A very challenging question on probability [closed]
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Suppose you invest $50$ dollars on a firm and on the next day it can either increase or decrease by $10%$. After a $10$ day period, what is the probability that you will have more than $50$ dollars?
probability
edited Jan 18 at 14:29
idriskameni
749321
asked Jan 18 at 13:21
Bishesh AdhikariBishesh Adhikari
41
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closed as off-topic by Did, lulu, Math1000, José Carlos Santos, RRL Jan 18 at 17:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, lulu, Math1000, José Carlos Santos, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose you invest $50$ dollars on a firm and on the next day it can either increase or decrease by $10%$. After a $10$ day period, what is the probability that you will have more than $50$ dollars?
probability
edited Jan 18 at 14:29
idriskameni
749321
asked Jan 18 at 13:21
Bishesh AdhikariBishesh Adhikari
41
$endgroup$
closed as off-topic by Did, lulu, Math1000, José Carlos Santos, RRL Jan 18 at 17:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, lulu, Math1000, José Carlos Santos, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
3
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What is the probability that it increases/decreases? Also your title is bad.
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– lightxbulb
Jan 18 at 13:23
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I think this is essentially a random walk question.
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– Matti P.
Jan 18 at 13:34
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Note: the order of the changes doesn't matter, so you just have to compute the probability that you get $n$ up moves for $nin {0,cdots, 10}$. Once you decide on the probability of an up move this is a straightforward computation.
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– lulu
Jan 18 at 13:40
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Can you clarify your question? As has been pointed out, the problem can't be answered without knowing the probability of the up and down moves. Also: what have you tried? This is a very straightforward binomial calculation (since $10$ is so small there's no need for any approximations to the distribution). Where are you getting stuck?
$endgroup$
– lulu
Jan 18 at 13:54
add a comment |
$begingroup$
Suppose you invest $50$ dollars on a firm and on the next day it can either increase or decrease by $10%$. After a $10$ day period, what is the probability that you will have more than $50$ dollars?
probability
edited Jan 18 at 14:29
idriskameni
749321
asked Jan 18 at 13:21
Bishesh AdhikariBishesh Adhikari
41
$endgroup$
Suppose you invest $50$ dollars on a firm and on the next day it can either increase or decrease by $10%$. After a $10$ day period, what is the probability that you will have more than $50$ dollars?
probability
probability
edited Jan 18 at 14:29
idriskameni
749321
asked Jan 18 at 13:21
Bishesh AdhikariBishesh Adhikari
41
edited Jan 18 at 14:29
idriskameni
749321
asked Jan 18 at 13:21
Bishesh AdhikariBishesh Adhikari
41
edited Jan 18 at 14:29
idriskameni
749321
edited Jan 18 at 14:29
idriskameni
749321
edited Jan 18 at 14:29
idriskameni
749321
749321
asked Jan 18 at 13:21
Bishesh AdhikariBishesh Adhikari
41
asked Jan 18 at 13:21
Bishesh AdhikariBishesh Adhikari
41
asked Jan 18 at 13:21
Bishesh AdhikariBishesh Adhikari
41
41
closed as off-topic by Did, lulu, Math1000, José Carlos Santos, RRL Jan 18 at 17:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, lulu, Math1000, José Carlos Santos, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, lulu, Math1000, José Carlos Santos, RRL Jan 18 at 17:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, lulu, Math1000, José Carlos Santos, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
What is the probability that it increases/decreases? Also your title is bad.
$endgroup$
– lightxbulb
Jan 18 at 13:23
$begingroup$
I think this is essentially a random walk question.
$endgroup$
– Matti P.
Jan 18 at 13:34
$begingroup$
Note: the order of the changes doesn't matter, so you just have to compute the probability that you get $n$ up moves for $nin {0,cdots, 10}$. Once you decide on the probability of an up move this is a straightforward computation.
$endgroup$
– lulu
Jan 18 at 13:40
$begingroup$
Can you clarify your question? As has been pointed out, the problem can't be answered without knowing the probability of the up and down moves. Also: what have you tried? This is a very straightforward binomial calculation (since $10$ is so small there's no need for any approximations to the distribution). Where are you getting stuck?
$endgroup$
– lulu
Jan 18 at 13:54
add a comment |
3
$begingroup$
What is the probability that it increases/decreases? Also your title is bad.
$endgroup$
– lightxbulb
Jan 18 at 13:23
$begingroup$
I think this is essentially a random walk question.
$endgroup$
– Matti P.
Jan 18 at 13:34
$begingroup$
Note: the order of the changes doesn't matter, so you just have to compute the probability that you get $n$ up moves for $nin {0,cdots, 10}$. Once you decide on the probability of an up move this is a straightforward computation.
$endgroup$
– lulu
Jan 18 at 13:40
$begingroup$
Can you clarify your question? As has been pointed out, the problem can't be answered without knowing the probability of the up and down moves. Also: what have you tried? This is a very straightforward binomial calculation (since $10$ is so small there's no need for any approximations to the distribution). Where are you getting stuck?
$endgroup$
– lulu
Jan 18 at 13:54
3
3
$begingroup$
What is the probability that it increases/decreases? Also your title is bad.
$endgroup$
– lightxbulb
Jan 18 at 13:23
$begingroup$
What is the probability that it increases/decreases? Also your title is bad.
$endgroup$
– lightxbulb
Jan 18 at 13:23
$begingroup$
I think this is essentially a random walk question.
$endgroup$
– Matti P.
Jan 18 at 13:34
$begingroup$
I think this is essentially a random walk question.
$endgroup$
– Matti P.
Jan 18 at 13:34
$begingroup$
Note: the order of the changes doesn't matter, so you just have to compute the probability that you get $n$ up moves for $nin {0,cdots, 10}$. Once you decide on the probability of an up move this is a straightforward computation.
$endgroup$
– lulu
Jan 18 at 13:40
$begingroup$
Note: the order of the changes doesn't matter, so you just have to compute the probability that you get $n$ up moves for $nin {0,cdots, 10}$. Once you decide on the probability of an up move this is a straightforward computation.
$endgroup$
– lulu
Jan 18 at 13:40
$begingroup$
Can you clarify your question? As has been pointed out, the problem can't be answered without knowing the probability of the up and down moves. Also: what have you tried? This is a very straightforward binomial calculation (since $10$ is so small there's no need for any approximations to the distribution). Where are you getting stuck?
$endgroup$
– lulu
Jan 18 at 13:54
$begingroup$
Can you clarify your question? As has been pointed out, the problem can't be answered without knowing the probability of the up and down moves. Also: what have you tried? This is a very straightforward binomial calculation (since $10$ is so small there's no need for any approximations to the distribution). Where are you getting stuck?
$endgroup$
– lulu
Jan 18 at 13:54
add a comment |
2 Answers
2
active
oldest
votes
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Let the probability of increasing be $p$ and let the probability of decreasing be $q$ in the understanding that $p+q=1$
Let $X$ denote the number of the $10$ times that there is increasing.
Let $E$ denote the event that you will have more than $50$ dollars after $10$ days.
You will have $50times1.1^Xtimes0.9^{10-X}$ dollars after $10$ days so that:$$P(E)=P(1.1^Xtimes0.9^{10-X}>1)=P(Xgeqcdots)=cdots$$
I leave the rest to you.
answered Jan 18 at 14:00
drhabdrhab
104k545136
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add a comment |
$begingroup$
An enumerative approach would be to consider all sequences of $+$ and $-$ of length 10, where $+$ means increase and $-$ means decrease. There are $2^{10}$ such sequences. All these sequences are equally probable by your hypothesis. You need to count all sequences which lead to an increase. This gives you immediately the result.
By the way, what happens to a decrease below zero? Does it stay at zero or can it become negative?
answered Jan 18 at 14:43
WuestenfuxWuestenfux
5,5561513
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let the probability of increasing be $p$ and let the probability of decreasing be $q$ in the understanding that $p+q=1$
Let $X$ denote the number of the $10$ times that there is increasing.
Let $E$ denote the event that you will have more than $50$ dollars after $10$ days.
You will have $50times1.1^Xtimes0.9^{10-X}$ dollars after $10$ days so that:$$P(E)=P(1.1^Xtimes0.9^{10-X}>1)=P(Xgeqcdots)=cdots$$
I leave the rest to you.
answered Jan 18 at 14:00
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Let the probability of increasing be $p$ and let the probability of decreasing be $q$ in the understanding that $p+q=1$
Let $X$ denote the number of the $10$ times that there is increasing.
Let $E$ denote the event that you will have more than $50$ dollars after $10$ days.
You will have $50times1.1^Xtimes0.9^{10-X}$ dollars after $10$ days so that:$$P(E)=P(1.1^Xtimes0.9^{10-X}>1)=P(Xgeqcdots)=cdots$$
I leave the rest to you.
answered Jan 18 at 14:00
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Let the probability of increasing be $p$ and let the probability of decreasing be $q$ in the understanding that $p+q=1$
Let $X$ denote the number of the $10$ times that there is increasing.
Let $E$ denote the event that you will have more than $50$ dollars after $10$ days.
You will have $50times1.1^Xtimes0.9^{10-X}$ dollars after $10$ days so that:$$P(E)=P(1.1^Xtimes0.9^{10-X}>1)=P(Xgeqcdots)=cdots$$
I leave the rest to you.
answered Jan 18 at 14:00
drhabdrhab
104k545136
$endgroup$
Let the probability of increasing be $p$ and let the probability of decreasing be $q$ in the understanding that $p+q=1$
Let $X$ denote the number of the $10$ times that there is increasing.
Let $E$ denote the event that you will have more than $50$ dollars after $10$ days.
You will have $50times1.1^Xtimes0.9^{10-X}$ dollars after $10$ days so that:$$P(E)=P(1.1^Xtimes0.9^{10-X}>1)=P(Xgeqcdots)=cdots$$
I leave the rest to you.
answered Jan 18 at 14:00
drhabdrhab
104k545136
answered Jan 18 at 14:00
drhabdrhab
104k545136
answered Jan 18 at 14:00
drhabdrhab
104k545136
answered Jan 18 at 14:00
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
$begingroup$
An enumerative approach would be to consider all sequences of $+$ and $-$ of length 10, where $+$ means increase and $-$ means decrease. There are $2^{10}$ such sequences. All these sequences are equally probable by your hypothesis. You need to count all sequences which lead to an increase. This gives you immediately the result.
By the way, what happens to a decrease below zero? Does it stay at zero or can it become negative?
answered Jan 18 at 14:43
WuestenfuxWuestenfux
5,5561513
$endgroup$
add a comment |
$begingroup$
An enumerative approach would be to consider all sequences of $+$ and $-$ of length 10, where $+$ means increase and $-$ means decrease. There are $2^{10}$ such sequences. All these sequences are equally probable by your hypothesis. You need to count all sequences which lead to an increase. This gives you immediately the result.
By the way, what happens to a decrease below zero? Does it stay at zero or can it become negative?
answered Jan 18 at 14:43
WuestenfuxWuestenfux
5,5561513
$endgroup$
add a comment |
$begingroup$
An enumerative approach would be to consider all sequences of $+$ and $-$ of length 10, where $+$ means increase and $-$ means decrease. There are $2^{10}$ such sequences. All these sequences are equally probable by your hypothesis. You need to count all sequences which lead to an increase. This gives you immediately the result.
By the way, what happens to a decrease below zero? Does it stay at zero or can it become negative?
answered Jan 18 at 14:43
WuestenfuxWuestenfux
5,5561513
$endgroup$
An enumerative approach would be to consider all sequences of $+$ and $-$ of length 10, where $+$ means increase and $-$ means decrease. There are $2^{10}$ such sequences. All these sequences are equally probable by your hypothesis. You need to count all sequences which lead to an increase. This gives you immediately the result.
By the way, what happens to a decrease below zero? Does it stay at zero or can it become negative?
answered Jan 18 at 14:43
WuestenfuxWuestenfux
5,5561513
answered Jan 18 at 14:43
WuestenfuxWuestenfux
5,5561513
answered Jan 18 at 14:43
WuestenfuxWuestenfux
5,5561513
answered Jan 18 at 14:43
WuestenfuxWuestenfux
5,5561513
5,5561513
add a comment |
add a comment |
3
$begingroup$
What is the probability that it increases/decreases? Also your title is bad.
$endgroup$
– lightxbulb
Jan 18 at 13:23
$begingroup$
I think this is essentially a random walk question.
$endgroup$
– Matti P.
Jan 18 at 13:34
$begingroup$
Note: the order of the changes doesn't matter, so you just have to compute the probability that you get $n$ up moves for $nin {0,cdots, 10}$. Once you decide on the probability of an up move this is a straightforward computation.
$endgroup$
– lulu
Jan 18 at 13:40
$begingroup$
Can you clarify your question? As has been pointed out, the problem can't be answered without knowing the probability of the up and down moves. Also: what have you tried? This is a very straightforward binomial calculation (since $10$ is so small there's no need for any approximations to the distribution). Where are you getting stuck?
$endgroup$
– lulu
Jan 18 at 13:54