Freedonia Rainy and Sunshine

0

$begingroup$

In Freedonia, every day is either cloudy or sunny (not both). If it's sunny on any given day, then the probability that the next day will be sunny is $frac 34$. If it's cloudy on any given day, then the probability that the next day will be cloudy is $frac 23$. Given that a consecutive Saturday and Sunday had the same weather in Freedonia, what is the probability that that weather was sunny?

So far, I have tried to do (Sunny)/(Sunny + Cloudy) which ended up giving me $9/17$. This though is wrong.

I know this question has been asked before but none of them gave me a complete answer.

probability combinatorics

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asked Jan 9 at 1:35

Math_GuyMath_Guy

457

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It might be easier to think about this as a Markov process ${X_n}$ with states $0=$cloudy and $1=$sunny. You can identify the transition matrix from the given probabilities. Let $n=0$ and $n=1$ represent Saturday and Sunday, respectively. Then you want $P(X_0=X_1=1 , | , X_0=X_1)$.
  $endgroup$
  – Just_to_Answer
  Jan 9 at 4:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not yet exposed to Matrix calculations. Is there another way
  $endgroup$
  – Math_Guy
  Jan 9 at 4:59
  

  
  
  
  

add a comment | 

0

$begingroup$

In Freedonia, every day is either cloudy or sunny (not both). If it's sunny on any given day, then the probability that the next day will be sunny is $frac 34$. If it's cloudy on any given day, then the probability that the next day will be cloudy is $frac 23$. Given that a consecutive Saturday and Sunday had the same weather in Freedonia, what is the probability that that weather was sunny?

So far, I have tried to do (Sunny)/(Sunny + Cloudy) which ended up giving me $9/17$. This though is wrong.

I know this question has been asked before but none of them gave me a complete answer.

probability combinatorics

share|cite|improve this question

asked Jan 9 at 1:35

Math_GuyMath_Guy

457

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It might be easier to think about this as a Markov process ${X_n}$ with states $0=$cloudy and $1=$sunny. You can identify the transition matrix from the given probabilities. Let $n=0$ and $n=1$ represent Saturday and Sunday, respectively. Then you want $P(X_0=X_1=1 , | , X_0=X_1)$.
  $endgroup$
  – Just_to_Answer
  Jan 9 at 4:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not yet exposed to Matrix calculations. Is there another way
  $endgroup$
  – Math_Guy
  Jan 9 at 4:59
  

  
  
  
  

add a comment | 

0

0

0

0

$begingroup$

In Freedonia, every day is either cloudy or sunny (not both). If it's sunny on any given day, then the probability that the next day will be sunny is $frac 34$. If it's cloudy on any given day, then the probability that the next day will be cloudy is $frac 23$. Given that a consecutive Saturday and Sunday had the same weather in Freedonia, what is the probability that that weather was sunny?

So far, I have tried to do (Sunny)/(Sunny + Cloudy) which ended up giving me $9/17$. This though is wrong.

I know this question has been asked before but none of them gave me a complete answer.

probability combinatorics

share|cite|improve this question

asked Jan 9 at 1:35

Math_GuyMath_Guy

457

$endgroup$

In Freedonia, every day is either cloudy or sunny (not both). If it's sunny on any given day, then the probability that the next day will be sunny is $frac 34$. If it's cloudy on any given day, then the probability that the next day will be cloudy is $frac 23$. Given that a consecutive Saturday and Sunday had the same weather in Freedonia, what is the probability that that weather was sunny?

So far, I have tried to do (Sunny)/(Sunny + Cloudy) which ended up giving me $9/17$. This though is wrong.

I know this question has been asked before but none of them gave me a complete answer.

probability combinatorics

probability combinatorics

share|cite|improve this question

asked Jan 9 at 1:35

Math_GuyMath_Guy

457

share|cite|improve this question

asked Jan 9 at 1:35

Math_GuyMath_Guy

457

share|cite|improve this question

share|cite|improve this question

asked Jan 9 at 1:35

Math_GuyMath_Guy

457

asked Jan 9 at 1:35

Math_GuyMath_Guy

457

asked Jan 9 at 1:35

Math_GuyMath_Guy

457

457

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It might be easier to think about this as a Markov process ${X_n}$ with states $0=$cloudy and $1=$sunny. You can identify the transition matrix from the given probabilities. Let $n=0$ and $n=1$ represent Saturday and Sunday, respectively. Then you want $P(X_0=X_1=1 , | , X_0=X_1)$.
  $endgroup$
  – Just_to_Answer
  Jan 9 at 4:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not yet exposed to Matrix calculations. Is there another way
  $endgroup$
  – Math_Guy
  Jan 9 at 4:59
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It might be easier to think about this as a Markov process ${X_n}$ with states $0=$cloudy and $1=$sunny. You can identify the transition matrix from the given probabilities. Let $n=0$ and $n=1$ represent Saturday and Sunday, respectively. Then you want $P(X_0=X_1=1 , | , X_0=X_1)$.
  $endgroup$
  – Just_to_Answer
  Jan 9 at 4:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not yet exposed to Matrix calculations. Is there another way
  $endgroup$
  – Math_Guy
  Jan 9 at 4:59
  

  
  
  
  

$begingroup$
It might be easier to think about this as a Markov process ${X_n}$ with states $0=$cloudy and $1=$sunny. You can identify the transition matrix from the given probabilities. Let $n=0$ and $n=1$ represent Saturday and Sunday, respectively. Then you want $P(X_0=X_1=1 , | , X_0=X_1)$.
$endgroup$
– Just_to_Answer
Jan 9 at 4:06

$begingroup$
It might be easier to think about this as a Markov process ${X_n}$ with states $0=$cloudy and $1=$sunny. You can identify the transition matrix from the given probabilities. Let $n=0$ and $n=1$ represent Saturday and Sunday, respectively. Then you want $P(X_0=X_1=1 , | , X_0=X_1)$.
$endgroup$
– Just_to_Answer
Jan 9 at 4:06

$begingroup$
I am not yet exposed to Matrix calculations. Is there another way
$endgroup$
– Math_Guy
Jan 9 at 4:59

$begingroup$
I am not yet exposed to Matrix calculations. Is there another way
$endgroup$
– Math_Guy
Jan 9 at 4:59

add a comment | 

1 Answer
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$begingroup$

Let $p=$ the probability that Freedonia is sunny on a day and $q=$ the probability that Freedonia is cloudy on a day. Then the given condition implies the following equilibrium condition holds:
$$begin{eqnarray}
p&=&frac{3}{4}p+frac{1}{3}q,\
q&=&frac{1}{4}p+frac{2}{3}q.
end{eqnarray}$$ Using $p+q=1$, we can solve $p=frac{4}{7}$ and $q=frac{3}{7}$. Now, the probability that consecutive days are sunny days is $pcdot frac{3}{4}=frac{3}{7}$, and probability that consecutive days are cloudy is $qcdotfrac{2}{3}=frac{2}{7}$. By the definition of conditional probability, this gives
$$begin{eqnarray}
&&P(text{consecutive days are sunny};|;text{consecutive days have the same weather})&\&&=frac{P(text{consecutive days are sunny})}{P(text{consecutive days are sunny})+P(text{consecutive days are cloudy})}\&&=frac{frac{3}{7}}{frac{3}{7}+frac{2}{7}}=frac{3}{5}.
end{eqnarray}$$

share|cite|improve this answer

edited Jan 9 at 5:01

answered Jan 9 at 4:37

SongSong

11.2k628

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add a comment | 

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1 Answer
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1 Answer
1

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active

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votes

active

oldest

votes

2

$begingroup$

Let $p=$ the probability that Freedonia is sunny on a day and $q=$ the probability that Freedonia is cloudy on a day. Then the given condition implies the following equilibrium condition holds:
$$begin{eqnarray}
p&=&frac{3}{4}p+frac{1}{3}q,\
q&=&frac{1}{4}p+frac{2}{3}q.
end{eqnarray}$$ Using $p+q=1$, we can solve $p=frac{4}{7}$ and $q=frac{3}{7}$. Now, the probability that consecutive days are sunny days is $pcdot frac{3}{4}=frac{3}{7}$, and probability that consecutive days are cloudy is $qcdotfrac{2}{3}=frac{2}{7}$. By the definition of conditional probability, this gives
$$begin{eqnarray}
&&P(text{consecutive days are sunny};|;text{consecutive days have the same weather})&\&&=frac{P(text{consecutive days are sunny})}{P(text{consecutive days are sunny})+P(text{consecutive days are cloudy})}\&&=frac{frac{3}{7}}{frac{3}{7}+frac{2}{7}}=frac{3}{5}.
end{eqnarray}$$

share|cite|improve this answer

edited Jan 9 at 5:01

answered Jan 9 at 4:37

SongSong

11.2k628

$endgroup$

add a comment | 

2

$begingroup$

Let $p=$ the probability that Freedonia is sunny on a day and $q=$ the probability that Freedonia is cloudy on a day. Then the given condition implies the following equilibrium condition holds:
$$begin{eqnarray}
p&=&frac{3}{4}p+frac{1}{3}q,\
q&=&frac{1}{4}p+frac{2}{3}q.
end{eqnarray}$$ Using $p+q=1$, we can solve $p=frac{4}{7}$ and $q=frac{3}{7}$. Now, the probability that consecutive days are sunny days is $pcdot frac{3}{4}=frac{3}{7}$, and probability that consecutive days are cloudy is $qcdotfrac{2}{3}=frac{2}{7}$. By the definition of conditional probability, this gives
$$begin{eqnarray}
&&P(text{consecutive days are sunny};|;text{consecutive days have the same weather})&\&&=frac{P(text{consecutive days are sunny})}{P(text{consecutive days are sunny})+P(text{consecutive days are cloudy})}\&&=frac{frac{3}{7}}{frac{3}{7}+frac{2}{7}}=frac{3}{5}.
end{eqnarray}$$

share|cite|improve this answer

edited Jan 9 at 5:01

answered Jan 9 at 4:37

SongSong

11.2k628

$endgroup$

add a comment | 

2

2

2

$begingroup$

Let $p=$ the probability that Freedonia is sunny on a day and $q=$ the probability that Freedonia is cloudy on a day. Then the given condition implies the following equilibrium condition holds:
$$begin{eqnarray}
p&=&frac{3}{4}p+frac{1}{3}q,\
q&=&frac{1}{4}p+frac{2}{3}q.
end{eqnarray}$$ Using $p+q=1$, we can solve $p=frac{4}{7}$ and $q=frac{3}{7}$. Now, the probability that consecutive days are sunny days is $pcdot frac{3}{4}=frac{3}{7}$, and probability that consecutive days are cloudy is $qcdotfrac{2}{3}=frac{2}{7}$. By the definition of conditional probability, this gives
$$begin{eqnarray}
&&P(text{consecutive days are sunny};|;text{consecutive days have the same weather})&\&&=frac{P(text{consecutive days are sunny})}{P(text{consecutive days are sunny})+P(text{consecutive days are cloudy})}\&&=frac{frac{3}{7}}{frac{3}{7}+frac{2}{7}}=frac{3}{5}.
end{eqnarray}$$

share|cite|improve this answer

edited Jan 9 at 5:01

answered Jan 9 at 4:37

SongSong

11.2k628

$endgroup$

Let $p=$ the probability that Freedonia is sunny on a day and $q=$ the probability that Freedonia is cloudy on a day. Then the given condition implies the following equilibrium condition holds:
$$begin{eqnarray}
p&=&frac{3}{4}p+frac{1}{3}q,\
q&=&frac{1}{4}p+frac{2}{3}q.
end{eqnarray}$$ Using $p+q=1$, we can solve $p=frac{4}{7}$ and $q=frac{3}{7}$. Now, the probability that consecutive days are sunny days is $pcdot frac{3}{4}=frac{3}{7}$, and probability that consecutive days are cloudy is $qcdotfrac{2}{3}=frac{2}{7}$. By the definition of conditional probability, this gives
$$begin{eqnarray}
&&P(text{consecutive days are sunny};|;text{consecutive days have the same weather})&\&&=frac{P(text{consecutive days are sunny})}{P(text{consecutive days are sunny})+P(text{consecutive days are cloudy})}\&&=frac{frac{3}{7}}{frac{3}{7}+frac{2}{7}}=frac{3}{5}.
end{eqnarray}$$

share|cite|improve this answer

edited Jan 9 at 5:01

answered Jan 9 at 4:37

SongSong

11.2k628

share|cite|improve this answer

share|cite|improve this answer

edited Jan 9 at 5:01

edited Jan 9 at 5:01

edited Jan 9 at 5:01

answered Jan 9 at 4:37

SongSong

11.2k628

answered Jan 9 at 4:37

SongSong

11.2k628

answered Jan 9 at 4:37

SongSong

11.2k628

11.2k628

add a comment | 

add a comment | 

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