Probability of a car accident given that a man is driving (Bayes formula)












0












$begingroup$


I came across the following exercise illustrating the Bayes formula:




Let's consider the events: F = {female driver}, G = {male driver}, and
A = {car accident}. We also know that P(A|F) = a, P(A|G) = b, and
P(F) = P(G) = 1/2. Compute P(A).




Then the solution simply gives the formula for computing P(A). But I don't understand the meaning of events such as $A cap F$. When we define the events F, G and A, what is the sample space $Omega$? Is it the cartesian product of two sample spaces $ Omega_1 = {$female driver, male driver$}$ and $ Omega_2 = {$accident, no accident$}$?



In this case, the events should be defined as follows: F = ${$ (female driver, accident), (female driver, no accident) $}$, and A= ${$ (female driver, accident), (male driver, accident) $}$. Therefore, events such as $A cap F$ can be properly defined: $A cap F$ = ${$ (female driver, accident) $}$.



Is my understanding correct?






probability conditional-probability bayes-theorem







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    the title is incredibly sexist-sounding.
    $endgroup$
    – Lost1
    Jan 15 '14 at 15:08










  • $begingroup$
    Your understanding looks correct to me.
    $endgroup$
    – Milo Brandt
    Nov 29 '14 at 1:37
















0












$begingroup$


I came across the following exercise illustrating the Bayes formula:




Let's consider the events: F = {female driver}, G = {male driver}, and
A = {car accident}. We also know that P(A|F) = a, P(A|G) = b, and
P(F) = P(G) = 1/2. Compute P(A).




Then the solution simply gives the formula for computing P(A). But I don't understand the meaning of events such as $A cap F$. When we define the events F, G and A, what is the sample space $Omega$? Is it the cartesian product of two sample spaces $ Omega_1 = {$female driver, male driver$}$ and $ Omega_2 = {$accident, no accident$}$?



In this case, the events should be defined as follows: F = ${$ (female driver, accident), (female driver, no accident) $}$, and A= ${$ (female driver, accident), (male driver, accident) $}$. Therefore, events such as $A cap F$ can be properly defined: $A cap F$ = ${$ (female driver, accident) $}$.



Is my understanding correct?






probability conditional-probability bayes-theorem







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    the title is incredibly sexist-sounding.
    $endgroup$
    – Lost1
    Jan 15 '14 at 15:08










  • $begingroup$
    Your understanding looks correct to me.
    $endgroup$
    – Milo Brandt
    Nov 29 '14 at 1:37














0












0








0





$begingroup$


I came across the following exercise illustrating the Bayes formula:




Let's consider the events: F = {female driver}, G = {male driver}, and
A = {car accident}. We also know that P(A|F) = a, P(A|G) = b, and
P(F) = P(G) = 1/2. Compute P(A).




Then the solution simply gives the formula for computing P(A). But I don't understand the meaning of events such as $A cap F$. When we define the events F, G and A, what is the sample space $Omega$? Is it the cartesian product of two sample spaces $ Omega_1 = {$female driver, male driver$}$ and $ Omega_2 = {$accident, no accident$}$?



In this case, the events should be defined as follows: F = ${$ (female driver, accident), (female driver, no accident) $}$, and A= ${$ (female driver, accident), (male driver, accident) $}$. Therefore, events such as $A cap F$ can be properly defined: $A cap F$ = ${$ (female driver, accident) $}$.



Is my understanding correct?






probability conditional-probability bayes-theorem







share|cite|improve this question











$endgroup$




I came across the following exercise illustrating the Bayes formula:




Let's consider the events: F = {female driver}, G = {male driver}, and
A = {car accident}. We also know that P(A|F) = a, P(A|G) = b, and
P(F) = P(G) = 1/2. Compute P(A).




Then the solution simply gives the formula for computing P(A). But I don't understand the meaning of events such as $A cap F$. When we define the events F, G and A, what is the sample space $Omega$? Is it the cartesian product of two sample spaces $ Omega_1 = {$female driver, male driver$}$ and $ Omega_2 = {$accident, no accident$}$?



In this case, the events should be defined as follows: F = ${$ (female driver, accident), (female driver, no accident) $}$, and A= ${$ (female driver, accident), (male driver, accident) $}$. Therefore, events such as $A cap F$ can be properly defined: $A cap F$ = ${$ (female driver, accident) $}$.



Is my understanding correct?






probability conditional-probability bayes-theorem




probability conditional-probability bayes-theorem






share|cite|improve this question















share|cite|improve this question













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edited Jan 15 '14 at 15:11







usual me

















asked Jan 15 '14 at 15:02









usual meusual me

378210




378210








  • 1




    $begingroup$
    the title is incredibly sexist-sounding.
    $endgroup$
    – Lost1
    Jan 15 '14 at 15:08










  • $begingroup$
    Your understanding looks correct to me.
    $endgroup$
    – Milo Brandt
    Nov 29 '14 at 1:37














  • 1




    $begingroup$
    the title is incredibly sexist-sounding.
    $endgroup$
    – Lost1
    Jan 15 '14 at 15:08










  • $begingroup$
    Your understanding looks correct to me.
    $endgroup$
    – Milo Brandt
    Nov 29 '14 at 1:37








1




1




$begingroup$
the title is incredibly sexist-sounding.
$endgroup$
– Lost1
Jan 15 '14 at 15:08




$begingroup$
the title is incredibly sexist-sounding.
$endgroup$
– Lost1
Jan 15 '14 at 15:08












$begingroup$
Your understanding looks correct to me.
$endgroup$
– Milo Brandt
Nov 29 '14 at 1:37




$begingroup$
Your understanding looks correct to me.
$endgroup$
– Milo Brandt
Nov 29 '14 at 1:37










2 Answers
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$begingroup$

P(A) = P(A,G) + P(A,F)               //Marginalization

P(A) = P(A|G) P(G) + P(A|F) P(F)     //Chain Rule

P(A) = b(1/2) + a(1/2)

P(A) = (a+b)/2


I'm guessing this would be the solution given? If not, please specify. It would help explain the solution better






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint -



    In this question intersection of two events not given. You have to find that first.



    For example -



    $P(A|F) = frac{P(A cap F)}{P(F)}$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






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      0












      $begingroup$

      P(A) = P(A,G) + P(A,F)               //Marginalization
      
P(A) = P(A|G) P(G) + P(A|F) P(F)     //Chain Rule
      
P(A) = b(1/2) + a(1/2)
      
P(A) = (a+b)/2


      I'm guessing this would be the solution given? If not, please specify. It would help explain the solution better






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        P(A) = P(A,G) + P(A,F)               //Marginalization
        
P(A) = P(A|G) P(G) + P(A|F) P(F)     //Chain Rule
        
P(A) = b(1/2) + a(1/2)
        
P(A) = (a+b)/2


        I'm guessing this would be the solution given? If not, please specify. It would help explain the solution better






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          P(A) = P(A,G) + P(A,F)               //Marginalization
          
P(A) = P(A|G) P(G) + P(A|F) P(F)     //Chain Rule
          
P(A) = b(1/2) + a(1/2)
          
P(A) = (a+b)/2


          I'm guessing this would be the solution given? If not, please specify. It would help explain the solution better






          share|cite|improve this answer









          $endgroup$



          P(A) = P(A,G) + P(A,F)               //Marginalization
          
P(A) = P(A|G) P(G) + P(A|F) P(F)     //Chain Rule
          
P(A) = b(1/2) + a(1/2)
          
P(A) = (a+b)/2


          I'm guessing this would be the solution given? If not, please specify. It would help explain the solution better







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '14 at 1:34









          Sunil D SSunil D S

          11




          11























              0












              $begingroup$

              Hint -



              In this question intersection of two events not given. You have to find that first.



              For example -



              $P(A|F) = frac{P(A cap F)}{P(F)}$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint -



                In this question intersection of two events not given. You have to find that first.



                For example -



                $P(A|F) = frac{P(A cap F)}{P(F)}$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint -



                  In this question intersection of two events not given. You have to find that first.



                  For example -



                  $P(A|F) = frac{P(A cap F)}{P(F)}$






                  share|cite|improve this answer









                  $endgroup$



                  Hint -



                  In this question intersection of two events not given. You have to find that first.



                  For example -



                  $P(A|F) = frac{P(A cap F)}{P(F)}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jun 2 '17 at 2:22









                  Kanwaljit SinghKanwaljit Singh

                  8,5201516




                  8,5201516






























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