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Let $A_1 = (0, 0), A_2 = (1, 0), A_3 = (1, 1)\ $and$\ A_4 = (0, 1)$ be the four vertices of a square. A particle starts from the point $A_1$ at time $0$ and moves either to $A_2$ or to $A_4$ with equal probability. Similarly, in each of the subsequent steps, it randomly chooses one of its adjacent vertices and moves there. Let $T$ be the minimum number of steps required to cover all four vertices. The probability $P(T = 4)$ is

  1. $0$
  2. $1/16$
  3. $1/8$
  4. $1/4$

2 Answers

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A1 A4
A2 A3

if you go A1 -> A2 or A1 -> A4 then there are two possibility

and total is 16 for ex. 12341,14321,14231,12431..............16possibility.

2/16 = 1/ 8

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In $0$ moves, it can go from $A_{1}$ to $A_{2}$ or $A_{4}$

Now, for $T=4$, i.e.  $4$ more moves are there and all vertices must be visited within $T=4$

And it can only visit adjacent vertex of square. 

So, moves will be $12143$ or $14123$

Only in these two moves, it can go all vertices at least once with minimum number of moves.

But, we know,in adjacent vertex movement, each vertex has $2$ possibility

So, total possibility will be $2^{4}$

Therefore, probability$P(T=4)=\frac{2}{2^{4}}=\frac{1}{8}$

Ref : https://math.stackexchange.com/questions/3058246/probability-of-covering-all-vertices-of-a-square

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