What is the value of $P(A \cap B\ |\ C)$? 
Let $A,B,C$ be three events such that $P(A)=0.4,P(B)=0.5,P(A \cup B)=0.6,P(C)=0.6$ and $P(A \cap B \cap C^c)=0.1$. Then $P(A \cap B\ |\ C)$ is equal to
$(1)$ $\frac {1} {2}.$
$(2)$ $\frac {1} {3}.$
$(3)$ $\frac {1} {4}.$
$(4)$ $\frac {1} {5}.$

My attempt $:$
We know that $A \cap B = (A \cap B\ |\ C) \cup (A \cap B\ |\ C^c)$. Let $E=(A \cap B\ |\ C)$ and $F=(A \cap B\ |\ C^c)$. Then clearly $E$ and $F$ are mutually exclusive events. So $P(A \cap B) = P(E)+P(F).$  We have to find out $P(E).$
Now $P(E) = P(A \cap B) - P(F) = P(A)+P(B) - P(A \cup B) - \frac  {P(A \cap B \cap C^c)} {P(C^c)}=0.4+0.5-0.6 - \frac {0.1} {0.4}=0.3-0.25=0.05=\frac {1} {20}.$
Which does not equal to any of the options given above. What are my mistakes? Please help me in showing my misconceptions.
Thank you in advance.
 A: It should be
$$A \cap B = (A \cap B \cap C) \cup (A \cap B \cap C^c)$$
Guide:
First find $P(A \cap B)$.
Then find $P(A \cap B \cap C)=P(A \cap B) - P(A \cap B \cap C^c)$
Then find $P(A \cap B|C)$ using definition of conditional probability.
A: $E$ and $F$ are not mutually exclusive events; there not even events.
One good idea is to compute $P(A \cap B)$:
$$\frac{6}{10} = P(A \cup B)  =P(A) + P(B) - P(A \cap B) = \frac{9}{10} - P(A \cap B)$$
and it follows that $P(A \cap B) = \frac{3}{10}$.
The law of total probabilities says in this case:
$$P(A \cap B) = P(A \cap B | C)P(C) + P(A \cap B| C^c)P(C^c)$$
as $C$ and $C^c$ are mutually exclusive events. Hence $P(C^c) = 1-P(C)$
Also by definition:
$$P(A \cap B | C^c)= \frac{P(A \cap B \cap C^c)}{P(C^c)}$$
Now you know all necessary values to solve for $P(A \cap B|C)$. 
A: We can get $$ P(A\cap B)=0.3$$ from the usual formula.
Then we have $$ P(A\cap B|C) = \frac{P(A\cap B\cap C)}{P(C)}.$$
And finally we have that $$P(A\cap B \cap C) + P(A\cap B\cap C^c) = P(A\cap B) $$
Combining these three equations with the known values gives one of the answers.
Your problem has origin in your first step when you wrote $$A\cap B = (A\cap B|C)\cup (A\cap B| C^c).$$ This makes no sense as $(A\cap B|C)$ is not an event. Instead you want $$A\cap B = (A\cap B\cap C) \cup (A\cap B\cap C^c).$$ Which, together with the fact that the two events whose union we take on the RHS are disjoint, gives my third equation 
