probability
Assume you have an uniformly distributed r.v. $U(0,1)$. Use the inversion method to generate random numbers from an exponential distribution with parameter $\lambda$.
I have no idea where to start, can someone please explain the inversion method and help me solve the problem?
I hope I am not repeating concepts you are already familiar with.
Roughly speaking, a random variable can take one of many values. Which exact value will appear as the outcome of your experiment is not known in advance to you. The probability distribution gives the frequency of these values if you were to perform a large number of experiments.
When you condition on a random variable, you fix a particular value for that variable and do the calculations you want. However, you are not supposed to know the value (otherwise, it will not be a random variable!), so you uncondition to get what you will get after a large number of experiments.
In your example, you first condition on $\Lambda$ taking the value of $\lambda$, and then compute the probability of three or more storms using $$ P(X\geq 3| \Lambda = \lambda). $$ Since $\Lambda$ is a random variable fixing it to one particular $\lambda$ does not give you the whole picture because $\Lambda$ can take any value between $0$ and $5$. Some of these values may appear more often than others, and you need to take this into account. And, this is done by unconditioning on $\Lambda$. That is, $$ P(X\geq 3) = \int_0^5 P(X\geq 3|\Lambda = \lambda)f_\Lambda(\lambda)d\lambda $$
One suggestion is to work with copulas. In a nutshell, a copula allows you to separate out the dependency structure of a distribution function. Say, $F_1,F_2,\ldots,F_n$ are the 1D marginals of a distribution $F$ then the copula $C$ is the function defined as
$$C(u_1,u_2,\ldots,u_n)=F(F^{-1}_1(u_1),F^{-1}_1(u_2),\ldots,F^{-1}_n(u_n))$$
This makes $C$ a function from $[0,1]^n$ to $[0,1]$. For instance, if you take the bivariate normal distribution, by doing the computation above, you'll find the Gaussian copula
$$C^{\text{Gauss}}_{\rho}=\int_{-\infty}^{\phi^{-1}(u_1)}\int_{-\infty}^{\phi^{-1}(u_2)}\frac{1}{2\pi\sqrt{1-\rho^2}}\exp\left(-\frac{s_1^2-2\rho s_1s_2+s_2^2}{2(1-\rho^2)}\right)ds_1ds_2$$
I used the package copula in R to illustrate. If you just take the copula as such, it is as if you constructed a probability distribution with the dependency structure of a bivariate normal, but with uniform marginals. So I generated 1000 random vectors from a Gaussian copula with $\rho=0.5$. Here's the code
library(copula);
norm.cop <- normalCopula(0.5); u <- rCopula(1000, norm.cop);
plot(u,col='blue',main='Random variables, uniform marginals, gaussian copula, > rho=0.5',xlab='X1',ylab='X2')
cor(u);
and the result 
I also computed the sample correlation which is $0.5060224$.
I also computed a plot to show you the marginals are indeed uniform
dom<-(1:length(u[,1]))/length(u[,1]);
par(mfrow=c(1,2));
plot(dom,sort(u[,1]),col='blue',main='marginal X1'); abline(0,1,col='red');
plot(dom,sort(u[,2]),col='blue',main='marginal X2'); abline(0,1,col='red');
This is all very nice, but there are a number of pitfalls that have to be discussed:
More can be said and I think the article I quoted in my last item is a nice starting point.
Best Answer
You 'invert' the cumulative probability of the uniform distribution to the exponential distribution by mapping $p$ in $U(0,1)$ to the $x$ that has the same cumulative distribution.
I found a nice document that explains it in more detail: http://www.columbia.edu/~ks20/4404-Sigman/4404-Notes-ITM.pdf
Example 1.1.1 is what you are looking for.