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Probability given in the form of $a^-x$. Is my answer correct?
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I'm having trouble solving this exercise about probability, as I'm new to the subject. It says:
There is a bus supposed to arrive at 8:00. The probability of it to set out with $m$ minutes of delay is $p_m = 3^-m (m = 1; 2; 3; ... ;m neq 0)$.
Due to regulatory issues, the set out is never advanced (i.e. it never sets out before 8:00).
1) What's the probability of the bus setting out at 8:00?
2) Prove that the probability of the bus to set out at 8:10 is double the probability that of after 8:10.
Here is what I think about it, but I have no answer to check with:
I assume all possible values go from 8:00 to infinity, so if I add all those values the probability should be 1. Now, $3^-m$ looks like a series I know, and I know it converges as 1/3 is less than 1, so the sum of all terms is $1*frac11-1/3=frac32$, but as that formula is when n starts at 0, then I have to do $frac32-3^0$, which is 0.5.
Then, I guess that 1-0.5 is the remaining probability, that is, of the bus setting out exactly at 8:00. Is what I've done correct? Because I don't think we're mixing calculus with probabability in the course.
For 2), my guess is calculate $3^-10$, and I have to prove that $3^-10 = 2*sum_m=11^infty 3^-m$. Is it correct? What I do next is calculate the sum, which I don't really know how to do, so I do $3/2 -sum_m=0^10 3^-m$, put that in the calculator and I get that it's true.
So, to wrap it all up... Have I done everything ok? Because I don't have the results. I suppose the second one is correct because I've proven it, but I've no idea about the first one, whether I'm meant to do that or there is another way.
I hope you can help me. Thank you!
probability
edited Jul 17 at 14:24
Key Flex
4,346425
add a comment |Â
up vote
4
down vote
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I'm having trouble solving this exercise about probability, as I'm new to the subject. It says:
There is a bus supposed to arrive at 8:00. The probability of it to set out with $m$ minutes of delay is $p_m = 3^-m (m = 1; 2; 3; ... ;m neq 0)$.
Due to regulatory issues, the set out is never advanced (i.e. it never sets out before 8:00).
1) What's the probability of the bus setting out at 8:00?
2) Prove that the probability of the bus to set out at 8:10 is double the probability that of after 8:10.
Here is what I think about it, but I have no answer to check with:
I assume all possible values go from 8:00 to infinity, so if I add all those values the probability should be 1. Now, $3^-m$ looks like a series I know, and I know it converges as 1/3 is less than 1, so the sum of all terms is $1*frac11-1/3=frac32$, but as that formula is when n starts at 0, then I have to do $frac32-3^0$, which is 0.5.
Then, I guess that 1-0.5 is the remaining probability, that is, of the bus setting out exactly at 8:00. Is what I've done correct? Because I don't think we're mixing calculus with probabability in the course.
For 2), my guess is calculate $3^-10$, and I have to prove that $3^-10 = 2*sum_m=11^infty 3^-m$. Is it correct? What I do next is calculate the sum, which I don't really know how to do, so I do $3/2 -sum_m=0^10 3^-m$, put that in the calculator and I get that it's true.
So, to wrap it all up... Have I done everything ok? Because I don't have the results. I suppose the second one is correct because I've proven it, but I've no idea about the first one, whether I'm meant to do that or there is another way.
I hope you can help me. Thank you!
probability
edited Jul 17 at 14:24
Key Flex
4,346425
@copper.hat I think you have to regard $p_0=0.5$ as well
– callculus
Jul 17 at 14:24
@callculus: Thanks! I realised that after reading more.
– copper.hat
Jul 17 at 14:25
$1 over 3^11 + 1 over 3^12 + cdots = 1 over 3^11(1+ 1 over 3^1 + cdots) = 1 over 3^11 3 over 2$. No need for electronics :-).
– copper.hat
Jul 17 at 14:26
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm having trouble solving this exercise about probability, as I'm new to the subject. It says:
There is a bus supposed to arrive at 8:00. The probability of it to set out with $m$ minutes of delay is $p_m = 3^-m (m = 1; 2; 3; ... ;m neq 0)$.
Due to regulatory issues, the set out is never advanced (i.e. it never sets out before 8:00).
1) What's the probability of the bus setting out at 8:00?
2) Prove that the probability of the bus to set out at 8:10 is double the probability that of after 8:10.
Here is what I think about it, but I have no answer to check with:
I assume all possible values go from 8:00 to infinity, so if I add all those values the probability should be 1. Now, $3^-m$ looks like a series I know, and I know it converges as 1/3 is less than 1, so the sum of all terms is $1*frac11-1/3=frac32$, but as that formula is when n starts at 0, then I have to do $frac32-3^0$, which is 0.5.
Then, I guess that 1-0.5 is the remaining probability, that is, of the bus setting out exactly at 8:00. Is what I've done correct? Because I don't think we're mixing calculus with probabability in the course.
For 2), my guess is calculate $3^-10$, and I have to prove that $3^-10 = 2*sum_m=11^infty 3^-m$. Is it correct? What I do next is calculate the sum, which I don't really know how to do, so I do $3/2 -sum_m=0^10 3^-m$, put that in the calculator and I get that it's true.
So, to wrap it all up... Have I done everything ok? Because I don't have the results. I suppose the second one is correct because I've proven it, but I've no idea about the first one, whether I'm meant to do that or there is another way.
I hope you can help me. Thank you!
probability
edited Jul 17 at 14:24
Key Flex
4,346425
I'm having trouble solving this exercise about probability, as I'm new to the subject. It says:
There is a bus supposed to arrive at 8:00. The probability of it to set out with $m$ minutes of delay is $p_m = 3^-m (m = 1; 2; 3; ... ;m neq 0)$.
Due to regulatory issues, the set out is never advanced (i.e. it never sets out before 8:00).
1) What's the probability of the bus setting out at 8:00?
2) Prove that the probability of the bus to set out at 8:10 is double the probability that of after 8:10.
Here is what I think about it, but I have no answer to check with:
I assume all possible values go from 8:00 to infinity, so if I add all those values the probability should be 1. Now, $3^-m$ looks like a series I know, and I know it converges as 1/3 is less than 1, so the sum of all terms is $1*frac11-1/3=frac32$, but as that formula is when n starts at 0, then I have to do $frac32-3^0$, which is 0.5.
Then, I guess that 1-0.5 is the remaining probability, that is, of the bus setting out exactly at 8:00. Is what I've done correct? Because I don't think we're mixing calculus with probabability in the course.
For 2), my guess is calculate $3^-10$, and I have to prove that $3^-10 = 2*sum_m=11^infty 3^-m$. Is it correct? What I do next is calculate the sum, which I don't really know how to do, so I do $3/2 -sum_m=0^10 3^-m$, put that in the calculator and I get that it's true.
So, to wrap it all up... Have I done everything ok? Because I don't have the results. I suppose the second one is correct because I've proven it, but I've no idea about the first one, whether I'm meant to do that or there is another way.
I hope you can help me. Thank you!
probability
edited Jul 17 at 14:24
Key Flex
4,346425
edited Jul 17 at 14:24
Key Flex
4,346425
edited Jul 17 at 14:24
Key Flex
4,346425
edited Jul 17 at 14:24
Key Flex
4,346425
4,346425
asked Jul 17 at 13:55
user577725
@copper.hat I think you have to regard $p_0=0.5$ as well
– callculus
Jul 17 at 14:24
@callculus: Thanks! I realised that after reading more.
– copper.hat
Jul 17 at 14:25
$1 over 3^11 + 1 over 3^12 + cdots = 1 over 3^11(1+ 1 over 3^1 + cdots) = 1 over 3^11 3 over 2$. No need for electronics :-).
– copper.hat
Jul 17 at 14:26
add a comment |Â
@copper.hat I think you have to regard $p_0=0.5$ as well
– callculus
Jul 17 at 14:24
@callculus: Thanks! I realised that after reading more.
– copper.hat
Jul 17 at 14:25
$1 over 3^11 + 1 over 3^12 + cdots = 1 over 3^11(1+ 1 over 3^1 + cdots) = 1 over 3^11 3 over 2$. No need for electronics :-).
– copper.hat
Jul 17 at 14:26
@copper.hat I think you have to regard $p_0=0.5$ as well
– callculus
Jul 17 at 14:24
@copper.hat I think you have to regard $p_0=0.5$ as well
– callculus
Jul 17 at 14:24
@callculus: Thanks! I realised that after reading more.
– copper.hat
Jul 17 at 14:25
@callculus: Thanks! I realised that after reading more.
– copper.hat
Jul 17 at 14:25
$1 over 3^11 + 1 over 3^12 + cdots = 1 over 3^11(1+ 1 over 3^1 + cdots) = 1 over 3^11 3 over 2$. No need for electronics :-).
– copper.hat
Jul 17 at 14:26
$1 over 3^11 + 1 over 3^12 + cdots = 1 over 3^11(1+ 1 over 3^1 + cdots) = 1 over 3^11 3 over 2$. No need for electronics :-).
– copper.hat
Jul 17 at 14:26
add a comment |Â
1 Answer
1
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up vote
1
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You have (1) correct. There is no calculus involved, just geometric series and probability.
From geometric series we indeed have: $mathsf P(mneq0)~=sum_m=1^infty 3^-m\=frac 12qquadcheckmark$
From the law of complementary probability: $mathsf P(m=0)~=1-mathsf P(mneq 0)\=tfrac12qquadcheckmark$
For (b) note that $mathsf P(m>10)~=sum_m=11^infty 3^-m\=3^-10sum_n=1^infty 3^-n\~~vdots$
answered Jul 17 at 14:46
Graham Kemp
80.1k43275
Is that a property or how did you do that in (b)?
– user577725
Jul 17 at 14:52
1
Distribution of common factor, $sum_m=11^infty 3^-m=3^-10sum_m=11^infty 3^10-m$, and a change of bound variable, $sum_m=11^infty 3^10-m=sum_n+10=11^infty 3^10-(n+10)$. @Esteban
– Graham Kemp
Jul 17 at 15:46
Oh, I understand. Thank you!
– user577725
Jul 17 at 15:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have (1) correct. There is no calculus involved, just geometric series and probability.
From geometric series we indeed have: $mathsf P(mneq0)~=sum_m=1^infty 3^-m\=frac 12qquadcheckmark$
From the law of complementary probability: $mathsf P(m=0)~=1-mathsf P(mneq 0)\=tfrac12qquadcheckmark$
For (b) note that $mathsf P(m>10)~=sum_m=11^infty 3^-m\=3^-10sum_n=1^infty 3^-n\~~vdots$
answered Jul 17 at 14:46
Graham Kemp
80.1k43275
Is that a property or how did you do that in (b)?
– user577725
Jul 17 at 14:52
1
Distribution of common factor, $sum_m=11^infty 3^-m=3^-10sum_m=11^infty 3^10-m$, and a change of bound variable, $sum_m=11^infty 3^10-m=sum_n+10=11^infty 3^10-(n+10)$. @Esteban
– Graham Kemp
Jul 17 at 15:46
Oh, I understand. Thank you!
– user577725
Jul 17 at 15:53
add a comment |Â
up vote
1
down vote
accepted
You have (1) correct. There is no calculus involved, just geometric series and probability.
From geometric series we indeed have: $mathsf P(mneq0)~=sum_m=1^infty 3^-m\=frac 12qquadcheckmark$
From the law of complementary probability: $mathsf P(m=0)~=1-mathsf P(mneq 0)\=tfrac12qquadcheckmark$
For (b) note that $mathsf P(m>10)~=sum_m=11^infty 3^-m\=3^-10sum_n=1^infty 3^-n\~~vdots$
answered Jul 17 at 14:46
Graham Kemp
80.1k43275
Is that a property or how did you do that in (b)?
– user577725
Jul 17 at 14:52
1
Distribution of common factor, $sum_m=11^infty 3^-m=3^-10sum_m=11^infty 3^10-m$, and a change of bound variable, $sum_m=11^infty 3^10-m=sum_n+10=11^infty 3^10-(n+10)$. @Esteban
– Graham Kemp
Jul 17 at 15:46
Oh, I understand. Thank you!
– user577725
Jul 17 at 15:53
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have (1) correct. There is no calculus involved, just geometric series and probability.
From geometric series we indeed have: $mathsf P(mneq0)~=sum_m=1^infty 3^-m\=frac 12qquadcheckmark$
From the law of complementary probability: $mathsf P(m=0)~=1-mathsf P(mneq 0)\=tfrac12qquadcheckmark$
For (b) note that $mathsf P(m>10)~=sum_m=11^infty 3^-m\=3^-10sum_n=1^infty 3^-n\~~vdots$
answered Jul 17 at 14:46
Graham Kemp
80.1k43275
You have (1) correct. There is no calculus involved, just geometric series and probability.
From geometric series we indeed have: $mathsf P(mneq0)~=sum_m=1^infty 3^-m\=frac 12qquadcheckmark$
From the law of complementary probability: $mathsf P(m=0)~=1-mathsf P(mneq 0)\=tfrac12qquadcheckmark$
For (b) note that $mathsf P(m>10)~=sum_m=11^infty 3^-m\=3^-10sum_n=1^infty 3^-n\~~vdots$
answered Jul 17 at 14:46
Graham Kemp
80.1k43275
answered Jul 17 at 14:46
Graham Kemp
80.1k43275
answered Jul 17 at 14:46
Graham Kemp
80.1k43275
answered Jul 17 at 14:46
Graham Kemp
80.1k43275
80.1k43275
Is that a property or how did you do that in (b)?
– user577725
Jul 17 at 14:52
1
Distribution of common factor, $sum_m=11^infty 3^-m=3^-10sum_m=11^infty 3^10-m$, and a change of bound variable, $sum_m=11^infty 3^10-m=sum_n+10=11^infty 3^10-(n+10)$. @Esteban
– Graham Kemp
Jul 17 at 15:46
Oh, I understand. Thank you!
– user577725
Jul 17 at 15:53
add a comment |Â
Is that a property or how did you do that in (b)?
– user577725
Jul 17 at 14:52
1
Distribution of common factor, $sum_m=11^infty 3^-m=3^-10sum_m=11^infty 3^10-m$, and a change of bound variable, $sum_m=11^infty 3^10-m=sum_n+10=11^infty 3^10-(n+10)$. @Esteban
– Graham Kemp
Jul 17 at 15:46
Oh, I understand. Thank you!
– user577725
Jul 17 at 15:53
Is that a property or how did you do that in (b)?
– user577725
Jul 17 at 14:52
Is that a property or how did you do that in (b)?
– user577725
Jul 17 at 14:52
1
1
Distribution of common factor, $sum_m=11^infty 3^-m=3^-10sum_m=11^infty 3^10-m$, and a change of bound variable, $sum_m=11^infty 3^10-m=sum_n+10=11^infty 3^10-(n+10)$. @Esteban
– Graham Kemp
Jul 17 at 15:46
Distribution of common factor, $sum_m=11^infty 3^-m=3^-10sum_m=11^infty 3^10-m$, and a change of bound variable, $sum_m=11^infty 3^10-m=sum_n+10=11^infty 3^10-(n+10)$. @Esteban
– Graham Kemp
Jul 17 at 15:46
Oh, I understand. Thank you!
– user577725
Jul 17 at 15:53
Oh, I understand. Thank you!
– user577725
Jul 17 at 15:53
add a comment |Â
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@copper.hat I think you have to regard $p_0=0.5$ as well
– callculus
Jul 17 at 14:24
@callculus: Thanks! I realised that after reading more.
– copper.hat
Jul 17 at 14:25
$1 over 3^11 + 1 over 3^12 + cdots = 1 over 3^11(1+ 1 over 3^1 + cdots) = 1 over 3^11 3 over 2$. No need for electronics :-).
– copper.hat
Jul 17 at 14:26