What is $P(A|B,C)$ when $B$ and $C$ are independent?

Question

Let's say that I know the following:

• $P(A|B)$ is the probability that a storm is coming given it's cloudy.
• $P(A|C)$ is the probability that a storm is coming given that the dogs bark.
• $P(B)$ and $P(C)$ are independent.

How do I compute the following?:

• $P(A|B,C)$, the probability of a storm coming given that it's cloudy AND the dogs are barking.

In layman's terms, I know that there is some likelihood that a storm is coming if it's cloudy. And, I know that there is some likelihood that a storm is coming if the dogs are barking. Therefore, shouldn't I have more confidence that a storm is coming if it's cloudy AND the dogs are barking? How do compute this?

The reason that I ask this question is because I am trying to combine measurements from two different sensors that measure the same thing. If I combine the measurements, shouldn't I expect greater confidence in my measurement?

This post and this post are related to my question, but the answers fall short in that I do not know the general probabilities of $P(B)$ and $P(C)$ to compute $P(A|B,C)$.

Answer 1

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 $\mathbb{P}(A|B,C)$.

Answer 2

The answer is the conflation of probabilities, explained here.

In the syntax that answers my question, the equation becomes:

$P(A|B,C)=\eta{P(A|B) P(A|C)}$

where $\eta$ is the normalization factor:

$\eta=\left({P(A|B)P(A|C) + P(\overline A|B)P(\overline A|C)}\right)^{-1}$

My question could be changed to a scenario that is more familiar to us, "How do I quantify the probability of having cancer if I got an opinion from two different doctors?" Surely getting a second opinion will give us more confidence about the prognosis! If $A$ indicates you having cancer and $B$ and $C$ are two different doctors' prognosis (thanks @josliber), where $P(A|B)=0.75$ and $P(A|C)=0.75$, then applying those values in the equation above gives us:

$P(A|B,C)=\frac{0.75 * 0.75}{0.75 * 0.75 + (1-0.75)*(1-0.75)}=0.9$

Two doctors were 75% confident about their prognosis. Combining those prognoses gives us 90% confidence that I have cancer.

Answer 3

given values of

P(A|B,C) P(A|B,!C) P(A|!B,C) P(A|!B,!C)

given

P(B), P(C) and P(A|B) and P(A|C)

given B and C are independent

Then:

P(A|B) = P(A|B,C) * P(C) + P(A|B,!C) * P(!C) P(A|C) = P(A|B,C) * P(B) + P(A|!B,C) * P(!B)

This only gives you two equations but three unknowns P(A|B,!C), P(A|!B,C) and P(A|B,C).