I was doing an example, and got the gist of the answer, but not the arithmetic.
Cabs leave with a group of passengers (1, 2, or 3) at rate λ = 10 (per hour). The question wants to know what the mean time is until one taxicab of each type has left.
- There's a 60% chance of a 1-passenger group
- 30% chance of a 2-passenger group
- 10% chance of a 3-passenger group
The answer multiplies the probabilities that at least one car of each group has left (which is $1- prob(0)$ for each type) which makes sense. But after multiplying the parenthesis out, he removes the 1 and reverses all the signs for the mean integral. I'm a little confused - how can he move it to the other side of the equation without the $1$ remaining?
There is a minor but unrelated typographical error in the solution. It should read:
That said, the solution does a poor job of explaining why these things are true. A more complete reasoning might be as follows:
The expected time it takes for at least one taxicab of each type to have left is equal to $$\operatorname{E}[X] = \int_{x=0}^\infty x f_X(x) \, dx = \int_{x=0}^\infty S_X(x) \, dx, \tag{1}$$ where $X$ is the random variable that measures the desired event time, $f_X$ is the probability density function of $X$, and $S_X = 1 - F_X = \Pr[X > x]$ is the survival function of the random event time. This identity $(1)$ holds when $X$ is a continuous random variable with nonnegative support.
In turn, we observe that $$F_X(x) = \Pr[X \le x] = \Pr[X_1 \le x]\Pr[X_2 \le x]\Pr[X_3 \le x], \tag{2}$$ where $X_i$ is the first departure time of a taxicab with $i$ passengers. This is because in order for all three types to have had at least one departure up to time $x$, each type must have had their first cab departing up to time $x$. This leads to the first equation in the solution.
Now we see how the expectation calculation works: we simply write $$S_X(x) = 1 - F_X(x) = e^{-6t} + e^{-3t} + e^{-t} - e^{-9t} - e^{-7t} - e^{-4t} + e^{-10t},$$ then integrate term by term.