EXAMPLE 5.10

When a certain method is used to collect a fixed volume of rock samples in a region, there are four resulting rock types. Let $X_1,X_2$, and $X_3$ denote the proportion by volume of rock types 1, 2, and 3 in a randomly selected sample (the proportion of rock type 4 is $1-X_1-X_2-X_3$, so a variable $X_4$ would be redundant). If the joint pdf of $X_1,X_2,X_3$ is

$$\begin{array}{rl}f\left(x_1,x_2,x_3\right)& =\{\begin{array}{ll}k{x}_1{x}_2\left(1-x_3\right)& \text{ if }0\le x_1\le 1,0\le x_2\le 1,0\le x_3\le 1,x_1+x_2+x_3\le 1\\ 0& \text{ otherwise }\end{array}\end{array}$$

then $k$ is determined by

$$\begin{array}{rl}1& ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f\left(x_1,x_2,x_3\right)d{x}_{3}d{x}_{2}d{x}_1\\ & ={\int }_0^1\left\{{\int }_0^{1-x_1}\left[{\int }_0^{1-x_1-x_2}k{x}_1{x}_2\left(1-x_3\right)d{x}_3\right]d{x}_2\right\}d{x}_1.\end{array}$$

This iterated integral has value $k/144$, so $k=144$. The probability that rocks of types 1 and 2 together account for at most $50\%$ of the sample is

$$\begin{array}{rl}& P\left(X_1+X_2\le 0.5\right)\\ & =\iiint f\left(x_1,x_2,x_3\right)d{x}_{3}d{x}_{2}d{x}_1\\ & \text{ subject to }\left\{\begin{array}{cc}0\le x_i\le 1& \text{ for }i=1,2,3\\ x_1+x_2+x_3\le 1,x_1+x_2\le 0.5& \end{array}\right\}\\ & ={\int }_0^{0.5}\left\{{\int }_0^{0.5-x_1}\left[{\int }_0^{1-x_1-x_2}144x_1{x}_2\left(1-x_3\right)d{x}_3\right]d{x}_2\right\}d{x}_1\\ & =\mathrm{0.6066.}\end{array}$$