Mixing
Simple problem on conditional geometric probability
Clash Royale CLAN TAG#URR8PPP
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The occurrence of the event $A$ is equally likely in every moment of the interval $[0, T]$. The probability of event $A$ occurring at all in this interval is $p$. Given that $A$ hasn't occurred in the interval $[0,t]$ what's the probability that $A$ will occur in $[t, T]$?
I am getting a different answer than the one given in the book. I wonder which one is correct. My answer is
$dfracTp-tpT-pt$
The book's answer is my answer divided by $p$.
probability
asked Aug 6 at 12:44
peter.petrov
5,321721
add a comment |Â
up vote
4
down vote
favorite
The occurrence of the event $A$ is equally likely in every moment of the interval $[0, T]$. The probability of event $A$ occurring at all in this interval is $p$. Given that $A$ hasn't occurred in the interval $[0,t]$ what's the probability that $A$ will occur in $[t, T]$?
I am getting a different answer than the one given in the book. I wonder which one is correct. My answer is
$dfracTp-tpT-pt$
The book's answer is my answer divided by $p$.
probability
asked Aug 6 at 12:44
peter.petrov
5,321721
Sorry for the bad formatting. I will improve it a bit later.
– peter.petrov
Aug 6 at 12:47
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The occurrence of the event $A$ is equally likely in every moment of the interval $[0, T]$. The probability of event $A$ occurring at all in this interval is $p$. Given that $A$ hasn't occurred in the interval $[0,t]$ what's the probability that $A$ will occur in $[t, T]$?
I am getting a different answer than the one given in the book. I wonder which one is correct. My answer is
$dfracTp-tpT-pt$
The book's answer is my answer divided by $p$.
probability
asked Aug 6 at 12:44
peter.petrov
5,321721
The occurrence of the event $A$ is equally likely in every moment of the interval $[0, T]$. The probability of event $A$ occurring at all in this interval is $p$. Given that $A$ hasn't occurred in the interval $[0,t]$ what's the probability that $A$ will occur in $[t, T]$?
I am getting a different answer than the one given in the book. I wonder which one is correct. My answer is
$dfracTp-tpT-pt$
The book's answer is my answer divided by $p$.
probability
asked Aug 6 at 12:44
peter.petrov
5,321721
edited Aug 6 at 13:13
asked Aug 6 at 12:44
peter.petrov
5,321721
asked Aug 6 at 12:44
peter.petrov
5,321721
asked Aug 6 at 12:44
peter.petrov
5,321721
5,321721
Sorry for the bad formatting. I will improve it a bit later.
– peter.petrov
Aug 6 at 12:47
add a comment |Â
Sorry for the bad formatting. I will improve it a bit later.
– peter.petrov
Aug 6 at 12:47
Sorry for the bad formatting. I will improve it a bit later.
– peter.petrov
Aug 6 at 12:47
Sorry for the bad formatting. I will improve it a bit later.
– peter.petrov
Aug 6 at 12:47
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
8
down vote
Look at the case when $t = 0$, it is the probability that the event $A$ occurs in $[0, T]$.
You see that the book is mistaken right?
edited Aug 7 at 12:44
pointguard0
1,152819
answered Aug 6 at 12:52
Pjonin
3206
Please use mathjax for your equation formatting. Thank you.
– Rumpelstiltskin
Aug 6 at 13:06
Oh right. I should have applied that trick. Thanks everyone for their answers. Much appreciated.
– peter.petrov
Aug 6 at 13:12
add a comment |Â
up vote
4
down vote
I concur with your answer. Â It looks like there is a typo in your book.
$beginalignmathsf P(Ain [t,T)mid Anotin [0,t)) &=dfracmathsf P(Anotin[0,t)cap Ain[t,T))mathsf P(Anotin[0,t)cap Ain[t,T))+mathsf P(Anotin[0,t)cap Anotin[t,T))\ &=dfracp(T-t)/Tp(T-t)/T+(1-p)\ &=dfracp~(T-t)T-ptendalign$
answered Aug 6 at 13:01
Graham Kemp
80.1k43275
add a comment |Â
up vote
3
down vote
We have $$Pr(Ain (0,T)mid Anotin(0,t))=fracPr(Ain (t,T))Pr(Anotin(0,t).$$
The top probability is $pfracT-tT$ and the bottom is $1-pfractT$; rearranging this gives the same answer you have.
answered Aug 6 at 13:00
Especially Lime
19.1k22252
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
Look at the case when $t = 0$, it is the probability that the event $A$ occurs in $[0, T]$.
You see that the book is mistaken right?
edited Aug 7 at 12:44
pointguard0
1,152819
answered Aug 6 at 12:52
Pjonin
3206
Please use mathjax for your equation formatting. Thank you.
– Rumpelstiltskin
Aug 6 at 13:06
Oh right. I should have applied that trick. Thanks everyone for their answers. Much appreciated.
– peter.petrov
Aug 6 at 13:12
add a comment |Â
up vote
8
down vote
Look at the case when $t = 0$, it is the probability that the event $A$ occurs in $[0, T]$.
You see that the book is mistaken right?
edited Aug 7 at 12:44
pointguard0
1,152819
answered Aug 6 at 12:52
Pjonin
3206
Please use mathjax for your equation formatting. Thank you.
– Rumpelstiltskin
Aug 6 at 13:06
Oh right. I should have applied that trick. Thanks everyone for their answers. Much appreciated.
– peter.petrov
Aug 6 at 13:12
add a comment |Â
up vote
8
down vote
up vote
8
down vote
Look at the case when $t = 0$, it is the probability that the event $A$ occurs in $[0, T]$.
You see that the book is mistaken right?
edited Aug 7 at 12:44
pointguard0
1,152819
answered Aug 6 at 12:52
Pjonin
3206
Look at the case when $t = 0$, it is the probability that the event $A$ occurs in $[0, T]$.
You see that the book is mistaken right?
edited Aug 7 at 12:44
pointguard0
1,152819
answered Aug 6 at 12:52
Pjonin
3206
edited Aug 7 at 12:44
pointguard0
1,152819
edited Aug 7 at 12:44
pointguard0
1,152819
edited Aug 7 at 12:44
pointguard0
1,152819
1,152819
answered Aug 6 at 12:52
Pjonin
3206
answered Aug 6 at 12:52
Pjonin
3206
answered Aug 6 at 12:52
Pjonin
3206
3206
Please use mathjax for your equation formatting. Thank you.
– Rumpelstiltskin
Aug 6 at 13:06
Oh right. I should have applied that trick. Thanks everyone for their answers. Much appreciated.
– peter.petrov
Aug 6 at 13:12
add a comment |Â
Please use mathjax for your equation formatting. Thank you.
– Rumpelstiltskin
Aug 6 at 13:06
Oh right. I should have applied that trick. Thanks everyone for their answers. Much appreciated.
– peter.petrov
Aug 6 at 13:12
Please use mathjax for your equation formatting. Thank you.
– Rumpelstiltskin
Aug 6 at 13:06
Please use mathjax for your equation formatting. Thank you.
– Rumpelstiltskin
Aug 6 at 13:06
Oh right. I should have applied that trick. Thanks everyone for their answers. Much appreciated.
– peter.petrov
Aug 6 at 13:12
Oh right. I should have applied that trick. Thanks everyone for their answers. Much appreciated.
– peter.petrov
Aug 6 at 13:12
add a comment |Â
up vote
4
down vote
I concur with your answer. Â It looks like there is a typo in your book.
$beginalignmathsf P(Ain [t,T)mid Anotin [0,t)) &=dfracmathsf P(Anotin[0,t)cap Ain[t,T))mathsf P(Anotin[0,t)cap Ain[t,T))+mathsf P(Anotin[0,t)cap Anotin[t,T))\ &=dfracp(T-t)/Tp(T-t)/T+(1-p)\ &=dfracp~(T-t)T-ptendalign$
answered Aug 6 at 13:01
Graham Kemp
80.1k43275
add a comment |Â
up vote
4
down vote
I concur with your answer. Â It looks like there is a typo in your book.
$beginalignmathsf P(Ain [t,T)mid Anotin [0,t)) &=dfracmathsf P(Anotin[0,t)cap Ain[t,T))mathsf P(Anotin[0,t)cap Ain[t,T))+mathsf P(Anotin[0,t)cap Anotin[t,T))\ &=dfracp(T-t)/Tp(T-t)/T+(1-p)\ &=dfracp~(T-t)T-ptendalign$
answered Aug 6 at 13:01
Graham Kemp
80.1k43275
add a comment |Â
up vote
4
down vote
up vote
4
down vote
I concur with your answer. Â It looks like there is a typo in your book.
$beginalignmathsf P(Ain [t,T)mid Anotin [0,t)) &=dfracmathsf P(Anotin[0,t)cap Ain[t,T))mathsf P(Anotin[0,t)cap Ain[t,T))+mathsf P(Anotin[0,t)cap Anotin[t,T))\ &=dfracp(T-t)/Tp(T-t)/T+(1-p)\ &=dfracp~(T-t)T-ptendalign$
answered Aug 6 at 13:01
Graham Kemp
80.1k43275
I concur with your answer. Â It looks like there is a typo in your book.
$beginalignmathsf P(Ain [t,T)mid Anotin [0,t)) &=dfracmathsf P(Anotin[0,t)cap Ain[t,T))mathsf P(Anotin[0,t)cap Ain[t,T))+mathsf P(Anotin[0,t)cap Anotin[t,T))\ &=dfracp(T-t)/Tp(T-t)/T+(1-p)\ &=dfracp~(T-t)T-ptendalign$
answered Aug 6 at 13:01
Graham Kemp
80.1k43275
answered Aug 6 at 13:01
Graham Kemp
80.1k43275
answered Aug 6 at 13:01
Graham Kemp
80.1k43275
answered Aug 6 at 13:01
Graham Kemp
80.1k43275
80.1k43275
add a comment |Â
add a comment |Â
up vote
3
down vote
We have $$Pr(Ain (0,T)mid Anotin(0,t))=fracPr(Ain (t,T))Pr(Anotin(0,t).$$
The top probability is $pfracT-tT$ and the bottom is $1-pfractT$; rearranging this gives the same answer you have.
answered Aug 6 at 13:00
Especially Lime
19.1k22252
add a comment |Â
up vote
3
down vote
We have $$Pr(Ain (0,T)mid Anotin(0,t))=fracPr(Ain (t,T))Pr(Anotin(0,t).$$
The top probability is $pfracT-tT$ and the bottom is $1-pfractT$; rearranging this gives the same answer you have.
answered Aug 6 at 13:00
Especially Lime
19.1k22252
add a comment |Â
up vote
3
down vote
up vote
3
down vote
We have $$Pr(Ain (0,T)mid Anotin(0,t))=fracPr(Ain (t,T))Pr(Anotin(0,t).$$
The top probability is $pfracT-tT$ and the bottom is $1-pfractT$; rearranging this gives the same answer you have.
answered Aug 6 at 13:00
Especially Lime
19.1k22252
We have $$Pr(Ain (0,T)mid Anotin(0,t))=fracPr(Ain (t,T))Pr(Anotin(0,t).$$
The top probability is $pfracT-tT$ and the bottom is $1-pfractT$; rearranging this gives the same answer you have.
answered Aug 6 at 13:00
Especially Lime
19.1k22252
answered Aug 6 at 13:00
Especially Lime
19.1k22252
answered Aug 6 at 13:00
Especially Lime
19.1k22252
answered Aug 6 at 13:00
Especially Lime
19.1k22252
19.1k22252
add a comment |Â
add a comment |Â
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Sorry for the bad formatting. I will improve it a bit later.
– peter.petrov
Aug 6 at 12:47