$P(x)=\frac{1}{2}, P(y)=\frac{1}{4}, P(z)=\frac{1}{6}$
$\text{all 3 are mutually exclusive so when P(x') at that time }$
$P(y)P(z') \;or \;P(y')P(z)$
so $P(x')=P(y)P(z')+P(y')P(z)$
$=(\frac{1}{4})(\frac{5}{6})+(\frac{3}{4})(\frac{1}{6})=.333$
what is wrong here?