Probability-Bayes Theorem

Problem:

The Belgian 20-Frank coin (B20), the Italian 500-Lire coin (L500), and the Hong Kong 5-Dollar (HK5) are approximately the same size. Coin purse one (C1) contains six of each of these coins. Coin purse two (C2) contains nine B20s, six I500s, and three HK5s. A fair four-sided die is rolled. If the outcome is {1}, a coin is selected randomly from C1. If the outcomes belong to {2, 3, 4}, a coin is selected randomly from C2.

a) Find P(HK5).

b) Find P(C2|HK5).

I've drawn a tree diagram.

Now for a), is this correct? $\displaystyle (3/4)(1/6)+(1/4)(1/3)=5/24 $

And for b), do I use Bayes Theorem? Is this correct? $\displaystyle ((3/4)(1/6))/((3/4)(1/6)+(1/4)(1/3))=3/5 $

Thanks for the help!