0
$\begingroup$

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

enter image description here

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: $$P(c \mid \neg d) = \frac{P(c,\neg d)}{P(\neg d)} = \frac{0.04}{0.288} \approx 0.139$$

d)

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.