Probability that coin $2$ is flipped third day Suppose that coin $1$ has probability $0.6$ of coming up heads, and coin $2$ has probability $0.3$ of coming up heads. If the coin flipped today comes up heads, then we select coin $1$ to flip tomorrow. If the coin flipped today comes up tails, then we select coin $1$ to flip tomorrow with probability $0.2$ and select coin $2$ to flip tomorrow with probability $0.8$.
$Q1$:If the coin initially flipped is equally likely to be coin $1$ or coin $2$ , what is the probability that the coin flipped on the third day after the initial flip is coin $2$?
$Q2$:What proportion of flips use coin $2$ in the long run?
For the first question what I did is the following:
Let $X_n$ denote the label of the coin that is flipped on the $n$th day after the initial flip. $X_n$ is a Markov chain.
The transition probability matrix is given by
$$P = \begin{pmatrix}0.6 & 0.4 \\ 0.2 & 0.8\end{pmatrix}$$
So $$P^{(2)}=P^2=\begin{pmatrix} 0.44 & 0.56 \\ 0.28 & 0.72\end{pmatrix}$$
Now $$P^{(3)}=P^3=\left(\begin{array}{ll}
0.376 & 0.624 \\
0.312 & 0.688
\end{array}\right)$$
Hence the required probability is
$$\mathbb{P}(X_3 = 2) \\= \mathbb{P}(X_3 = 2|X_0 = 1)\mathbb{P}(X_0 = 1) + \mathbb{P}(X_3 = 2|X_0 = 2)\mathbb{P}(X_0 = 2)\\ = \frac{1}{2}(P^{(3)}_{12} + P^{(3)}_{22}) = 0.656$$
I am unable to answer Question $2$?
 A: As I hinted, you haven't used part of the  given data.
On day $0$, when each coin is tossed with equal probability, $P(H) = 0.5*0.6 + 0.5*0.3 = 0.45, P(T) = 0.55$
So P(coin $1$ on day $1) = 0.45+0.55*0.2 = 0.56$
and P(coin $2$ on day $1) = 0.55*0.8 = 0.44$
You can re-compute the required values for days $2$ and $3$ accordingly.

For part $2$, let the steady state probabilities (row vector) for coins $1$ and $2$ be $p$ and $q$ respectively, then
$p*0.6 + q*0.2 = p$
$p*0.4 + q*0.8 = q$
$p+q=1$
Solving, $p= \frac13, \boxed{ q=\frac23}$
A: My understanding of question $1$ gives a different answer. Let $P_t[C=1]$ be the probability that coin $1$ is flipped on day $t$ (initial flip is $t=0$) and $P_t[H]$ the probability of obtaining heads in day $t$. Then
$$P_t[C=1]=P_{t-1}[H]+(1-P_{t-1}[H]).2$$
This means that
\begin{aligned}P_1[C=1]&=\frac{1}{2}.6+\frac{1}{2}.3+\left(1-\left(\frac{1}{2}.6+\frac{1}{2}.3\right)\right).2=.56\\
P_2[C=1]&=.56\times.6+.44\times.3+\left(1-\left(.56\times.6+.44\times.3\right)\right).2=.5744\\
\end{aligned}
and $$P_3[C=1]=.5744\times.6+.4356\times.3+\left(1-\left(.5744\times.6+.4356\times.3\right)\right).2=.577856$$
I can't help you with the second question
