Probability of covering all vertices of a square












3












$begingroup$


Let A1 = (0,0), A2 = (1,0), A3 = (1,1) and A4 = (0,1) be the four vertices of a square. A particle starts from the point A1 at time 0 and moves either to A2 or to A4 withequalprobability. Similarly, in eachof 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 (A) 0 (B) 1/16 (C) 1/8 (D) 1/4.



I am getting it as 3/4 but answer is 1/8 please help!






probability







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$endgroup$








  • 1




    $begingroup$
    Hint: What is the probability of the cycle $1234$?
    $endgroup$
    – Sean Roberson
    Jan 1 at 6:30






  • 1




    $begingroup$
    @SeanRoberson don't forget $1432$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 1 at 6:37










  • $begingroup$
    Could the particle go from (0,0) to (0,1) then go back to (0,0)? When at A3 could it go back to A1? A similar problem is here:nrich.maths.org/2370
    $endgroup$
    – NoChance
    Jan 1 at 7:16


















3












$begingroup$


Let A1 = (0,0), A2 = (1,0), A3 = (1,1) and A4 = (0,1) be the four vertices of a square. A particle starts from the point A1 at time 0 and moves either to A2 or to A4 withequalprobability. Similarly, in eachof 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 (A) 0 (B) 1/16 (C) 1/8 (D) 1/4.



I am getting it as 3/4 but answer is 1/8 please help!






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: What is the probability of the cycle $1234$?
    $endgroup$
    – Sean Roberson
    Jan 1 at 6:30






  • 1




    $begingroup$
    @SeanRoberson don't forget $1432$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 1 at 6:37










  • $begingroup$
    Could the particle go from (0,0) to (0,1) then go back to (0,0)? When at A3 could it go back to A1? A similar problem is here:nrich.maths.org/2370
    $endgroup$
    – NoChance
    Jan 1 at 7:16
















3












3








3


0



$begingroup$


Let A1 = (0,0), A2 = (1,0), A3 = (1,1) and A4 = (0,1) be the four vertices of a square. A particle starts from the point A1 at time 0 and moves either to A2 or to A4 withequalprobability. Similarly, in eachof 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 (A) 0 (B) 1/16 (C) 1/8 (D) 1/4.



I am getting it as 3/4 but answer is 1/8 please help!






probability







share|cite|improve this question









$endgroup$




Let A1 = (0,0), A2 = (1,0), A3 = (1,1) and A4 = (0,1) be the four vertices of a square. A particle starts from the point A1 at time 0 and moves either to A2 or to A4 withequalprobability. Similarly, in eachof 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 (A) 0 (B) 1/16 (C) 1/8 (D) 1/4.



I am getting it as 3/4 but answer is 1/8 please help!






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 1 at 6:27









debduttdebdutt

525




525








  • 1




    $begingroup$
    Hint: What is the probability of the cycle $1234$?
    $endgroup$
    – Sean Roberson
    Jan 1 at 6:30






  • 1




    $begingroup$
    @SeanRoberson don't forget $1432$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 1 at 6:37










  • $begingroup$
    Could the particle go from (0,0) to (0,1) then go back to (0,0)? When at A3 could it go back to A1? A similar problem is here:nrich.maths.org/2370
    $endgroup$
    – NoChance
    Jan 1 at 7:16
















  • 1




    $begingroup$
    Hint: What is the probability of the cycle $1234$?
    $endgroup$
    – Sean Roberson
    Jan 1 at 6:30






  • 1




    $begingroup$
    @SeanRoberson don't forget $1432$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 1 at 6:37










  • $begingroup$
    Could the particle go from (0,0) to (0,1) then go back to (0,0)? When at A3 could it go back to A1? A similar problem is here:nrich.maths.org/2370
    $endgroup$
    – NoChance
    Jan 1 at 7:16










1




1




$begingroup$
Hint: What is the probability of the cycle $1234$?
$endgroup$
– Sean Roberson
Jan 1 at 6:30




$begingroup$
Hint: What is the probability of the cycle $1234$?
$endgroup$
– Sean Roberson
Jan 1 at 6:30




1




1




$begingroup$
@SeanRoberson don't forget $1432$
$endgroup$
– Mohammad Zuhair Khan
Jan 1 at 6:37




$begingroup$
@SeanRoberson don't forget $1432$
$endgroup$
– Mohammad Zuhair Khan
Jan 1 at 6:37












$begingroup$
Could the particle go from (0,0) to (0,1) then go back to (0,0)? When at A3 could it go back to A1? A similar problem is here:nrich.maths.org/2370
$endgroup$
– NoChance
Jan 1 at 7:16






$begingroup$
Could the particle go from (0,0) to (0,1) then go back to (0,0)? When at A3 could it go back to A1? A similar problem is here:nrich.maths.org/2370
$endgroup$
– NoChance
Jan 1 at 7:16












2 Answers
2






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2












$begingroup$

To have $4$ steps, it means it visited exactly one node twice and we can check that the repeated node must be $1$.



It could have been $12143$ or $14123$.



Hence $frac{2}{2^4}=frac18$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How are you labelling the nodes?
    $endgroup$
    – coffeemath
    Jan 1 at 9:26






  • 1




    $begingroup$
    oops, I see. i made a mistake. thanks.
    $endgroup$
    – Siong Thye Goh
    Jan 1 at 9:30





















1












$begingroup$

The only two favorable outcomes are: $12143$ and $14123$. Each has the probability of $1/16$, hence their sum is $1/8$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    To have $4$ steps, it means it visited exactly one node twice and we can check that the repeated node must be $1$.



    It could have been $12143$ or $14123$.



    Hence $frac{2}{2^4}=frac18$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How are you labelling the nodes?
      $endgroup$
      – coffeemath
      Jan 1 at 9:26






    • 1




      $begingroup$
      oops, I see. i made a mistake. thanks.
      $endgroup$
      – Siong Thye Goh
      Jan 1 at 9:30


















    2












    $begingroup$

    To have $4$ steps, it means it visited exactly one node twice and we can check that the repeated node must be $1$.



    It could have been $12143$ or $14123$.



    Hence $frac{2}{2^4}=frac18$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How are you labelling the nodes?
      $endgroup$
      – coffeemath
      Jan 1 at 9:26






    • 1




      $begingroup$
      oops, I see. i made a mistake. thanks.
      $endgroup$
      – Siong Thye Goh
      Jan 1 at 9:30
















    2












    2








    2





    $begingroup$

    To have $4$ steps, it means it visited exactly one node twice and we can check that the repeated node must be $1$.



    It could have been $12143$ or $14123$.



    Hence $frac{2}{2^4}=frac18$






    share|cite|improve this answer











    $endgroup$



    To have $4$ steps, it means it visited exactly one node twice and we can check that the repeated node must be $1$.



    It could have been $12143$ or $14123$.



    Hence $frac{2}{2^4}=frac18$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 1 at 9:31

























    answered Jan 1 at 6:47









    Siong Thye GohSiong Thye Goh

    103k1468119




    103k1468119












    • $begingroup$
      How are you labelling the nodes?
      $endgroup$
      – coffeemath
      Jan 1 at 9:26






    • 1




      $begingroup$
      oops, I see. i made a mistake. thanks.
      $endgroup$
      – Siong Thye Goh
      Jan 1 at 9:30




















    • $begingroup$
      How are you labelling the nodes?
      $endgroup$
      – coffeemath
      Jan 1 at 9:26






    • 1




      $begingroup$
      oops, I see. i made a mistake. thanks.
      $endgroup$
      – Siong Thye Goh
      Jan 1 at 9:30


















    $begingroup$
    How are you labelling the nodes?
    $endgroup$
    – coffeemath
    Jan 1 at 9:26




    $begingroup$
    How are you labelling the nodes?
    $endgroup$
    – coffeemath
    Jan 1 at 9:26




    1




    1




    $begingroup$
    oops, I see. i made a mistake. thanks.
    $endgroup$
    – Siong Thye Goh
    Jan 1 at 9:30






    $begingroup$
    oops, I see. i made a mistake. thanks.
    $endgroup$
    – Siong Thye Goh
    Jan 1 at 9:30













    1












    $begingroup$

    The only two favorable outcomes are: $12143$ and $14123$. Each has the probability of $1/16$, hence their sum is $1/8$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The only two favorable outcomes are: $12143$ and $14123$. Each has the probability of $1/16$, hence their sum is $1/8$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The only two favorable outcomes are: $12143$ and $14123$. Each has the probability of $1/16$, hence their sum is $1/8$.






        share|cite|improve this answer









        $endgroup$



        The only two favorable outcomes are: $12143$ and $14123$. Each has the probability of $1/16$, hence their sum is $1/8$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 6:50









        farruhotafarruhota

        21.6k2842




        21.6k2842






























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