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.

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 $P(A \cap B \cap C)=(7/8)(2/9)(3/10)=42/720=7/120$.
