Mixing
Curious habit, will I be late for work in n-days ? [Probability]
$begingroup$
Here is an interesting problem of probability :
Everyday, I can either go to work with a car or with the sub. If I
take the car, I will be late half of the time, whereas with the metro,
only a fourth (1/4 of the time). If today I'm late, tomorrow, I'll
change my way of commuting. So if one day I'm late using the car, the
next day I'll use the subway.
I start this habit tomorrow, and tomorrow I'll have a chance $p$ to
use the car.
1. What is the probability that I'm using the car the n-th day ?
2. What is the probability that I'm late on the n-th day ?
Here comes my answer. Can you please tell me if there are any mistakes ?
I ve set the event $C_n$ as taking the car on n-th day, $S_n$ for the sub.
$R$ is the event "being late".
The event $R^c$ is the contrary, beeing on time.
$$ P(C_n) = P( S_{n-1} cap R ) + P( C_{n-1} cap R^c )$$ by hypothesis.
$$ begin{split}
P(C_n) = & P( S_{n-1} ) P ( R vert S_{n-1} ) + P( C_{n-1} ) P ( R^c vert C_{n-1} ) \
=& frac 1 4 ( 1 - P( C_{n-1} ) ) + frac 1 2 P( C_{n-1} ) \
=& frac 1 4 ( P( C_{n-1} ) + 1 )
end{split} $$
that because I can either take the car of the subway, so the two event $S_n, C_n$ are creating a partition for each day
(Can you maybe clarify this statement ? I don't know how to precisely justify this point ; This is for this equality : $1 - P( C_{n-1} ) = P( S_{n-1} ) $).
Getting to this point, we recognise an arithmetico / geometrico sequence.
I think that we get this general expression :
$$ P(C_n) = (frac 1 4)^n ( p- frac 1 3 ) + frac 1 3 $$
So this would be the answer to the first point, and for the second point, we simply have to use the formula of total probability :
$$ P( R) = P( S_{n} ) frac 1 4 + P( C_{n} ) frac 1 2 = frac 1 4 ( 1 + P( C_n ))$$
probability puzzle conditional-probability
asked Jan 27 at 19:08
Marine GalantinMarine Galantin
940419
$endgroup$
add a comment |
$begingroup$
Here is an interesting problem of probability :
Everyday, I can either go to work with a car or with the sub. If I
take the car, I will be late half of the time, whereas with the metro,
only a fourth (1/4 of the time). If today I'm late, tomorrow, I'll
change my way of commuting. So if one day I'm late using the car, the
next day I'll use the subway.
I start this habit tomorrow, and tomorrow I'll have a chance $p$ to
use the car.
1. What is the probability that I'm using the car the n-th day ?
2. What is the probability that I'm late on the n-th day ?
Here comes my answer. Can you please tell me if there are any mistakes ?
I ve set the event $C_n$ as taking the car on n-th day, $S_n$ for the sub.
$R$ is the event "being late".
The event $R^c$ is the contrary, beeing on time.
$$ P(C_n) = P( S_{n-1} cap R ) + P( C_{n-1} cap R^c )$$ by hypothesis.
$$ begin{split}
P(C_n) = & P( S_{n-1} ) P ( R vert S_{n-1} ) + P( C_{n-1} ) P ( R^c vert C_{n-1} ) \
=& frac 1 4 ( 1 - P( C_{n-1} ) ) + frac 1 2 P( C_{n-1} ) \
=& frac 1 4 ( P( C_{n-1} ) + 1 )
end{split} $$
that because I can either take the car of the subway, so the two event $S_n, C_n$ are creating a partition for each day
(Can you maybe clarify this statement ? I don't know how to precisely justify this point ; This is for this equality : $1 - P( C_{n-1} ) = P( S_{n-1} ) $).
Getting to this point, we recognise an arithmetico / geometrico sequence.
I think that we get this general expression :
$$ P(C_n) = (frac 1 4)^n ( p- frac 1 3 ) + frac 1 3 $$
So this would be the answer to the first point, and for the second point, we simply have to use the formula of total probability :
$$ P( R) = P( S_{n} ) frac 1 4 + P( C_{n} ) frac 1 2 = frac 1 4 ( 1 + P( C_n ))$$
probability puzzle conditional-probability
asked Jan 27 at 19:08
Marine GalantinMarine Galantin
940419
$endgroup$
add a comment |
$begingroup$
Here is an interesting problem of probability :
Everyday, I can either go to work with a car or with the sub. If I
take the car, I will be late half of the time, whereas with the metro,
only a fourth (1/4 of the time). If today I'm late, tomorrow, I'll
change my way of commuting. So if one day I'm late using the car, the
next day I'll use the subway.
I start this habit tomorrow, and tomorrow I'll have a chance $p$ to
use the car.
1. What is the probability that I'm using the car the n-th day ?
2. What is the probability that I'm late on the n-th day ?
Here comes my answer. Can you please tell me if there are any mistakes ?
I ve set the event $C_n$ as taking the car on n-th day, $S_n$ for the sub.
$R$ is the event "being late".
The event $R^c$ is the contrary, beeing on time.
$$ P(C_n) = P( S_{n-1} cap R ) + P( C_{n-1} cap R^c )$$ by hypothesis.
$$ begin{split}
P(C_n) = & P( S_{n-1} ) P ( R vert S_{n-1} ) + P( C_{n-1} ) P ( R^c vert C_{n-1} ) \
=& frac 1 4 ( 1 - P( C_{n-1} ) ) + frac 1 2 P( C_{n-1} ) \
=& frac 1 4 ( P( C_{n-1} ) + 1 )
end{split} $$
that because I can either take the car of the subway, so the two event $S_n, C_n$ are creating a partition for each day
(Can you maybe clarify this statement ? I don't know how to precisely justify this point ; This is for this equality : $1 - P( C_{n-1} ) = P( S_{n-1} ) $).
Getting to this point, we recognise an arithmetico / geometrico sequence.
I think that we get this general expression :
$$ P(C_n) = (frac 1 4)^n ( p- frac 1 3 ) + frac 1 3 $$
So this would be the answer to the first point, and for the second point, we simply have to use the formula of total probability :
$$ P( R) = P( S_{n} ) frac 1 4 + P( C_{n} ) frac 1 2 = frac 1 4 ( 1 + P( C_n ))$$
probability puzzle conditional-probability
asked Jan 27 at 19:08
Marine GalantinMarine Galantin
940419
$endgroup$
Here is an interesting problem of probability :
Everyday, I can either go to work with a car or with the sub. If I
take the car, I will be late half of the time, whereas with the metro,
only a fourth (1/4 of the time). If today I'm late, tomorrow, I'll
change my way of commuting. So if one day I'm late using the car, the
next day I'll use the subway.
I start this habit tomorrow, and tomorrow I'll have a chance $p$ to
use the car.
1. What is the probability that I'm using the car the n-th day ?
2. What is the probability that I'm late on the n-th day ?
Here comes my answer. Can you please tell me if there are any mistakes ?
I ve set the event $C_n$ as taking the car on n-th day, $S_n$ for the sub.
$R$ is the event "being late".
The event $R^c$ is the contrary, beeing on time.
$$ P(C_n) = P( S_{n-1} cap R ) + P( C_{n-1} cap R^c )$$ by hypothesis.
$$ begin{split}
P(C_n) = & P( S_{n-1} ) P ( R vert S_{n-1} ) + P( C_{n-1} ) P ( R^c vert C_{n-1} ) \
=& frac 1 4 ( 1 - P( C_{n-1} ) ) + frac 1 2 P( C_{n-1} ) \
=& frac 1 4 ( P( C_{n-1} ) + 1 )
end{split} $$
that because I can either take the car of the subway, so the two event $S_n, C_n$ are creating a partition for each day
(Can you maybe clarify this statement ? I don't know how to precisely justify this point ; This is for this equality : $1 - P( C_{n-1} ) = P( S_{n-1} ) $).
Getting to this point, we recognise an arithmetico / geometrico sequence.
I think that we get this general expression :
$$ P(C_n) = (frac 1 4)^n ( p- frac 1 3 ) + frac 1 3 $$
So this would be the answer to the first point, and for the second point, we simply have to use the formula of total probability :
$$ P( R) = P( S_{n} ) frac 1 4 + P( C_{n} ) frac 1 2 = frac 1 4 ( 1 + P( C_n ))$$
probability puzzle conditional-probability
probability puzzle conditional-probability
asked Jan 27 at 19:08
Marine GalantinMarine Galantin
940419
asked Jan 27 at 19:08
Marine GalantinMarine Galantin
940419
edited Jan 28 at 0:42
Marine Galantin
asked Jan 27 at 19:08
Marine GalantinMarine Galantin
940419
asked Jan 27 at 19:08
Marine GalantinMarine Galantin
940419
asked Jan 27 at 19:08
Marine GalantinMarine Galantin
940419
940419
add a comment |
add a comment |
1 Answer
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$begingroup$
Your argument is sound. The equality you are worried about is find because you have to take either the car or the subway, so both sides are the probability you take the car. I didn't check your solution to the recurrence, but it looks good. It should converge to $frac 13$ and it does.
answered Jan 30 at 18:33
Ross MillikanRoss Millikan
300k24200375
$endgroup$
add a comment |
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$begingroup$
Your argument is sound. The equality you are worried about is find because you have to take either the car or the subway, so both sides are the probability you take the car. I didn't check your solution to the recurrence, but it looks good. It should converge to $frac 13$ and it does.
answered Jan 30 at 18:33
Ross MillikanRoss Millikan
300k24200375
$endgroup$
add a comment |
$begingroup$
Your argument is sound. The equality you are worried about is find because you have to take either the car or the subway, so both sides are the probability you take the car. I didn't check your solution to the recurrence, but it looks good. It should converge to $frac 13$ and it does.
answered Jan 30 at 18:33
Ross MillikanRoss Millikan
300k24200375
$endgroup$
add a comment |
$begingroup$
Your argument is sound. The equality you are worried about is find because you have to take either the car or the subway, so both sides are the probability you take the car. I didn't check your solution to the recurrence, but it looks good. It should converge to $frac 13$ and it does.
answered Jan 30 at 18:33
Ross MillikanRoss Millikan
300k24200375
$endgroup$
Your argument is sound. The equality you are worried about is find because you have to take either the car or the subway, so both sides are the probability you take the car. I didn't check your solution to the recurrence, but it looks good. It should converge to $frac 13$ and it does.
answered Jan 30 at 18:33
Ross MillikanRoss Millikan
300k24200375
answered Jan 30 at 18:33
Ross MillikanRoss Millikan
300k24200375
answered Jan 30 at 18:33
Ross MillikanRoss Millikan
300k24200375
answered Jan 30 at 18:33
Ross MillikanRoss Millikan
300k24200375
300k24200375
add a comment |
add a comment |
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