Probability of covering all vertices of a square
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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
asked Jan 1 at 6:27
debduttdebdutt
525
$endgroup$
add a comment |
$begingroup$
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
asked Jan 1 at 6:27
debduttdebdutt
525
$endgroup$
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
add a comment |
$begingroup$
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
asked Jan 1 at 6:27
debduttdebdutt
525
$endgroup$
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
asked Jan 1 at 6:27
debduttdebdutt
525
asked Jan 1 at 6:27
debduttdebdutt
525
asked Jan 1 at 6:27
debduttdebdutt
525
asked Jan 1 at 6:27
debduttdebdutt
525
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $frac{2}{2^4}=frac18$
answered Jan 1 at 6:47
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
The only two favorable outcomes are: $12143$ and $14123$. Each has the probability of $1/16$, hence their sum is $1/8$.
answered Jan 1 at 6:50
farruhotafarruhota
21.6k2842
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $frac{2}{2^4}=frac18$
answered Jan 1 at 6:47
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $frac{2}{2^4}=frac18$
answered Jan 1 at 6:47
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $frac{2}{2^4}=frac18$
answered Jan 1 at 6:47
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
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 $frac{2}{2^4}=frac18$
answered Jan 1 at 6:47
Siong Thye GohSiong Thye Goh
103k1468119
edited Jan 1 at 9:31
answered Jan 1 at 6:47
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 1 at 6:47
Siong Thye GohSiong Thye Goh
103k1468119
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
The only two favorable outcomes are: $12143$ and $14123$. Each has the probability of $1/16$, hence their sum is $1/8$.
answered Jan 1 at 6:50
farruhotafarruhota
21.6k2842
$endgroup$
add a comment |
$begingroup$
The only two favorable outcomes are: $12143$ and $14123$. Each has the probability of $1/16$, hence their sum is $1/8$.
answered Jan 1 at 6:50
farruhotafarruhota
21.6k2842
$endgroup$
add a comment |
$begingroup$
The only two favorable outcomes are: $12143$ and $14123$. Each has the probability of $1/16$, hence their sum is $1/8$.
answered Jan 1 at 6:50
farruhotafarruhota
21.6k2842
$endgroup$
The only two favorable outcomes are: $12143$ and $14123$. Each has the probability of $1/16$, hence their sum is $1/8$.
answered Jan 1 at 6:50
farruhotafarruhota
21.6k2842
answered Jan 1 at 6:50
farruhotafarruhota
21.6k2842
answered Jan 1 at 6:50
farruhotafarruhota
21.6k2842
answered Jan 1 at 6:50
farruhotafarruhota
21.6k2842
21.6k2842
add a comment |
add a comment |
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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