city B

$city A\begin{align} \ P (city B|city A)&& B=sunny && B=Rain \\A=sunny &&\frac{4}{5} && \frac{1}{5} \\ A=Rain &&\frac{1}{2}&&\frac{1}{2} \\ \end{align}$

i need to calculate the entropy for $H(City A)$ , $H(city B|City A=sunny)$, $H(city A|City b)$ and $H(city a,city b)$

but im not sure i how can i find $p(A,B)$ from the table such as $p(A=sunny,B=sunny)$, $p(A=sunny,B=rain)$ ,$p(A=rain,B=sunny)$, $p(A=rain,B=rain)$

i know that $P(city A=sunny)=p(B=rain).p(A=sunny|B=rain)+p(B=sunny).p(A=sunny|B=sunny)$

is this related to bayes rule in joint conditional probability? im used to tabel that is given joint probability such as $p(A,B)$ but in this table it is given the conditional probability what rule should i use

and also $P(B=sunny|A=sunny)=0.8=\frac{p(city A=sunny|city B=sunny).p(city B=sunny)}{P(city A=sunny)}$

i try to draw the tree diagram but im not sure how can i relate two probability of two city?

EDIT: there is similar problem in book that said $a=P(city A=sunny)=P(city B=sunny=b$ but im not sure is this right or not

$\color{brown}{\frac4{5}\cdot a+\frac1{2}\cdot (1-a)=b }$

from a=b

$\color{brown}{\frac4{5}\cdot a+\frac1{2}\cdot (1-a)=a}$

solving $a=\frac{5}{7}$

$P(a=S,b=S)=\frac{5}{7} * 0.8=\frac{4}{7}$





We know that $P(X=x|Y=y)\cdot P(Y=y)=P(X=x\cap Y=y)$. Then let me first define some denotations.

$A$=City a is sunny ($a$), $\overline A$=City a is sunny $((1-a))$, $B$=City a is sunny ($b$), $\overline B$=City a is sunny $((1-b))$.

$\small{\textrm{The small letters in the brackets will replace the corresponding probabilities}}$

1.1 The proability that City A and City B are sunny is

$P(B|A)\cdot P(A)=P(A\cap B)\Rightarrow \frac4{5}\cdot a=P(A\cap B)$

1.2 The proability that City A is rainy and City B is sunny is

$P(B|\overline A)\cdot P(\overline A)=P(\overline A\cap B)\Rightarrow \frac1{2}\cdot (1-a)=P(\overline A\cap B)$

We can sum up both equations and get $$\color{brown}{\frac4{5}\cdot a+\frac1{2}\cdot (1-a)=b \quad (1)}$$

2.1 The proability that City A and City B are sunny is

$P(B|A)\cdot P(A)=P(A\cap B)\Rightarrow \frac4{5}\cdot a =P(A\cap B)$

2.2 The proability that City A is sunny and City B is rainy is

$P(\overline B|A)\cdot P( A)=P( A\cap \overline B)\Rightarrow \frac1{5}\cdot (1-b)=P( A\cap \overline B)$

Summing up both equations again and get

$$\color{brown}{\frac4{5}\cdot a+\frac1{5}\cdot (1-b)=a \quad (2)}$$

You just have to solve this little equation system and then use the very first equation to calculate the joint distribution. I got

$$\large{\begin{array}{|c|c|c|} \hline P(B \cap A) & B & \overline B \\ \hline A &\frac{4}{13} & \frac{1}{13} \\ \hline \overline A &\frac{4}{13}&\frac{4}{13} \\ \hline \end{array}}$$

  • $\begingroup$ thankyou so much, i edit my post, there is similar probelm in my book, that said $a=P(city A=sunny)=P(city B=sunny=b)$, but im not sure why is that.. i tried to solve the equation that lead to different result from your answer, i dont know which one is right(?) $\endgroup$ – fiksx Jul 11 '19 at 15:46
  • $\begingroup$ Have you posted the similar problem? $\endgroup$ – callculus Jul 11 '19 at 15:49
  • $\begingroup$ posted the problem here? $\endgroup$ – fiksx Jul 11 '19 at 15:51
  • $\begingroup$ If you ask me, then it is good for me to know what the problem is. $\endgroup$ – callculus Jul 11 '19 at 15:53
  • $\begingroup$ Is still not clear what your problem is. Please state clear what the (new) problem is. A new question would be the best place. $\endgroup$ – callculus Jul 11 '19 at 16:14

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