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A probability problem..

We have a regular tetrahedron with vertices ABCD.Length of each side being 1 unit.An insect starts at vertex A. At any vertex his probability of continuing on any of the three paths is 1/3 (including returning from the path from where it came). The insect starts at vertex A and travels 5 units.What is the probability that after travelling 5 untis he is still at vertex A?? Answer [20/81] .Tell me how to do it??

Note by Divyaanand Sinha
2 years, 11 months ago

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This is an example of a Markov chain with (basically) two states. At any time the insect is either at vertex \(A\) (state 1) or else at one of the other three vertices (state 2).

If the insect is in state 1, he will be in state 2 after one move with probability \(1\). On the other hand, if the insect is in state 2, then after one move he will be in state 1 with probability \(\tfrac13\) and in state 2 with probability \(\tfrac23\).

If we let \(p_{n,j}\) be the probability that the insect is in state \(j\) after \(n\) moves, then we can write the information in the previous paragraph in matrix form: \[ \left( \begin{array}{c} p_{n,1} \\ p_{n,2}\end{array} \right) \; = \; \left( \begin{array}{cc} 0 & \tfrac13 \\ 1 & \tfrac23 \end{array} \right) \left(\begin{array}{c} p_{n-1,1} \\ p_{n-1,2}\end{array} \right) \] and hence (since \(p_{0,1} = 1\) and \(p_{0,2} = 0\)): \[ \left( \begin{array}{c} p_{n,1} \\ p_{n,2}\end{array} \right) \; = \; \left( \begin{array}{cc} 0 & \tfrac13 \\ 1 & \tfrac23 \end{array} \right)^n \left( \begin{array}{c} p_{0,1} \\ p_{0,2} \end{array} \right) \; = \; \left( \begin{array}{cc} 0 & \tfrac13 \\ 1 & \tfrac23 \end{array} \right)^n \left( \begin{array}{c} 1 \\ 0 \end{array} \right) \] for any \(n \ge 0\). Since \[ \left( \begin{array}{c} p_{5,1} \\ p_{5,2} \end{array} \right) \; = \; \left(\begin{array}{c} \tfrac{20}{81} \\ \tfrac{60}{81} \end{array} \right) \] and your question asks for \(p_{5,1}\), we are done. Mark Hennings · 2 years, 11 months ago

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We can denote the complete path the insect takes by a sequence of 5 letters where no adjacent letters are the same. For instance \(BCADB\) means that it goes to B, then C, then A, then D, then finally B.

We first calculate the sample size of the probability, namely how many different paths can he take. He has 3 choices for the first journey, 3 choices again for the second, and so on. So we have \(3^5\)

Next, we count the number of paths where he eventually stops at A. The paths can be represented by: \( _ _ _ _ A\), where the first and fourth spot can't have an A. We will count this by case-working as follow:

Case 1, we don't have any As for the second and third spot, which gives \(3*2*2*2=24\)

Case 2, we have an A for either second or third spot, which gives \(2(3*3*2)=36\)

This gives a total of \(36+24=60\) paths.

Therefore our probability is \(\frac {60}{3^5}=\frac {20}{81}\)

I don't think it was necessary to introduce the denotion of paths after all......oh well it helps me see things more clearly, I even drew a square with vertices labeled ABCD. :) Xuming Liang · 2 years, 11 months ago

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