Science:Math Exam Resources/Courses/MATH152/April 2017/Question B 05 (b)

MATH152 April 2017

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Question B 05 (b)

Suppose four towns are connected by roads in the configuration shown below. A random driver wakes up every morning and flips a coin. If the coin is heads, she stays where she is for the day. If the coin is tails, she drives to the next town, choosing one of the roads with no preference. (For example, if she leaves Town A, she is equally likely to go to Town B or Town C, but she will not go to Town D that day.) (b) If the driver starts in Town A, what is the probability she will be in Town A two days later?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Consider all the possibilities after one day, then for each one find the possibility that goes to A at the second day.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. At the first day, from matrix $P$, driver can end up with A, with chance A to A is (p=1/2). At the second day, A move back to A is also 1/2. This routine (A-A-A)has chance of $1/2\cdot 1/2=1/4$. 1st day, driver can also end up with B, A to B is (p=1/4), under this case, if driver wanna go back to A at second day, the chance from B to A is (p=1/6). This routine (A-B-A)has chance of $1/4\cdot 1/6=1/24$. Same, the routine (A-C-A)has chance of $1/4\cdot 1/6=1/24$. The routine (A-D-A)has chance of $0\cdot 0=0$. answer: $\color {blue}1/4+1/24+1/24+0=1/3$