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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:

enter image description here

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?

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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.

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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 .

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    $\begingroup$ I think there are other possibilities: BUU and BBU $\endgroup$ – Angina Seng Jan 22 '19 at 16:58
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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.

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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

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  • $\begingroup$ Your answer adds nothing new to the accepted answer. $\endgroup$ – José Carlos Santos Feb 14 '19 at 16:23

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