.  The probability that you do not occour an earthquake in 2n half-year intervals is 
$P(X\geq n)=(1-p/2)^{2n}$
The intervals can be further reduced. 
$P(X\geq n)=(1-p/3)^{3n}$
$P(X\geq n)=(1-p/4)^{4n}$
$P(X\geq n)=(1-p/5)^{5n}$
...
More generally it is 
$P(X\geq n)=(1-p/m)^{m\cdot n}$
m=Number of intervals per year.
Let $m \to \infty$. The random variable X becomes continuous.
$$P(X\geq n)=\lim_{m \to \infty}(1-p/m)^{m\cdot n}=e^{-p\cdot n}$$
The converse probability is $$P(X\leq n)=1-e^{-p\cdot n}$$
Replacing p by $\lambda$ and $n$ by $x$ we have
$P(X\leq x)=1-e^{-\lambda \cdot x} \ \ \forall \ x>0$
This is the probability of an exponential distributed random variable. The expected value is $\frac{1}{\lambda}$. In your case $\lambda=p=0.04$. Therefore $E(X)=\frac{1}{0.04}=\frac{100}{4}=25$
