probability of frog moving exactly 6 meters away
A confused frog has lost its way. Every 10 seconds it decides where to turn to. With probability $frac{1}{3} $ it jumps one meter to the right, with probability $frac{1} {3}$it jumps on meter to the left, otherwise it says in its place. Assume that all decisions are mutually independent. Find the probability that after 1 minute and 40 seconds the frog will be at the distance of exactly 6 meters from its initial location.
The way I approach this problem is that I see in the time there can be 10 jumps, and to be 6 meters to one direction, there are 6 possibilities, either 2 jumps to the left and 8 to the right, 1 to the left, 2 pauses and 7 to the right, or 4 pauses and 6 to the left. All of this can be for the opposite directions as well. With each jump being a 1/3 probability, of going in a certain direction the total probability of being exactly 6 meters away would be
$$ (6) times left( frac{1}{3}right)^{!10} $$
Is this right?
probability
edited Jan 4 at 1:00
jayant98
476115
asked Jan 4 at 0:33
Edgar Smith
61
add a comment |
A confused frog has lost its way. Every 10 seconds it decides where to turn to. With probability $frac{1}{3} $ it jumps one meter to the right, with probability $frac{1} {3}$it jumps on meter to the left, otherwise it says in its place. Assume that all decisions are mutually independent. Find the probability that after 1 minute and 40 seconds the frog will be at the distance of exactly 6 meters from its initial location.
The way I approach this problem is that I see in the time there can be 10 jumps, and to be 6 meters to one direction, there are 6 possibilities, either 2 jumps to the left and 8 to the right, 1 to the left, 2 pauses and 7 to the right, or 4 pauses and 6 to the left. All of this can be for the opposite directions as well. With each jump being a 1/3 probability, of going in a certain direction the total probability of being exactly 6 meters away would be
$$ (6) times left( frac{1}{3}right)^{!10} $$
Is this right?
probability
edited Jan 4 at 1:00
jayant98
476115
asked Jan 4 at 0:33
Edgar Smith
61
1
I suggest you check this logic on a much smaller problem where you can write out all the possible sequences of jumps and count the ones that succeed. Perhaps four jumps with a goal of landing two meters away.
– Ethan Bolker
Jan 4 at 0:46
add a comment |
A confused frog has lost its way. Every 10 seconds it decides where to turn to. With probability $frac{1}{3} $ it jumps one meter to the right, with probability $frac{1} {3}$it jumps on meter to the left, otherwise it says in its place. Assume that all decisions are mutually independent. Find the probability that after 1 minute and 40 seconds the frog will be at the distance of exactly 6 meters from its initial location.
The way I approach this problem is that I see in the time there can be 10 jumps, and to be 6 meters to one direction, there are 6 possibilities, either 2 jumps to the left and 8 to the right, 1 to the left, 2 pauses and 7 to the right, or 4 pauses and 6 to the left. All of this can be for the opposite directions as well. With each jump being a 1/3 probability, of going in a certain direction the total probability of being exactly 6 meters away would be
$$ (6) times left( frac{1}{3}right)^{!10} $$
Is this right?
probability
edited Jan 4 at 1:00
jayant98
476115
asked Jan 4 at 0:33
Edgar Smith
61
A confused frog has lost its way. Every 10 seconds it decides where to turn to. With probability $frac{1}{3} $ it jumps one meter to the right, with probability $frac{1} {3}$it jumps on meter to the left, otherwise it says in its place. Assume that all decisions are mutually independent. Find the probability that after 1 minute and 40 seconds the frog will be at the distance of exactly 6 meters from its initial location.
The way I approach this problem is that I see in the time there can be 10 jumps, and to be 6 meters to one direction, there are 6 possibilities, either 2 jumps to the left and 8 to the right, 1 to the left, 2 pauses and 7 to the right, or 4 pauses and 6 to the left. All of this can be for the opposite directions as well. With each jump being a 1/3 probability, of going in a certain direction the total probability of being exactly 6 meters away would be
$$ (6) times left( frac{1}{3}right)^{!10} $$
Is this right?
probability
probability
edited Jan 4 at 1:00
jayant98
476115
asked Jan 4 at 0:33
Edgar Smith
61
edited Jan 4 at 1:00
jayant98
476115
asked Jan 4 at 0:33
Edgar Smith
61
edited Jan 4 at 1:00
jayant98
476115
edited Jan 4 at 1:00
jayant98
476115
edited Jan 4 at 1:00
jayant98
476115
476115
asked Jan 4 at 0:33
Edgar Smith
61
asked Jan 4 at 0:33
Edgar Smith
61
asked Jan 4 at 0:33
Edgar Smith
61
61
1
I suggest you check this logic on a much smaller problem where you can write out all the possible sequences of jumps and count the ones that succeed. Perhaps four jumps with a goal of landing two meters away.
– Ethan Bolker
Jan 4 at 0:46
add a comment |
1
I suggest you check this logic on a much smaller problem where you can write out all the possible sequences of jumps and count the ones that succeed. Perhaps four jumps with a goal of landing two meters away.
– Ethan Bolker
Jan 4 at 0:46
1
1
I suggest you check this logic on a much smaller problem where you can write out all the possible sequences of jumps and count the ones that succeed. Perhaps four jumps with a goal of landing two meters away.
– Ethan Bolker
Jan 4 at 0:46
I suggest you check this logic on a much smaller problem where you can write out all the possible sequences of jumps and count the ones that succeed. Perhaps four jumps with a goal of landing two meters away.
– Ethan Bolker
Jan 4 at 0:46
add a comment |
2 Answers
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That is close, but you have to account for the number of ways each of those different possibilities can occur. Two jumps to the left and eight to the right can occur in ${10choose{2}} = 45$ ways. 4 pauses and 6 to the left can happen in ${10choose{4}} = 210$ ways. For 1 to the right, 2 pauses and 7 to the left, there are ${10choose{3}}$ ways to order the left jumps, and then ${3choose{2}}$ ways to order the other three jumps, giving ${10choose{3}}{3choose{2}} = 360$ ways. So there are 615 ways to end up 6 meters to the left and thus 1230 ways to end up 6 meters away. So the answer is $1230times(frac13)^{10}$
answered Jan 4 at 0:59
Erik Parkinson
9159
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It's not right.
Guide:
You forgot to take order of action into considerations. For example, if we know that there are $2$ jumps to the left and $8$ to the right. We will have to decide when to make the left jump and when to make the right jump. There are $binom{10}2$ such choices.
Take order of actions into consideration for the other possibilities as well.
answered Jan 4 at 0:52
Siong Thye Goh
99.6k1464117
add a comment |
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2 Answers
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2 Answers
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That is close, but you have to account for the number of ways each of those different possibilities can occur. Two jumps to the left and eight to the right can occur in ${10choose{2}} = 45$ ways. 4 pauses and 6 to the left can happen in ${10choose{4}} = 210$ ways. For 1 to the right, 2 pauses and 7 to the left, there are ${10choose{3}}$ ways to order the left jumps, and then ${3choose{2}}$ ways to order the other three jumps, giving ${10choose{3}}{3choose{2}} = 360$ ways. So there are 615 ways to end up 6 meters to the left and thus 1230 ways to end up 6 meters away. So the answer is $1230times(frac13)^{10}$
answered Jan 4 at 0:59
Erik Parkinson
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
That is close, but you have to account for the number of ways each of those different possibilities can occur. Two jumps to the left and eight to the right can occur in ${10choose{2}} = 45$ ways. 4 pauses and 6 to the left can happen in ${10choose{4}} = 210$ ways. For 1 to the right, 2 pauses and 7 to the left, there are ${10choose{3}}$ ways to order the left jumps, and then ${3choose{2}}$ ways to order the other three jumps, giving ${10choose{3}}{3choose{2}} = 360$ ways. So there are 615 ways to end up 6 meters to the left and thus 1230 ways to end up 6 meters away. So the answer is $1230times(frac13)^{10}$
answered Jan 4 at 0:59
Erik Parkinson
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
That is close, but you have to account for the number of ways each of those different possibilities can occur. Two jumps to the left and eight to the right can occur in ${10choose{2}} = 45$ ways. 4 pauses and 6 to the left can happen in ${10choose{4}} = 210$ ways. For 1 to the right, 2 pauses and 7 to the left, there are ${10choose{3}}$ ways to order the left jumps, and then ${3choose{2}}$ ways to order the other three jumps, giving ${10choose{3}}{3choose{2}} = 360$ ways. So there are 615 ways to end up 6 meters to the left and thus 1230 ways to end up 6 meters away. So the answer is $1230times(frac13)^{10}$
answered Jan 4 at 0:59
Erik Parkinson
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That is close, but you have to account for the number of ways each of those different possibilities can occur. Two jumps to the left and eight to the right can occur in ${10choose{2}} = 45$ ways. 4 pauses and 6 to the left can happen in ${10choose{4}} = 210$ ways. For 1 to the right, 2 pauses and 7 to the left, there are ${10choose{3}}$ ways to order the left jumps, and then ${3choose{2}}$ ways to order the other three jumps, giving ${10choose{3}}{3choose{2}} = 360$ ways. So there are 615 ways to end up 6 meters to the left and thus 1230 ways to end up 6 meters away. So the answer is $1230times(frac13)^{10}$
answered Jan 4 at 0:59
Erik Parkinson
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 4 at 0:59
Erik Parkinson
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 4 at 0:59
Erik Parkinson
9159
answered Jan 4 at 0:59
Erik Parkinson
9159
9159
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
It's not right.
Guide:
You forgot to take order of action into considerations. For example, if we know that there are $2$ jumps to the left and $8$ to the right. We will have to decide when to make the left jump and when to make the right jump. There are $binom{10}2$ such choices.
Take order of actions into consideration for the other possibilities as well.
answered Jan 4 at 0:52
Siong Thye Goh
99.6k1464117
add a comment |
It's not right.
Guide:
You forgot to take order of action into considerations. For example, if we know that there are $2$ jumps to the left and $8$ to the right. We will have to decide when to make the left jump and when to make the right jump. There are $binom{10}2$ such choices.
Take order of actions into consideration for the other possibilities as well.
answered Jan 4 at 0:52
Siong Thye Goh
99.6k1464117
add a comment |
It's not right.
Guide:
You forgot to take order of action into considerations. For example, if we know that there are $2$ jumps to the left and $8$ to the right. We will have to decide when to make the left jump and when to make the right jump. There are $binom{10}2$ such choices.
Take order of actions into consideration for the other possibilities as well.
answered Jan 4 at 0:52
Siong Thye Goh
99.6k1464117
It's not right.
Guide:
You forgot to take order of action into considerations. For example, if we know that there are $2$ jumps to the left and $8$ to the right. We will have to decide when to make the left jump and when to make the right jump. There are $binom{10}2$ such choices.
Take order of actions into consideration for the other possibilities as well.
answered Jan 4 at 0:52
Siong Thye Goh
99.6k1464117
answered Jan 4 at 0:52
Siong Thye Goh
99.6k1464117
answered Jan 4 at 0:52
Siong Thye Goh
99.6k1464117
answered Jan 4 at 0:52
Siong Thye Goh
99.6k1464117
99.6k1464117
add a comment |
add a comment |
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1
I suggest you check this logic on a much smaller problem where you can write out all the possible sequences of jumps and count the ones that succeed. Perhaps four jumps with a goal of landing two meters away.
– Ethan Bolker
Jan 4 at 0:46