1. ## Conditional Probability #3

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)?

2. the first bag contains totally 5 coins,
if u take the a gold coin from it ,
p(C|A)=3/5

3. Why is $\displaystyle p(A)=\frac{1}{2}$?

4. there are two bags ,
one is bag A, one is bag B,
so P(A) = 1/2

5. p(A) is $\displaystyle \frac{1}{2}$ because there are 2 bags???

6. Originally Posted by chengbin
p(A) is $\displaystyle \frac{1}{2}$ because there are 2 bags???
exactly