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?
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.
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&amp;rep=rep1&amp;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