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?"


  • 1
    $\begingroup$ $A$ and $B$ being independent does not mean $A$ and $B\cap C$ are independent. For an extreme case, consider $C = A$. $\endgroup$
    – Arthur
    Sep 13, 2021 at 21:41
  • $\begingroup$ Right ya, I saw that mistake after my final answer didn't make sense. But I'm still not sure how to solve the problem. $\endgroup$
    – BeepBoop
    Sep 13, 2021 at 21:45
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    $\begingroup$ Wouldn't we need to know something about the probability someone who ordered neither steak nor hotdog likes their food? $\endgroup$
    – aschepler
    Sep 13, 2021 at 21:48
  • $\begingroup$ Unsure about that, but this was the problem I was given. I've pasted it pretty much word for word, so I don't think I've left out a detail (unless the question itself is missing a detail making it unsolvable). $\endgroup$
    – BeepBoop
    Sep 13, 2021 at 21:49

1 Answer 1


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?

  • $\begingroup$ So I'll have to find P(someone likes food | they ordered steak and they order food)? In that case will I just use the numerator from each of your results, and multiply that by the probability they like steak? Or am I finding P(someone likes food | they ordered steak | they order food)? $\endgroup$
    – BeepBoop
    Sep 14, 2021 at 0:32

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