EXAMPLE 5.2 (Example 5.1 continued)

EXAMPLE 5.1 continued

Possible $X$ values are $x=100,500$, and 1000. Computing row totals from the joint probability table yields

$$p_X\left(100\right)=p\left(100,500\right)+p\left(100,1000\right)+p\left(100,5000\right)=.30+.05+0=.35$$

$p_X\left(500\right)=.15+.20+.05=.40$ $p_X\left(1000\right)=1-\left(.35+.40\right)=.25$

The marginal pmf of $X$ is then

$$p_X\left(x\right)=\{\begin{array}{ll}.35& x=100\\ .40& x=500\\ .25& x=1000\\ 0& \text{ otherwise }\end{array}$$

From this pmf, $P\left(X\ge 500\right)=.40+.25=.65$, which we already calculated in Example 5.1. Similarly, the marginal pmf of $Y$ is obtained from the column totals as

$$p_Y\left(y\right)=\{\begin{array}{cc}.55& y=500\\ .35& y=1000\\ .10& y=5000\\ 0& \text{ otherwise }\end{array}$$