Compute the probability that you will have to wait between $2$ to $4$ hours before you see $4$ leopards.

Question

Suppose you are watching wild life and you expect to see a leopard once every $\frac{1}{2}$ hours. Compute the probability that you will have to wait between $2$ to $4$ hours before you see $4$ leopards.

$\underline{Attempt}$

Exponential distribution with $\lambda =2$, so $f(x)=2e^{-2x}$

then I found $P(2<x<4)= 0.018$

Then the probability we want is $(0.018)^4$ Is it correct?

Answer 1

I do not agree with the solution $X\sim Exp\left(\frac{1}{2}\right)$ is the distribution to see 1 leopard. The distribution to see 4 leopards is a Sum of 4 independent exponentials, thus it is

$$Y\sim Gamma\left(4;\frac{1}{2}\right)=\chi_{(8)}^2$$

Thus

$$\mathbb{P}[2<Y<4]=0.9810-0.8751=0.1239$$

Answer 2

I concede that it is actually Gamma/Erlang. $X\sim Erlang(4, \frac 1 2)$. $P(2<X<4)\approx .12389$