Two Conditional Probability In a city, every day is either cloudy or sunny (not both). If it's sunny on any given day, then the probability that the next day will be sunny is $\frac 34$. If it's cloudy on any given day, then the probability that the next day will be cloudy is $\frac 23$.
a) In the long run, what fraction of days are sunny? 
b) Given that a consecutive Saturday and Sunday had the same weather in the city, what is the probability that the weather was sunny?

For part a, I was thinking that the answer would be $(\frac{3}{4}+ \frac{1}{3})/2$ to get $\frac{13}{24}$ because the probability if it is sunny followed by another sunny day is $\frac34$ and the probability that it is cloudy followed by a sunny day is $\frac13$.
For part b, I'm thinking of using casework, but I'm not sure. 
 A: Let's denote the given probabilities as $p_{s\mid s}=3/4$ and $p_{c\mid c}=2/3$. Then also $p_{s\mid c}=1/3$, which is the probability that a day is sunny if it follows a cloudy day.
For part (a)
Let's denote the equilibrium probability of that a day is sunny as $p_s$.  At equilibrium this is a constant from day to day; so we want to solve $p_s= p_{s\mid s}\,p_s +p_{s\mid c}\,(1-p_s)$ for $p_s$.
$$p_s = \tfrac 3 4 p_s +\tfrac 1 3(1-p_s)$$

For part (b) use this equilibrium constant as the probability that Saturday was sunny, and work out $p_w$: the probability that the weekend was sunny given that both days had the same weather.
$$p_{w} = \frac{p_{s\mid s} p_s}{p_{s\mid s}p_s+p_{c\mid c}(1-p_s)} $$
A: Ad a)
The tansition-matrix is 
$$P=\begin{pmatrix} & S_2 & C_2  \\ S_1&\frac{3}{4} & \frac{1}{4} \\ C_1 & \frac{1}{3} & \frac{2}{3} \end{pmatrix} $$
$C_1$: Cloudy on a given day.
$S_2$: Sunny on the consecutive day.
To get the steady-state distribution one has to solve the following equation:
$\begin{pmatrix} S & C \end{pmatrix}\cdot P=\begin{pmatrix} S & C \end{pmatrix}$
Thus the equation system is 
$\frac{3}{4}\cdot S+\frac{1}{3}\cdot C=S$
$\frac{1}{2}\cdot S+\frac{2}{3}\cdot C=C$
Can you proceed from here ?
