Two particles are located somewhere between a left and right partition 4 inches apart as shown in the diagram below. The probability of each particle being located a distance x from the left partition is given by the following two equations:
$$P_a = 0.5 - \frac{x}{8}$$
$$P_b = \frac{x}{8}$$
What is the probability that particle a will be found to the left of particle b?
My attempt uses an approximation method much like finding the area under a graph by the trapezoid method. I divide the 4 inch region between the partitions into n equal widths and sum all the probabilities of particle a being in each region with particle b being to the right. The more divisions there are the more accurate the answer.
Example 1:
$n=4: P = \frac{(7\cdot 15)+(5\cdot 12)+(3\cdot 7)+(1\cdot 0)}{16^2} = .7266$
Eample 2:
For $n=8: P = \frac{(15\cdot 63)+(13\cdot 60)+(11\cdot 55)+(9\cdot 48)(7\cdot 39)+(5\cdot 28)+(3\cdot 15)+(1\cdot 0)}{64^2} = .7861$
As $n$ increases the accuracy increases and will be precise as $n\to \infty$
My question, is there a mathematically more elegant way of caculating this?
So, I think your $P_a, \ P_b$ are the probability density functions of the position of your particles, since each integrates to 1 over the interval $[0, 4]$. I'm also going to assume that positions of the particles are independent. To compute the probability simply evaluate the integral
$$ \int_0^4 \int_0^y \Big(\frac{1}{2} - \frac{x}{8} \Big)\Big(\frac{y}{8}\Big) \ dx \ dy = 4 \Big(\frac{1}{3} - \frac{1}{8}\Big) \approx 0.8333. $$