Conditional probability with 3 events i'm struggling to understand how this answer for a past paper is correct.
Question:
In a lake there are 10 fish, 3 of which are tagged. 3 fish are caught randomly from the lake without replacement. What is the probability that the first two fish that are caught are tagged but not the third?
It also gives us that P(A int B int C) = P(C given A int B)*P(B given A)*P(A)
I understand that we need to use the above formula, but in the answers it suggests that P(C given A int B) is 1/7, which I cannot deduce why.
Any help is greatly appreciated.
 A: I disagree with you.  Think about it this way.
The probability that the first fish caught is tagged is $\frac{3}{10}$.
Given that the first fish was tagged, there are now $9$ fish total, of which $2$ are tagged; so, the probability that the second fish is tagged, GIVEN that the first was, is $\frac{2}{9}$.
Finally, GIVEN that the first two caught were tagged, there are now $8$ fish left, $1$ of which is tagged. So, the conditional probability of catching an untagged fish is $\frac{7}{8}$.
So, all told, you get a probability of
$$
\frac{3}{10}\cdot\frac{2}{9}\cdot\frac{7}{8}=\frac{7}{120}.
$$
A: Define $A=\mbox{the first fish is tagged}$, $B = \mbox{the second fish is tagged}$ and $C = \mbox{the third fish isn't tagged}$. We are willing to calculate $P(A \cap B \cap C)$. Now, $P(A)=3/10$, because there are $3$ tagged fishes of $10$. Similarly, $P(B\mid A) = 2/9$, because there are now $2$ tagged fishes out of $9$. Finally, $P(C \mid A \cap B) = 7/8$, because there are now $7$ non-tagged fishes out of $8$. This gives the result $P(A \cap B \cap C)=(7/8)(2/9)(3/10)=42/720=7/120$.
