Finding $P(A \cap B | C)$, knowing A and B are Independent. So suppose I know $P(A), P(B), P(C|A), P(C|B)$, and I know that $A, B$ are independent. I'm trying to find $P(A \cap B | C)$. I first thought that, since $A$ and $B$ are independent, the $A$ and $B \cap C$ must be independent. However, I was wrong about this, and I can't seem to find a solution. What I've tried is:
$P(A \cap B|C) = \frac{P(A \cap B \cap C)}{P(C)} = \frac{P(C|A \cap B)P(A)P(B)}{P(C)}$. However, I still don't know what $P(C|A \cap B)$ would be.
EDIT: If some context helps, here's the full problem:
"The probability that that a person orders a steak is $1\times10^{-10}$. The probability that a person orders a hotdog is $1\times10^{-11}$. The probability of a person liking their food given they ordered a steak is 0.7, and the probability of a person liking their food if they ordered a hotdog is 0.9.
What is the probability of a person liking their food, assuming that a person ordering steak and a person ordering hotdogs are independent events?
Given that a person liked their food, what's the probability they ordered hotdogs?
What is the probability someone ordered a hotdog and a steak given they liked their food?"
Thanks!
 A: I think you have posted two different questions, I assume in your notation that $C$ is the event someone likes their food and $A,B$ are respectively them ordering a hot dog and a steak. And you seek the probability they like their food.
For the worded problem the probability they like their food is a bit tricky, what happens if they don't order food? Let's make this super simplistic and assume we are actually after, the probability they liked their food given they ordered something. They could have either ordered a steak, a hotdog or both. With respective probabilities: $10^{-10} \cdot (1-10^{-11}) , 10^{-11} \cdot (1-10^{-10}) $ and $10^{-10} 10^{-11}$ Hence given someone ordered food the probability of their orders are steak, hotdog, both : $\frac{10^{-10} \cdot (1-10^{-11})}{10^{-10} \cdot (1-10^{-11})+ 10^{-11} \cdot (1-10^{-10})+10^{-10} 10^{-11}} , \frac{10^{-11} \cdot (1-10^{-10})}{10^{-10} \cdot (1-10^{-11})+ 10^{-11} \cdot (1-10^{-10})+10^{-10} 10^{-11}}, \frac{10^{-10} 10^{-11}}{10^{-10} \cdot (1-10^{-11})+ 10^{-11} \cdot (1-10^{-10})+10^{-10} 10^{-11}}$
And from here you can use the law of total probability, conditioning on what they ordered to find out the probability they liked their order.
Your symbol question is slightly different, it states the probability they ordered both a hotdog and steak given that they enjoyed their food. Do you also want the answer to this?
