Given two bags, where the first bag contains 3 gold coins and two silver coins and the second bag contains 4 gold coins and 6 coins, determine the probability that a coin chosen randomly between the two bags in gold.

Let A be the event of choosing from the first bag

B be the event of choosing from the second bag

C be the event of choosing a gold coin.

The probability of choosing a gold coin between the 2 bags is

$\displaystyle p(A\cap C) + p(B\cap C)=p(A)\cdot p(C|A) + p(B)\cdot p(C|B)$

Why does $\displaystyle p(C|A)=\frac{3}{5}?$, isn't it $\displaystyle \frac {1}{2}$? Or did they just write the answer reversed (which they shouldn't do)?