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












0














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?










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  • 1




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










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
















0














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?










share|cite|improve this question




















  • 1




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










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














0












0








0







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?










share|cite|improve this question















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






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edited Jan 15 '14 at 15:11







usual me

















asked Jan 15 '14 at 15:02









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  • 1




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










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














  • 1




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










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








1




1




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




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












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




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










2 Answers
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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






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    0














    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





















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      2 Answers
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      active

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      2 Answers
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      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


























        0














        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
























          0












          0








          0






          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












          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














              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


























                0














                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
























                  0












                  0








                  0






                  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












                  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

                  8,5101516




                  8,5101516






























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