You actually had your answer right there.
$P(H=hit)$ is the marginal probability. It reads "The probability of getting hit.". It is the proportion of people that got hit crossing the street, irrespective of traffic light.
$P(H=hit|L=red)$ is the conditional probability. It reads "The probability that you get hit, given that the light is red". It is the proportion of hits among the people that cross the street in red light.
Finally, $P(H=hit, L=red)$ is the joint probability. It reads "the probability that a person gets hit by a car and that the light is red". It is the proportion of hits in red light among all people.
You certainly know the relationship
$P(H=hit, L=red) = P(H=hit | L=red) * P(L=red)$
In "layman's parlance", we can look at it as follows. Assume that the probability of having a red light is extremely small, but that people always get hit when crossing in red light. Let us assume you are an observer at the side of the street. You will see people getting hit, and rarely will you see the light turning red. Out of all people that cross the street, the chance they will get hit in red light is very tiny, since they almost never have that opportunity ($P(H=hit,L=red)$ is small because a red light is rare). However, if you observe long enough, you will eventually see people getting hit in red light, and notice that whenever the light is red, people crossing the street will get hit for sure ($P(H=hit|L=red)=1$).
Let's say we have a problem of predicting whether a storm is coming or not.
So we'd like to predict whether a storm is coming or not (event $A$), and we have some clues available to us, namely the amount of clouds in the sky (event $B$) and how scared your dogs are (event $C$).
We can visualise the problem at hand using a Venn diagram:
We are interested in calculating the probability of a storm given the clues, $P(A|B,C)$. That quantity isn't represented directly in the diagram; instead, we can get $P(A,B,C)$ (a.k.a $P(A \cap B \cap C)$) from the white central area in the diagram. Fortunately, the relationship between $P(A|B,C)$ and $P(A,B,C)$ is simple:
$$P(A,B,C) = P(A|B,C) \cdot P(B,C)$$
where $P(B,C)$ corresponds to the conjunction between the magenta and white areas of the diagram.
We have a model P(storm is coming | how many clouds are outside), and have another model P(storm is coming | how scared the dogs are)
So we know $P(A|B)$ and $P(A|C)$. Like before, these two quantities are not represented directly in the diagram. Instead, we have $P(A,B)$, which corresponds to the yellow and white areas, and $P(A,C)$, which corresponds to the cyan and white areas. As before, we know the relationship between $P(A,B)$ and $P(A|B)$:
$$P(A,B) = P(A|B) \cdot P(B)$$
Same goes for $P(A,C)$ and $P(A|C)$.
To recap, we would like to know $P(A|B,C)$, which is related to the white area in the Venn diagram. So what happens if we add $P(A)$, $P(B)$ and $P(C)$? We are counting the magenta, yellow and cyan areas twice each, and the white central area three times. So we subtract the magenta, yellow and cyan areas once:
$$P(A) + P(B) + P(C) - P(A,B) - P(A,C) - P(B,C)$$
Except now we removed the white area from the summation; we added the white area three times when we summed up $A$, $B$, and $C$, but we removed it three times when we subtracted $(A,B)$, $(A,C)$ and $(B,C)$. So we add it back:
$$P(A) + P(B) + P(C) - P(A,B) - P(A,C) - P(B,C) + P(A,B,C)$$
We didn't account for the area outside all the circles, which corresponds to $P(\tilde{} A, \tilde{} B, \tilde{} C)$, which is the chance that there is no storm AND there are no clouds AND the dogs aren't scared.
$$P(A) + P(B) + P(C) - P(A,B) - P(A,C) - P(B,C) + P(A,B,C) + P(\tilde{} A, \tilde{} B, \tilde{} C) = 1$$
Let's assume that a storm ocurring with a spotless sky is very unlikely; $P(\tilde{} A, \tilde{} B, \tilde{} C) \approx 0$. In that case,
$$P(A) + P(B) + P(C) - P(A,B) - P(A,C) - P(B,C) + P(A,B,C) = 1$$
Let's apply the transformations we saw before:
$\begin{align}
P(A|B,C) \cdot P(B,C) &= 1 - [P(A) + P(B) + P(C) - P(A|B) \cdot P(B) - P(A|C) \cdot P(C) - P(B,C)]\\
P(A|B,C) &= \dfrac{1 - P(A) - P(B) - P(C) + P(A|B) \cdot P(B) + P(A|C) \cdot P(C)}{P(B,C)} + 1
\end{align}
$
As you can see, you would need more information if you want to calculate the probability of a storm given your clues. Namely:
- The probability of a storm in general;
- The probability of a cloudy sky in general;
- The probability of your dogs being scared in general; and
- The probability that your dogs will be scared AND the sky will be cloudy.
If you think about it, numbers 1-3 make sense:
- The clues may increase the probability of a storm, but if there aren't many storms to begin with, then the probability of a storm given your clues will still be small (albeit larger than the baseline probability of a storm);
- If you live in a typically cloudy area, the amount of clouds in the sky will probably be a poor predictor of a storm (because it's always cloudy, storm or no storm);
- Ditto for your dogs being scared.
Number 4 is a bit trickier. If either your dogs or the sky (or both) are perfect predictors of a storm, then there is no need for the other.
Now all this math assumes that your model outputs $P(\mathrm{storm} | \mathrm{clouds})$ ($P(A | B)$) and $P(\mathrm{storm} | \mathrm{scared\ dogs})$ ($P(A|C)$). However, it is typically easier to observe $P(\mathrm{clouds} | \mathrm{storm})$ ($P(B | A)$) and $P(\mathrm{scared\ dogs} | \mathrm{storm})$ ($P(C|A)$). In that case, we must note that
$$P(A,B) = P(A|B) \cdot P(B) = P(B|A) \cdot P(A)$$
so our previous model becomes
$$P(A|B,C) = \dfrac{1 - P(A) - P(B) - P(C) + P(B|A) \cdot P(A) + P(C|A) \cdot P(A)}{P(B,C)} + 1$$
Best Answer
Note: This answer is to the original version of the question, which asked whether $\mathbb{P}(B|A)$ and $\mathbb{P}(C|A)$ can be used with no other information to obtain $\mathbb{P}(A|B,C)$.
Consider two scenarios, both of which have $\mathbb{P}(B|A) = \mathbb{P}(C|A) = 0.5$:
Scenario 1:
\begin{align*} \mathbb{P}(A,B,\overline C) &= 0.2 \\ \mathbb{P}(A,\overline B, C) &= 0.2 \\ \mathbb{P}(\overline A, B, C) &= 0.2 \\ \mathbb{P}(\overline A, \overline B, \overline C) &= 0.4 \end{align*}
Scenario 2:
\begin{align*} \mathbb{P}(A,B,C) &= 0.1 \\ \mathbb{P}(A,B,\overline C) &= 0.1 \\ \mathbb{P}(A,\overline B, C) &= 0.1 \\ \mathbb{P}(A,\overline B,\overline C) &= 0.1 \\ \mathbb{P}(\overline A, \overline B, \overline C) &= 0.6 \end{align*}
In scenario 1, $\mathbb{P}(A|B,C) = 0$. In scenario 2, $\mathbb{P}(A|B,C) = 1$. Clearly, you'll need more than just $\mathbb{P}(B|A)$ and $\mathbb{P}(C|A)$ to compute $\mathbb{P}(A|B,C)$.