[Math] Bayes Theorum with Multiple conditions with independent ancenstors

bayes-theoremprobability

I have the following Bayes' Net:
{D}->{C}<-{A}->{B}

I need to find P(A|B,C) with B and C both being true. I have calculated the probability of A given B using the formula

$$
P(A|B)=\frac{P(B|A)P(A)}{P(B)}
$$

but I am unsure how to add C into this equation since it links to D.

Would I use the same method as described in Extended Bayes' theorem: p(A | B, C, D) – Constructing a Bayesian Network ?

The nodes in my diagram have the following values:

P(A)=0.01

P(D)=0.02

+---+---+----------+
| A | D | P(C|A,D) |
+---+---+----------+
| T | T | 0.9      |
| F | T | 0.5      |
| T | F | 0.8      |
| F | F | 0.05     |
+---+---+----------+

+---+---+--------+
| A | B | P(B|A) |
+---+---+--------+
| T | T | 0.6    |
| F | T | 0.02   |
+---+---+--------+

Best Answer

You have $P(C)=0.02$ but I assume you mean $P(D)=0.02$. We can proceed as follows:

\begin{align} P(A\mid B,C) &= \dfrac{P(A,B,C)}{P(B,C)} = \dfrac{P(A,B,C)}{P(A,B,C) + P(A^c,B,C)}. \\ \end{align}

\begin{align} & \\ P(A,B,C) &= P(B,C\mid A)\cdot P(A) \\ &= P(B\mid A)\cdot P(C\mid A)\cdot P(A) \qquad\text{since $B,C$ are conditionally independent given $A$} \\ &= P(B\mid A)\cdot P(A)\cdot \left[P(C\mid D,A)\cdot P(D\mid A) + P(C\mid D^c,A)\cdot P(D^c\mid A)\right] \\ &= P(B\mid A)\cdot P(A)\cdot \left[P(C\mid D,A)\cdot P(D) + P(C\mid D^c,A)\cdot P(D^c)\right] \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{since $A,D$ are independent} \\ &= 0.6\times 0.01\times [0.9\times 0.02 + 0.8\times 0.98] \\ &= 0.004812. \\ \end{align}

Similarly, \begin{align} P(A^c,B,C) &= P(B,C\mid A^c)\cdot P(A^c) \\ &= P(B\mid A^c)\cdot P(C\mid A^c)\cdot P(A^c) \qquad\text{since $B,C$ are conditionally independent given $A^c$} \\ &= P(B\mid A^c)\cdot P(A^c)\cdot \left[P(C\mid D,A^c)\cdot P(D\mid A^c) + P(C\mid D^c,A^c)\cdot P(D^c\mid A^c)\right] \\ &= P(B\mid A^c)\cdot P(A^c)\cdot \left[P(C\mid D,A^c)\cdot P(D) + P(C\mid D^c,A^c)\cdot P(D^c)\right] \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{since $A^c,D$ are independent} \\ &= 0.02\times 0.99\times [0.5\times 0.02 + 0.05\times 0.98] \\ &= 0.0011682. \\ \end{align}

Therefore,

\begin{align} P(A\mid B,C) &= \dfrac{0.004812}{0.004812 + 0.0011682} \approx 0.80466. \\ \end{align}

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