The probability of not raining on a weekend I've been trying to solve the following problem:

After years of observations it has been found out that if it rains on
given day, there is a 60% chance that it will rain on the next day
too. If it is not raining, the chance of rain on the next day is only
25%. The weather forecast for Friday predicts the chance of rain is
75%. What is the probability that at least one day of the weekend will
have no rain?

I calculate the probability that at least one day of the weekend will have no rain as
$$P(\overline{Saturday} \cup  \overline{Sunday}) = 1 - P(Saturday \cap  Sunday) = 1 - P(Saturday)*P(Sunday)$$
where $P(Saturday)$ and $P(Sunday)$ are the probabilities of raining on Saturday and Sunday respectively.
I compute the probability of raining on Saturday by adding the likelihoods of two exclusive events: raining on Saturday after raining on Friday and raining on Saturday after not raining on Friday.
$$P(Saturday)=P(Saturday \cap Friday)+P(Saturday \cap \overline{Friday}) =P(Saturday|Friday)*P(Friday)+P(Saturday|\overline{Friday})*P(\overline{Friday})=0.6*0.75+0.25*(1-0.75)=0.5125$$
Similarly, I compute the probability of raining on Sunday:
$$P(Sunday)=P(Sunday \cap Saturday)+P(Sunday \cap \overline{Saturday})=P(Sunday|Saturday)*P(Saturday)+P(Sunday|\overline{Saturday})*P(\overline{Saturday})=0.6*0.5125+0.25*(1-0.5125)=0.429375$$
Now I can find the result:
$$P(\overline{Saturday} \cup  \overline{Sunday}) = 1 - P(Saturday)*P(Sunday)=1-0.5125*0.429375=0.7799453125$$
Unfortunately, my result doesn't match the expected answer 0.69. Does anyone see a mistake in my calculations?
 A: A natural solution to this problem is to consider a Markov chain. Define the random variable $X(n)$ which is equal to 1 when it rains on day $n$ and 0 otherwise.
Based on the data given in the problem, we have $P(X(n+1)=1|X(n)=1)=0.6$, $P(X(n+1)=1|X(n)=0)=0.25$. We can then build the dynamical system
$$p(n+1)=p(n)\begin{bmatrix}
 P(X(n+1)=1|X(n)=1) & P(X(n+1)=0|X(n)=1)\\
P(X(n+1)=1|X(n)=0) & P(X(n+1)=0|X(n)=0)
\end{bmatrix}=p(n)\begin{bmatrix}
 0.6 & 0.4\\0.25 & 0.75
\end{bmatrix},$$
where $p(n)=\begin{bmatrix}
 P(X(n)=1) & P(X(n)=0)
\end{bmatrix}$. The other entries of the matrix come from the fact that $$P(X(n+1)=1|X(n)=1)+P(X(n+1)=0|X(n)=1)=1$$ and $$P(X(n+1)=1|X(n)=0)+P(X(n+1)=0|X(n)=0)=1.$$
Assume that day 0 is Friday. Then, we have that
$$p(0) = \begin{bmatrix}
0.75 & 0.25
\end{bmatrix}.$$
We then have that $$p(1) = \begin{bmatrix}
41/80 & 39/80
\end{bmatrix}.$$
Now, the probability that there will be no rain at least on one day of the weekend is given by $1-P(X(1)=1 \cap X(2)=1)$ as it is the complementary event of "it will rain all weekend". Moreover, we have that
$$P(X(2)=1 \cap X(3)=1)=P(X(3)=1|X(2)=1)P(X(2)=1)$$
which results in $1-P(X(1)=1 \cap X(2)=1)=1-0.5125*0.6=0.6925.$
A: Denoting Friday, Saturday, Sunday, by the suffixes $1,2,3$ respectively,
P(Rain on both Saturday and Sunday)
$= P(R_1).P(R_2|R_1).P(R_3|R2) +P(R^c_1).P(R_2|R^c_1).P(R_3|R_2)$
$= 0.75*06*0.6 + 0.25*0.25*0.6 = 0.3075$
Thus P(at least one day of weekend has no rain) $=1-0.3075 = 0.6925$
