The Bayesian network below contains only binary states. The conditional probability for each state is listed. From the Bayesian network, calculate the following probabilities:

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a) $P(b)$

b) $P(d)$

c) $P(c \mid \neg d)$

d) $P(a \mid \neg c, d)$

For a) I calculated this to be $ P(b) = \sum_a P(b \mid a) \cdot P(a) = P(b \mid a) \cdot p(a) + P(b \mid \neg a) \cdot P(\neg a) = 0.44$

b) Exact same method as in a: $P(d) = \sum_b P(d \mid b) \cdot P(b) = P(d \mid b) \cdot P(b) + P(d \mid \neg b) \cdot P(\neg b) = 0.712 $

Now for c) and d) Im not so sure.

c) I first tried calculating $P(c,\neg d)$ using the formula $P(x_1,...x_n) = \prod_{i = 1}^n P(x_i \mid Parents(x_i))$ This gives $P(c,\neg d) = P(c \mid b) \cdot P(\neg d \mid b) = 0.1 \times 0.4 = 0.04.$ Then Applying Bayes rule gives: $$P(c \mid \neg d) = \frac{P(c,\neg d)}{P(\neg d)} = \frac{0.04}{0.288} \approx 0.139$$


Same approach as in c), I first tried to calculate the joint distribution with the same formula: $P(a,\neg c, d) = P(a) \cdot P(\neg c \mid b) \cdot P(d \mid b) = 0.8 \times 0.9 \times 0.6 = 0.432$ Finally, applying Baye's rule again this gives $$P(a \mid \neg c, d) = \frac{P(a,\neg c, d))}{P(\neg c, d)} = \frac{0.432}{0.04} = 10.8$$ which is clearly wrong.

Is anyone able to see what Im doing incorrectly?


1 Answer 1


You have a problem in c) and d), you need to use the same marginalization procedures that you correctly use in a) and b)

c) $P(c,\neg d)=\sum_b P(c,\neg d, b) = \sum_b P(b) P(c,\neg d | b) = \sum_b P(b) P(c |b) P(\neg d|b)$. You should then be able to proceed with the numerical application.

d) $P(a, \neg c, d) = \sum_b P(a, \neg c, d, b) = P(a) \sum_b P(b|a)P(\neg c|b)P(d|b)$. You then have everything for the numerical application.


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