Chapter 39 - Wave Functions and Uncertainty - Exercises and Problems - Page 1174: 1

$20\%,\;10\%$

We know that $$P_A+P_B+P_C+P_D=1$$ Plug the known; $$0.4+0.3+P_C+P_D=1$$ Hence, $$ P_C+P_D=0.3\tag 1$$ We are told that $P_C=2P_D$, plug that into (1) $$ 2P_D+P_D=0.3$$ Thus, $$P_D=0.1$$ Plug into (1); And hence, $$P_D=0.2$$ Therefore, the probabilities of C $$P_C=\color{red}{\bf 20}\%$$ and the probabilities of C $$P_D=\color{red}{\bf 10}\%$$