Let $D_n$ be the event that it is dry on day $n$ (and $\overline{D_n}$ that it is wet)

$$\mathrm{P} \left( \overline{D_{n+1}} \ | \ \overline{D_n} \right) = a; \ \ \mathrm{P} \left( \overline{D_{n+1}} \ | \ D_n \right) = b$$ $$ \begin{align} \Rightarrow \mathrm{P} \left( D_{n+1} \ | \ \overline{D_n} \right) = & 1 - \mathrm{P} \left(\overline{D_{n+1}} \ | \ \overline{D_n} \right) = 1 - a \\ \mathrm{P} \left( D_{n+1} \ | \ D_n \right) = & 1 - \mathrm{P} \left(\overline{D_{n+1}} \ | \ D_n \right) = 1 - b \\ \end{align} $$ $$ \begin{align} \mathrm{P} \left( D_{n+1} \right) = & \mathrm{P} \left(D_{n+1} \ | \ \overline{D_n} \right) \mathrm{P} \left( \overline{D_n} \right) + \mathrm{P} \left( D_{n+1} \ | \ D_n \right) \mathrm{P} \left( D_n \right) \\ & = (1-a) \ \mathrm{P} \left( \overline{D_n} \right) + (1-b) \ \mathrm{P} \left( D_n \right)\\ \end{align} $$ $$\mathrm{P} \left( \overline{D_n} \right) = 1 - \mathrm{P} \left( D_n \right)$$ $$ \begin{align} \Rightarrow \mathrm{P} \left( D_{n+1} \right) = & (1-a) \left( 1 - \mathrm{P} \left( D_n \right) \right) + (1-b) \ \mathrm{P} \left( D_n \right) \\ & = 1 - \mathrm{P} \left( D_n \right) - a + a \mathrm{P} \left( D_n \right) + \mathrm{P} \left( D_n \right) - b \mathrm{P} \left( D_n \right) \\ & = 1 - a + (a-b)\ \mathrm{P} \left( D_n \right) \\ \end{align} $$

Let $p_n = \mathrm{P} \left( D_n \right)$

$$\Rightarrow p_{n+1} = (1-a) + (a-b)\ p_n$$

This is what is called an 'inhomogeneous linear recurrence relationship' (a.k.a. linear difference equation) in $p_n$ of order 1. It is a discrete analogue of a differential equation and the methods for solving linear recurrence relationships are very similar to those for linear differential equations: the equivalent homogeneous equation is first solved and then a particular solution to account for the inhomogenity then found.

The homogeneous equivalent is:

$$q_{n+1} - (a-b)\ q_n = 0$$

Analogously to the initial assumption that the solution of an homogeneous second order linear differential equation $\left( a\frac{\mathrm{d^2}y}{\mathrm{d}x^2} + b\frac{\mathrm{d}y}{\mathrm{d}x} + cy = 0 \right)$ is of the form $y = \alpha e^{\lambda x}$ (where $\alpha$ is a constant), the solution to a linear difference equation is assumed to be of the form $y = \alpha \lambda^x$.

Therefore assuming $q_n = \alpha \lambda^n$ where $\alpha$ is some constant $\Rightarrow q_{n+1} = \alpha \lambda^{n+1}$

$$ \begin{align} \Rightarrow & \alpha \lambda^{n+1} - (a-b) \alpha \lambda^n = 0 \\ & \alpha \lambda^n \left[ \lambda - (a-b) \right] = 0 \\ \end{align} $$

Non-trivial solution $\Rightarrow \alpha \lambda^n \ne 0$

$$\therefore \ \lambda = a-b \Rightarrow q_n = \alpha (a-b)^n$$

To find the particular solution that corresponds to the inhomogeneous terms, let $p_n = \beta$ were $\beta$ is some constant

$$ \begin{align} \Rightarrow \beta = & (1-a) + (a-b) \ \beta \\ \beta (1 + b - a) = & 1 - a \\ \beta = & \frac{1 - a}{1 + b - a} \\ \end{align} $$

By superposition, the general solution is $p_n = q_n + \beta$

$$\therefore \ p_n = \frac{1-a}{1+b-a} + \alpha (a-b)^n$$

Using inital condition, $p_0 = p$

$$\Rightarrow p = \frac{1-a}{1+b-a} + \alpha \ \Rightarrow \ \alpha = p - \frac{1-a}{1+b-a}$$

Therefore the probablility of day $n$ being dry (where day 0 is June 1st) is:

$$\mathrm{P} \left( D_n \right) = \frac{1-a}{1+b-a} + \left( p - \frac{1-a}{1+b-a} \right) (a-b)^n$$ The values $a$ and $b$ are both non-zero, non-unity probabilities: $$0 < a < 1 \textrm{ and } 0 < b < 1 \Rightarrow |a-b| < 1$$ $$\therefore \ \lim_{n \to \infty} (a - b)^n = 0$$

Therefore for a large number of days $n$, the probability of it being a dry day tends to

$$\mathrm{P} \left( D_n \right) \approx \frac{1-a}{1+b-a}$$

and so becomes increasingly independent of the probability of it raining on the first day.

Substituting in the values given in the question $a=0.7$ and $b=0.2$

$$\Rightarrow \mathrm{P} \left( D_n \right) = \frac{1-0.7}{1+0.2-0.7} = \frac{0.3}{0.5} = 0.6$$ >