Problem on total probability Let us conisder the following question from the textbook named Introduction to Probability

Alice is taking a probability class and at the end of each week she
  can be either up-to-date or she may have fallen behind. If she is
  up-to-date in a given week, the probability that she will be
  up-to-date (or behind) in the next week is 0.8 (or 0.2, respectively).
  If she is behind in a given week, the probability that she will be
  up-to-date (or behind) in the next week is 0.6 (or 0.4, respectively).
  Alice is (by default) up-to-date when she starts the class. What is
  the probability that she is up-to-date after three weeks?

My solution: 

Therefore my answer is $$0.8 \times 0.8 \times 0.8+0.8 \times 0.2 \times 0.6+0.2 \times 0.6 \times 0.8+0.2 \times 0.4 \times 0.6 = \bf{0.752}$$
but according to textbook, it is $\bf{0.688}$.
Where I went wrong?
 A: Let U be uptodate, B be behind.
The four possible paths are from U to {UUU,UBU,BUU,BBU}
The required probability $= 0.8^3+(0.8)(0.2)(0.6) + (0.2)(0.6)(0.8) + (0.2)(0.4)((0.6) = 0.752$
The book answer is wrong.
A: So I wanted to opt in as I see that all the answers to this question are wrong. I agree that the way that the book arrives at their solution is a bit unintuitive, but it is nontheless, correct. 
I have attached an image that, graphically, shows how the probability that Alice is up-to-date after 3 weeks is indeed: 0.688
Representing the probabilities as a full binary tree, we can clearly see the paths (marked in red) that lead to Alice being up-to-date after 3 weeks. 
Starting from the left-most path, the calculations are as follows: 
$$ 
\begin{align}
\mathbb{P}(U_3) &= 0.8 \cdot 0.8 \cdot 0.8 + 0.8 \cdot 0.2 \cdot 0.4 + 0.2 \cdot 0.4 \cdot 0.8 + 0.2 \cdot 0.6 \cdot 0.4 \\
&= 0.512 + 0.064 + 0.064 + 0.048\\
&= 0.688
\end{align}
$$
Hence the probability that Alice is up-to-date after 3 weeks is indeed 0.688. 
A: I think 
The condition is after the first week,  alice Is uptodate.  Then we only get 2 possible ways { (UUU), (UBU) }
SO,  the answer 0,8 *0,8 *0,8 + 0,8*0,2*0,8 = 0,608 .
A: The answer in the book is correct. The purpose of the example was to demonstrate how the total probability theorem can be used to determine the probability of an event that might happen 20 or 30 nodes down. Making a tree for 30 or 40 nodes is quite cumbersome. A formula can easily be forged in a computer to solve such complex trees. The original question mixed up the use of (B|U) and (U|B), which caused the answer to be different from the book.
Assumptions

*

*start is from week 0, not week 1.

*there are two distinct ways to end up with updated status in the end, either(B)(U|B) or (U)(U|U).

*use the above baseline to move up the tree in inverse, which will expand the equation to include all the possible four paths.

the following  will make the solution clear.

A: Way accepted answer is arrived at is little incorrect. Looking at Reference, it is clear that solution arrived in original question is correct - http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.706.2591&rep=rep1&type=pdf
This is a question of total probability. So here our possible paths are only two - {UUU, UBU}. Given first week is U (by default up-to-date). We need to calculate probability of week2 and week3 given week1
Let Ui be up-to-date in given week and Bi behind in given week. Given week1 is by default up-to-date, represented as (U1). 
So calculation will be -
P(U3) = P(U3|U2)*P(U2) + P(U3|B2)*P(B2)
P(U3) = 0.8*P(U2) + 0.6*P(B2)
Now, we need to find values of P(U2) and P(B2)
P(U2) = P(U2|U1)*P(U1) + P(U2|B1)*P(B1)
P(U2) = 0.8*0.8 + 0.6*0.2 = 0.76
Also,
P(B2) = P(B2|U1)*P(U1) + P(B2|B1)*P(B1)
P(B2) = 0.2*0.8 + 0.4*0.2 = 0.24
So, replacing values in our equation 1 -
P(U3) = 0.8*P(U2) + 0.6*P(B2)
P(U3) = 0.8*0.76 + 0.6*0.24 = 0.752
