Probability about three independent exponential random variables Suppose we have three independent exponential random variables $A$, $B$ and $C$  with respective parameters $1$, $2$ and $3$. 
Calculate $P(A<B<C)$.
The hint says this problem could be solved with calculus and without calculus. I am really curious how to approach it with different methods.
 A: Here is how to solve it without much calculus. Remember that for exponential variable $X$ with parameter $\lambda$, $\mathbb{P}(X > x) = \exp(-\lambda x)$ for $x \ge 0$.
We will make use of the following result, that relies on memoryless property of exponential distribution, i.e. $\mathbb{E}(f(X-a) ; X > a) = \mathbb{E}(f(X))$:
$$
  \begin{eqnarray}
   \mathbb{E}( \mathbf{1}_{X > a} \exp(-\mu X)) &=& \mathbb{E}(\exp(-\mu X); X > a) \mathbb{P}(X > a) \\ 
  &=& \exp(-\mu a) \cdot \mathbb{E}(\exp(-\mu (X-a)); X > a) \cdot \exp(-\lambda a) \\
  &=& \exp(-(\mu+\lambda) a) \cdot \mathbb{E}(\exp(-\mu X) )  \\ 
  &=& \exp(-(\mu+\lambda) a) \cdot \frac{\lambda}{\lambda+\mu}
  \end{eqnarray}
$$
Now, use conditioning:
$$ 
\begin{eqnarray}
  \mathbb{P}(C > B > A) &=& \mathbb{E}_A( \mathbb{E}_B( \mathbf{1}_{B>a} \mathbb{P}(C > b ; B=b, A=a) ; A=a)) \\
   &=& \mathbb{E}_A( \mathbb{E}_B( \mathbf{1}_{B>a} \exp(-\lambda_c b) ; A=a)) \\
   &=& \mathbb{E}_A( \frac{\lambda_b}{\lambda_b+\lambda_c} \exp(-a(\lambda_b+\lambda_c)   ) \\  &=& \frac{\lambda_b}{\lambda_b+\lambda_c} \cdot \frac{\lambda_a}{\lambda_a + \lambda_b+\lambda_c}
\end{eqnarray}
$$
With $\lambda_a=1$, $\lambda_b = 2$, $\lambda_c=3$, the answer comes to $\frac{1}{15}$.
A: This can be done from first principles using minimal Calculus (just differentiation and integration of monomials).  By one definition, when $X$ has an Exponential distribution with parameter $\lambda\gt 0$, $\Pr(X\le x) = 1 - \exp(-\lambda x)$.  Therefore, for any $0\lt x\le 1$,
$$\Pr(\exp(-X)\lt x) = \Pr(X \gt -\log(x)) = \exp(\lambda\log(x)) = x^\lambda.$$
By differentiation we find the probability density of $\exp(-X)$ is $\lambda x^{\lambda-1}\ dx$. The solution is now obtained by integrating the joint density of $(A,B,C)$ over the event $A\lt B\lt C$ via this transformation:
$$\eqalign{
\Pr(A \lt B \lt C) &= \Pr(\exp(-C)\lt \exp(-B)\lt \exp(-A)) \\
&= \int_0^1 da \int_0^a 2b\ db\int_0^b 3c^2\ dc \\
&= \int_0^1 da \int_0^a 2b\ b^3\ db\\
&= \frac{2}{5}\int_0^1 a^5\ da \\
&= \frac{1}{15}.
}$$
A: It is rather well known that $$P(A<B)=\frac{\lambda_a}{\lambda_a+\lambda_b}.$$
You should also know the fact that $$\min(A,B)\sim \mathcal E(\lambda_a+\lambda_b),$$ think about the first arrival time of the addition of two Poisson processes, for example.
From there, you could decompose the event $\{A<B<C\}$ into the intersection of the two independent events $$\{A<\min(B,C)\}$$ and $$\{B<C\}$$ to get an intuition of the result.
A: The naive way would be to use conditional probability:
$\mathrm P(A<B<C) = \int_0^\infty \mathrm P(A < B < x) f_C(x) \mathrm d x = \int_0^\infty \int_0^x \mathrm P(A<y) f_B(y) f_C(x) \mathrm d y \mathrm d x$.
A: I have written a program to test the probabilities for three independent exponential random variables A, B and C with respective parameters λa = 1, λb = 2, λc = 3. Here are the results:
P (A < B < C) = 0.3275
P (A < C < B) = 0.2181
P (B < A < C) = 0.2047
P (B < C < A) = 0.0681
P (C < A < B) = 0.1211
P (C < B < A) = 0.0603

P (A < B < C) is not 1/15.
Here is the source code of the program:
Random Rand = new Random(2015);

public Double GetExponential(Double mean)
{
    return -mean * Math.Log(Rand.NextDouble());
}

var totalCase = 0;
for (int i = 0; i < 1000000; ++i)
{
    var p1 = GetExponential(1);
    var p2 = GetExponential(2);
    var p3 = GetExponential(3);

    if (p1 < p2 && p2 < p3)
        ++totalCase;
}
Console.WriteLine(totalCase / 1000000.0);

