## Precalculus (10th Edition)

$P(1)=P(3)=P(5)=\frac{2}{9}$, $P(2)=P(4)=P(6)=\frac{1}{9}.$
Here if the probability of getting an even number is $x$, then $P(2)=P(4)=P(6)=x$ and $P(1)=P(3)=P(5)=2x$. We know that $P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1\\2x+x+2x+x+2x+x=1\\9x=1\\x=\frac{1}{9}.$ Therefore $P(1)=P(3)=P(5)=\frac{2}{9}$, $P(2)=P(4)=P(6)=\frac{1}{9}.$