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?
Best Answer
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.