Mixing
Determine $P(X > n +kmid X > n)$
$begingroup$
Bob is at the shooting range.
With probability $frac{1}{3}$ Bob hits the target.
Every shot is independent of the previous ones.
Bob starts and keeps shooting until he hits the target.
Let the random variable $X$ be the number of the shot that first hits the target.
For integers $n$ and $k$, determine $P(X > n +kmid X > n)$.
I start working on the conditional probability that becomes: $P(X > n +kmid X > n)=frac{P(X > n + k:cap:X > n)}{P(X > n)}=frac{P(X > n+ k)}{P(X > n)}$
Now how can I compute the probabilities from the data that I know, which distribution should I use, or how can I use the probabilities given in the question?
probability
asked Jan 19 at 17:07
FTACFTAC
2649
$endgroup$
add a comment |
$begingroup$
Bob is at the shooting range.
With probability $frac{1}{3}$ Bob hits the target.
Every shot is independent of the previous ones.
Bob starts and keeps shooting until he hits the target.
Let the random variable $X$ be the number of the shot that first hits the target.
For integers $n$ and $k$, determine $P(X > n +kmid X > n)$.
I start working on the conditional probability that becomes: $P(X > n +kmid X > n)=frac{P(X > n + k:cap:X > n)}{P(X > n)}=frac{P(X > n+ k)}{P(X > n)}$
Now how can I compute the probabilities from the data that I know, which distribution should I use, or how can I use the probabilities given in the question?
probability
asked Jan 19 at 17:07
FTACFTAC
2649
$endgroup$
1
$begingroup$
What does "$X gt n$" mean in words? Can you find $P(X gt n)$?
$endgroup$
– Henry
Jan 19 at 17:09
1
$begingroup$
Is the probability that the number of the shot that first hits the target is greater than a positive integer $n$. No, I don't.. :(
$endgroup$
– FTAC
Jan 19 at 17:14
1
$begingroup$
Does Alice shoot too? This is not clear.
$endgroup$
– Bernard
Jan 19 at 17:23
$begingroup$
No, only Bob, I'll edit the question
$endgroup$
– FTAC
Jan 19 at 17:26
add a comment |
$begingroup$
Bob is at the shooting range.
With probability $frac{1}{3}$ Bob hits the target.
Every shot is independent of the previous ones.
Bob starts and keeps shooting until he hits the target.
Let the random variable $X$ be the number of the shot that first hits the target.
For integers $n$ and $k$, determine $P(X > n +kmid X > n)$.
I start working on the conditional probability that becomes: $P(X > n +kmid X > n)=frac{P(X > n + k:cap:X > n)}{P(X > n)}=frac{P(X > n+ k)}{P(X > n)}$
Now how can I compute the probabilities from the data that I know, which distribution should I use, or how can I use the probabilities given in the question?
probability
asked Jan 19 at 17:07
FTACFTAC
2649
$endgroup$
Bob is at the shooting range.
With probability $frac{1}{3}$ Bob hits the target.
Every shot is independent of the previous ones.
Bob starts and keeps shooting until he hits the target.
Let the random variable $X$ be the number of the shot that first hits the target.
For integers $n$ and $k$, determine $P(X > n +kmid X > n)$.
I start working on the conditional probability that becomes: $P(X > n +kmid X > n)=frac{P(X > n + k:cap:X > n)}{P(X > n)}=frac{P(X > n+ k)}{P(X > n)}$
Now how can I compute the probabilities from the data that I know, which distribution should I use, or how can I use the probabilities given in the question?
probability
probability
asked Jan 19 at 17:07
FTACFTAC
2649
asked Jan 19 at 17:07
FTACFTAC
2649
edited Jan 19 at 17:26
FTAC
asked Jan 19 at 17:07
FTACFTAC
2649
asked Jan 19 at 17:07
FTACFTAC
2649
asked Jan 19 at 17:07
FTACFTAC
2649
2649
1
$begingroup$
What does "$X gt n$" mean in words? Can you find $P(X gt n)$?
$endgroup$
– Henry
Jan 19 at 17:09
1
$begingroup$
Is the probability that the number of the shot that first hits the target is greater than a positive integer $n$. No, I don't.. :(
$endgroup$
– FTAC
Jan 19 at 17:14
1
$begingroup$
Does Alice shoot too? This is not clear.
$endgroup$
– Bernard
Jan 19 at 17:23
$begingroup$
No, only Bob, I'll edit the question
$endgroup$
– FTAC
Jan 19 at 17:26
add a comment |
1
$begingroup$
What does "$X gt n$" mean in words? Can you find $P(X gt n)$?
$endgroup$
– Henry
Jan 19 at 17:09
1
$begingroup$
Is the probability that the number of the shot that first hits the target is greater than a positive integer $n$. No, I don't.. :(
$endgroup$
– FTAC
Jan 19 at 17:14
1
$begingroup$
Does Alice shoot too? This is not clear.
$endgroup$
– Bernard
Jan 19 at 17:23
$begingroup$
No, only Bob, I'll edit the question
$endgroup$
– FTAC
Jan 19 at 17:26
1
1
$begingroup$
What does "$X gt n$" mean in words? Can you find $P(X gt n)$?
$endgroup$
– Henry
Jan 19 at 17:09
$begingroup$
What does "$X gt n$" mean in words? Can you find $P(X gt n)$?
$endgroup$
– Henry
Jan 19 at 17:09
1
1
$begingroup$
Is the probability that the number of the shot that first hits the target is greater than a positive integer $n$. No, I don't.. :(
$endgroup$
– FTAC
Jan 19 at 17:14
$begingroup$
Is the probability that the number of the shot that first hits the target is greater than a positive integer $n$. No, I don't.. :(
$endgroup$
– FTAC
Jan 19 at 17:14
1
1
$begingroup$
Does Alice shoot too? This is not clear.
$endgroup$
– Bernard
Jan 19 at 17:23
$begingroup$
Does Alice shoot too? This is not clear.
$endgroup$
– Bernard
Jan 19 at 17:23
$begingroup$
No, only Bob, I'll edit the question
$endgroup$
– FTAC
Jan 19 at 17:26
$begingroup$
No, only Bob, I'll edit the question
$endgroup$
– FTAC
Jan 19 at 17:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint:
Observe that event ${X>m}$ is the same as the event that the first $m$ attempts did not succeed.
Find the probability of this event for $m=n$ and for $m=n+k$.
Google on "geometric distribution".
answered Jan 19 at 17:32
drhabdrhab
102k545136
$endgroup$
$begingroup$
Thanks for the hint, so for a geometric distribution $P(X=m)=(1-p)^{m-1}$ and $P(X>m)=(1-p)^{m-1}$ right?
$endgroup$
– FTAC
Jan 19 at 17:53
1
$begingroup$
$P(X>m)=P(text{first fails and …and }mtext{-th fails})=(1-p)^m$
$endgroup$
– drhab
Jan 19 at 17:59
$begingroup$
What you write is not Bernoulli distribution? Should I use Geometric for this example?
$endgroup$
– FTAC
Jan 19 at 18:05
1
$begingroup$
$X$ has geometric distribution here and consequently $P(X>m)=(1-p)^m$. Further $P(X=m)=(1-p)^{m-1}p$ but that is not relevant. You only need $P(X>m)$.
$endgroup$
– drhab
Jan 19 at 18:08
$begingroup$
Thanks for the help, now it's clear!
$endgroup$
– FTAC
Jan 19 at 18:15
|
show 1 more comment
$begingroup$
The way I understand this is:
What's the probability that it will take at least n+k shots to git the target given that it will take at least n shots?
So I imagine this scenario:
Bob already took n shots (and missed), what's the probability that he'll have to take at least k more shots before he hits the target?
Note the condition given: Every shot is independent of the previous ones.
So to me it is clear that it doesn't matter that Bob already took n shots and missed. The chances he'll need k more shots now is the same as if he had just now started trying. In other words:
P(X > n + k | X > n) = P(X > k)
answered Jan 19 at 17:38
WalterWalter
662
$endgroup$
$begingroup$
This makes sense, thanks for the hint!
$endgroup$
– FTAC
Jan 19 at 17:54
$begingroup$
This is the memoryless property of the geometric distribution
$endgroup$
– Henry
Jan 19 at 19:01
add a comment |
$begingroup$
The probability of Bob not hitting the target is $2/3$, so the probability that he misses $x$ times is $(2/3)^x$, since the individual shots are independent. How could you express the statement "Bob misses the target $n$ times" symbolically?
answered Jan 19 at 17:31
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
$begingroup$
$P(X>n)=(1-frac{1}{3})^{n-1} right?$
$endgroup$
– FTAC
Jan 19 at 17:51
$begingroup$
not quite: for $X>n$ he has to miss $n$ times
$endgroup$
– Henry
Jan 19 at 19:00
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Observe that event ${X>m}$ is the same as the event that the first $m$ attempts did not succeed.
Find the probability of this event for $m=n$ and for $m=n+k$.
Google on "geometric distribution".
answered Jan 19 at 17:32
drhabdrhab
102k545136
$endgroup$
$begingroup$
Thanks for the hint, so for a geometric distribution $P(X=m)=(1-p)^{m-1}$ and $P(X>m)=(1-p)^{m-1}$ right?
$endgroup$
– FTAC
Jan 19 at 17:53
1
$begingroup$
$P(X>m)=P(text{first fails and …and }mtext{-th fails})=(1-p)^m$
$endgroup$
– drhab
Jan 19 at 17:59
$begingroup$
What you write is not Bernoulli distribution? Should I use Geometric for this example?
$endgroup$
– FTAC
Jan 19 at 18:05
1
$begingroup$
$X$ has geometric distribution here and consequently $P(X>m)=(1-p)^m$. Further $P(X=m)=(1-p)^{m-1}p$ but that is not relevant. You only need $P(X>m)$.
$endgroup$
– drhab
Jan 19 at 18:08
$begingroup$
Thanks for the help, now it's clear!
$endgroup$
– FTAC
Jan 19 at 18:15
|
show 1 more comment
$begingroup$
Hint:
Observe that event ${X>m}$ is the same as the event that the first $m$ attempts did not succeed.
Find the probability of this event for $m=n$ and for $m=n+k$.
Google on "geometric distribution".
answered Jan 19 at 17:32
drhabdrhab
102k545136
$endgroup$
$begingroup$
Thanks for the hint, so for a geometric distribution $P(X=m)=(1-p)^{m-1}$ and $P(X>m)=(1-p)^{m-1}$ right?
$endgroup$
– FTAC
Jan 19 at 17:53
1
$begingroup$
$P(X>m)=P(text{first fails and …and }mtext{-th fails})=(1-p)^m$
$endgroup$
– drhab
Jan 19 at 17:59
$begingroup$
What you write is not Bernoulli distribution? Should I use Geometric for this example?
$endgroup$
– FTAC
Jan 19 at 18:05
1
$begingroup$
$X$ has geometric distribution here and consequently $P(X>m)=(1-p)^m$. Further $P(X=m)=(1-p)^{m-1}p$ but that is not relevant. You only need $P(X>m)$.
$endgroup$
– drhab
Jan 19 at 18:08
$begingroup$
Thanks for the help, now it's clear!
$endgroup$
– FTAC
Jan 19 at 18:15
|
show 1 more comment
$begingroup$
Hint:
Observe that event ${X>m}$ is the same as the event that the first $m$ attempts did not succeed.
Find the probability of this event for $m=n$ and for $m=n+k$.
Google on "geometric distribution".
answered Jan 19 at 17:32
drhabdrhab
102k545136
$endgroup$
Hint:
Observe that event ${X>m}$ is the same as the event that the first $m$ attempts did not succeed.
Find the probability of this event for $m=n$ and for $m=n+k$.
Google on "geometric distribution".
answered Jan 19 at 17:32
drhabdrhab
102k545136
answered Jan 19 at 17:32
drhabdrhab
102k545136
answered Jan 19 at 17:32
drhabdrhab
102k545136
answered Jan 19 at 17:32
drhabdrhab
102k545136
102k545136
$begingroup$
Thanks for the hint, so for a geometric distribution $P(X=m)=(1-p)^{m-1}$ and $P(X>m)=(1-p)^{m-1}$ right?
$endgroup$
– FTAC
Jan 19 at 17:53
1
$begingroup$
$P(X>m)=P(text{first fails and …and }mtext{-th fails})=(1-p)^m$
$endgroup$
– drhab
Jan 19 at 17:59
$begingroup$
What you write is not Bernoulli distribution? Should I use Geometric for this example?
$endgroup$
– FTAC
Jan 19 at 18:05
1
$begingroup$
$X$ has geometric distribution here and consequently $P(X>m)=(1-p)^m$. Further $P(X=m)=(1-p)^{m-1}p$ but that is not relevant. You only need $P(X>m)$.
$endgroup$
– drhab
Jan 19 at 18:08
$begingroup$
Thanks for the help, now it's clear!
$endgroup$
– FTAC
Jan 19 at 18:15
|
show 1 more comment
$begingroup$
Thanks for the hint, so for a geometric distribution $P(X=m)=(1-p)^{m-1}$ and $P(X>m)=(1-p)^{m-1}$ right?
$endgroup$
– FTAC
Jan 19 at 17:53
1
$begingroup$
$P(X>m)=P(text{first fails and …and }mtext{-th fails})=(1-p)^m$
$endgroup$
– drhab
Jan 19 at 17:59
$begingroup$
What you write is not Bernoulli distribution? Should I use Geometric for this example?
$endgroup$
– FTAC
Jan 19 at 18:05
1
$begingroup$
$X$ has geometric distribution here and consequently $P(X>m)=(1-p)^m$. Further $P(X=m)=(1-p)^{m-1}p$ but that is not relevant. You only need $P(X>m)$.
$endgroup$
– drhab
Jan 19 at 18:08
$begingroup$
Thanks for the help, now it's clear!
$endgroup$
– FTAC
Jan 19 at 18:15
$begingroup$
Thanks for the hint, so for a geometric distribution $P(X=m)=(1-p)^{m-1}$ and $P(X>m)=(1-p)^{m-1}$ right?
$endgroup$
– FTAC
Jan 19 at 17:53
$begingroup$
Thanks for the hint, so for a geometric distribution $P(X=m)=(1-p)^{m-1}$ and $P(X>m)=(1-p)^{m-1}$ right?
$endgroup$
– FTAC
Jan 19 at 17:53
1
1
$begingroup$
$P(X>m)=P(text{first fails and …and }mtext{-th fails})=(1-p)^m$
$endgroup$
– drhab
Jan 19 at 17:59
$begingroup$
$P(X>m)=P(text{first fails and …and }mtext{-th fails})=(1-p)^m$
$endgroup$
– drhab
Jan 19 at 17:59
$begingroup$
What you write is not Bernoulli distribution? Should I use Geometric for this example?
$endgroup$
– FTAC
Jan 19 at 18:05
$begingroup$
What you write is not Bernoulli distribution? Should I use Geometric for this example?
$endgroup$
– FTAC
Jan 19 at 18:05
1
1
$begingroup$
$X$ has geometric distribution here and consequently $P(X>m)=(1-p)^m$. Further $P(X=m)=(1-p)^{m-1}p$ but that is not relevant. You only need $P(X>m)$.
$endgroup$
– drhab
Jan 19 at 18:08
$begingroup$
$X$ has geometric distribution here and consequently $P(X>m)=(1-p)^m$. Further $P(X=m)=(1-p)^{m-1}p$ but that is not relevant. You only need $P(X>m)$.
$endgroup$
– drhab
Jan 19 at 18:08
$begingroup$
Thanks for the help, now it's clear!
$endgroup$
– FTAC
Jan 19 at 18:15
$begingroup$
Thanks for the help, now it's clear!
$endgroup$
– FTAC
Jan 19 at 18:15
|
show 1 more comment
$begingroup$
The way I understand this is:
What's the probability that it will take at least n+k shots to git the target given that it will take at least n shots?
So I imagine this scenario:
Bob already took n shots (and missed), what's the probability that he'll have to take at least k more shots before he hits the target?
Note the condition given: Every shot is independent of the previous ones.
So to me it is clear that it doesn't matter that Bob already took n shots and missed. The chances he'll need k more shots now is the same as if he had just now started trying. In other words:
P(X > n + k | X > n) = P(X > k)
answered Jan 19 at 17:38
WalterWalter
662
$endgroup$
$begingroup$
This makes sense, thanks for the hint!
$endgroup$
– FTAC
Jan 19 at 17:54
$begingroup$
This is the memoryless property of the geometric distribution
$endgroup$
– Henry
Jan 19 at 19:01
add a comment |
$begingroup$
The way I understand this is:
What's the probability that it will take at least n+k shots to git the target given that it will take at least n shots?
So I imagine this scenario:
Bob already took n shots (and missed), what's the probability that he'll have to take at least k more shots before he hits the target?
Note the condition given: Every shot is independent of the previous ones.
So to me it is clear that it doesn't matter that Bob already took n shots and missed. The chances he'll need k more shots now is the same as if he had just now started trying. In other words:
P(X > n + k | X > n) = P(X > k)
answered Jan 19 at 17:38
WalterWalter
662
$endgroup$
$begingroup$
This makes sense, thanks for the hint!
$endgroup$
– FTAC
Jan 19 at 17:54
$begingroup$
This is the memoryless property of the geometric distribution
$endgroup$
– Henry
Jan 19 at 19:01
add a comment |
$begingroup$
The way I understand this is:
What's the probability that it will take at least n+k shots to git the target given that it will take at least n shots?
So I imagine this scenario:
Bob already took n shots (and missed), what's the probability that he'll have to take at least k more shots before he hits the target?
Note the condition given: Every shot is independent of the previous ones.
So to me it is clear that it doesn't matter that Bob already took n shots and missed. The chances he'll need k more shots now is the same as if he had just now started trying. In other words:
P(X > n + k | X > n) = P(X > k)
answered Jan 19 at 17:38
WalterWalter
662
$endgroup$
The way I understand this is:
What's the probability that it will take at least n+k shots to git the target given that it will take at least n shots?
So I imagine this scenario:
Bob already took n shots (and missed), what's the probability that he'll have to take at least k more shots before he hits the target?
Note the condition given: Every shot is independent of the previous ones.
So to me it is clear that it doesn't matter that Bob already took n shots and missed. The chances he'll need k more shots now is the same as if he had just now started trying. In other words:
P(X > n + k | X > n) = P(X > k)
answered Jan 19 at 17:38
WalterWalter
662
answered Jan 19 at 17:38
WalterWalter
662
answered Jan 19 at 17:38
WalterWalter
662
answered Jan 19 at 17:38
WalterWalter
662
662
$begingroup$
This makes sense, thanks for the hint!
$endgroup$
– FTAC
Jan 19 at 17:54
$begingroup$
This is the memoryless property of the geometric distribution
$endgroup$
– Henry
Jan 19 at 19:01
add a comment |
$begingroup$
This makes sense, thanks for the hint!
$endgroup$
– FTAC
Jan 19 at 17:54
$begingroup$
This is the memoryless property of the geometric distribution
$endgroup$
– Henry
Jan 19 at 19:01
$begingroup$
This makes sense, thanks for the hint!
$endgroup$
– FTAC
Jan 19 at 17:54
$begingroup$
This makes sense, thanks for the hint!
$endgroup$
– FTAC
Jan 19 at 17:54
$begingroup$
This is the memoryless property of the geometric distribution
$endgroup$
– Henry
Jan 19 at 19:01
$begingroup$
This is the memoryless property of the geometric distribution
$endgroup$
– Henry
Jan 19 at 19:01
add a comment |
$begingroup$
The probability of Bob not hitting the target is $2/3$, so the probability that he misses $x$ times is $(2/3)^x$, since the individual shots are independent. How could you express the statement "Bob misses the target $n$ times" symbolically?
answered Jan 19 at 17:31
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
$begingroup$
$P(X>n)=(1-frac{1}{3})^{n-1} right?$
$endgroup$
– FTAC
Jan 19 at 17:51
$begingroup$
not quite: for $X>n$ he has to miss $n$ times
$endgroup$
– Henry
Jan 19 at 19:00
add a comment |
$begingroup$
The probability of Bob not hitting the target is $2/3$, so the probability that he misses $x$ times is $(2/3)^x$, since the individual shots are independent. How could you express the statement "Bob misses the target $n$ times" symbolically?
answered Jan 19 at 17:31
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
$begingroup$
$P(X>n)=(1-frac{1}{3})^{n-1} right?$
$endgroup$
– FTAC
Jan 19 at 17:51
$begingroup$
not quite: for $X>n$ he has to miss $n$ times
$endgroup$
– Henry
Jan 19 at 19:00
add a comment |
$begingroup$
The probability of Bob not hitting the target is $2/3$, so the probability that he misses $x$ times is $(2/3)^x$, since the individual shots are independent. How could you express the statement "Bob misses the target $n$ times" symbolically?
answered Jan 19 at 17:31
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
The probability of Bob not hitting the target is $2/3$, so the probability that he misses $x$ times is $(2/3)^x$, since the individual shots are independent. How could you express the statement "Bob misses the target $n$ times" symbolically?
answered Jan 19 at 17:31
Teddan the TerranTeddan the Terran
1,206210
answered Jan 19 at 17:31
Teddan the TerranTeddan the Terran
1,206210
answered Jan 19 at 17:31
Teddan the TerranTeddan the Terran
1,206210
answered Jan 19 at 17:31
Teddan the TerranTeddan the Terran
1,206210
1,206210
$begingroup$
$P(X>n)=(1-frac{1}{3})^{n-1} right?$
$endgroup$
– FTAC
Jan 19 at 17:51
$begingroup$
not quite: for $X>n$ he has to miss $n$ times
$endgroup$
– Henry
Jan 19 at 19:00
add a comment |
$begingroup$
$P(X>n)=(1-frac{1}{3})^{n-1} right?$
$endgroup$
– FTAC
Jan 19 at 17:51
$begingroup$
not quite: for $X>n$ he has to miss $n$ times
$endgroup$
– Henry
Jan 19 at 19:00
$begingroup$
$P(X>n)=(1-frac{1}{3})^{n-1} right?$
$endgroup$
– FTAC
Jan 19 at 17:51
$begingroup$
$P(X>n)=(1-frac{1}{3})^{n-1} right?$
$endgroup$
– FTAC
Jan 19 at 17:51
$begingroup$
not quite: for $X>n$ he has to miss $n$ times
$endgroup$
– Henry
Jan 19 at 19:00
$begingroup$
not quite: for $X>n$ he has to miss $n$ times
$endgroup$
– Henry
Jan 19 at 19:00
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1
$begingroup$
What does "$X gt n$" mean in words? Can you find $P(X gt n)$?
$endgroup$
– Henry
Jan 19 at 17:09
1
$begingroup$
Is the probability that the number of the shot that first hits the target is greater than a positive integer $n$. No, I don't.. :(
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– FTAC
Jan 19 at 17:14
1
$begingroup$
Does Alice shoot too? This is not clear.
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– Bernard
Jan 19 at 17:23
$begingroup$
No, only Bob, I'll edit the question
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– FTAC
Jan 19 at 17:26