# Bayes Network 2 parents one child

I have the following Bayes network:

S R
\ /
H


I know that: $$P(s) = .7$$$$P(r) = .01$$$$P(h|s,r) = 1$$$$P(h|!s,r) = .9$$$$P(h|s,!r) = .7$$$$P(h|!s,!r) = .1$$ and I would like to find $p(r|h,s), p(r|h), p(r|h,!s)$.

I can imitate the explanation I received to solve this problem but it doesn't feel like I really understand it, so I decided to figure it out for myself. I think I get a reasonable solution but for some reason it is not the right one. This suggests that I misunderstand something and I am hoping somebody can nudge me in the right direction.

My first goal was to explore the space:
$$P(!s) = .3$$$$P(!r) = .99$$ Then to relate the conditional probabilities to the likelihood of them happening:
$$P(h,s,r) = p(h|s,r) * P(s) * P(r) = 0.007$$$$P(h,!s,r) = p(h|!s,r) * P(!s) * P(r) = 0.0027$$$$P(h,s,!r) = p(h|s,!r) * P(s) * P(!r) = 0.4851$$$$P(h,!s,!r) = p(h|!s,!r) * P(!s) * P(!r) = 0.00099$$ and to sum them up to find all the probabilities of h $p(h): .49579$
(I use an excel sheet for my calculation to limit impreciseness)
$$a = P(h,s,r)/P(h) = 0.01412$$$$b = P(h,!s,r)/P(h) = 0.00545$$$$c = P(h,s,!r)/P(h)= 0.97844$$$$d = P(h,!s,!r)/P(h)= 0.001997$$ now I know the likelihood of each scenario happening. I used variables a,b,c,d because it seems clearer:
$$P(r|h,s) = a/(a+c) = 0.14225$$$$P(r|h) = (a+b)/(a+b+c+d) = 0.01956$$$$P(r|h,!s)= b/(b+d) = 0.7317$$ I would really appreciate it if somebody can point out the flaw in my thinking,
Thank you,

Boaz

ps. This question is taken from a udacity course intro to AI where it is solved using Bayes Theorem.

Your notation of p'(h|s,r) is confusing. Simply use p(h,s,r)= p(h|s,r)*p(s)*p(r)
Then $p(r|h,s)$ $= p(h,s,r)/p(h,s) \\ = p(h,s,r)/[p(h,s,r)+p(h,s,!r)] \\ = p(h|s,r)\cdot p(r)\Big/[p(h|s,r)\cdot p(r)+p(h|s,!r)\cdot p(!r)] \\ = 1\cdot 0.01 / [1\cdot 0.01 + 0.7\cdot 0.99] \\ = 0.01422475106685633001422475106686...$
And $p(r|h,!s)$ $= p(h|!s,r)\cdot p(r)\Big/[p(h|!s,r)\cdot p(r)+p(h|!s,!r)\cdot p(!r)] \\ = 0.9\cdot 0.01 / [0.9\cdot 0.01 + 0.1\cdot 0.99] \\ = 0.08333333333333333333333333333333...$
And $p(r|h) = \frac{p(h|s,r)p(s)p(r)+p(h|!s,r)p(!s)p(r)}{p(h|s,r)p(s)p(r)+p(h|!s,r)p(!s)p(r)+p(h|s,!r)p(s)p(!r)+p(h|!s,!r)p(!s)p(!r)}$ ... and so on.
• @BJR Look to your fourth calculation. $p(h\mid !s,!r)\cdot p(!s)\cdot p(!r) = 0.1\cdot 0.3\cdot 0.99 = 0.0297 \;\color{red}{\neq 0.00099}$ Nov 18, 2015 at 0:30