Probability of a car accident given that a man is driving (Bayes formula)
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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 thatP(A|F) = a,P(A|G) = b, and
P(F) = P(G) = 1/2. ComputeP(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
asked Jan 15 '14 at 15:02
usual meusual me
378210
$endgroup$
add a comment |
$begingroup$
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 thatP(A|F) = a,P(A|G) = b, and
P(F) = P(G) = 1/2. ComputeP(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
asked Jan 15 '14 at 15:02
usual meusual me
378210
$endgroup$
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
add a comment |
$begingroup$
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 thatP(A|F) = a,P(A|G) = b, and
P(F) = P(G) = 1/2. ComputeP(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
asked Jan 15 '14 at 15:02
usual meusual me
378210
$endgroup$
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 thatP(A|F) = a,P(A|G) = b, and
P(F) = P(G) = 1/2. ComputeP(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
asked Jan 15 '14 at 15:02
usual meusual me
378210
asked Jan 15 '14 at 15:02
usual meusual me
378210
edited Jan 15 '14 at 15:11
usual me
asked Jan 15 '14 at 15:02
usual meusual me
378210
asked Jan 15 '14 at 15:02
usual meusual me
378210
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
add a comment |
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
add a comment |
2 Answers
2
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oldest
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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
answered Nov 29 '14 at 1:34
Sunil D SSunil D S
11
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add a comment |
$begingroup$
Hint -
In this question intersection of two events not given. You have to find that first.
For example -
$P(A|F) = frac{P(A cap F)}{P(F)}$
answered Jun 2 '17 at 2:22
Kanwaljit SinghKanwaljit Singh
8,5201516
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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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
answered Nov 29 '14 at 1:34
Sunil D SSunil D S
11
$endgroup$
add a comment |
$begingroup$
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
answered Nov 29 '14 at 1:34
Sunil D SSunil D S
11
$endgroup$
add a comment |
$begingroup$
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
answered Nov 29 '14 at 1:34
Sunil D SSunil D S
11
$endgroup$
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
answered Nov 29 '14 at 1:34
Sunil D SSunil D S
11
answered Nov 29 '14 at 1:34
Sunil D SSunil D S
11
answered Nov 29 '14 at 1:34
Sunil D SSunil D S
11
answered Nov 29 '14 at 1:34
Sunil D SSunil D S
11
11
add a comment |
add a comment |
$begingroup$
Hint -
In this question intersection of two events not given. You have to find that first.
For example -
$P(A|F) = frac{P(A cap F)}{P(F)}$
answered Jun 2 '17 at 2:22
Kanwaljit SinghKanwaljit Singh
8,5201516
$endgroup$
add a comment |
$begingroup$
Hint -
In this question intersection of two events not given. You have to find that first.
For example -
$P(A|F) = frac{P(A cap F)}{P(F)}$
answered Jun 2 '17 at 2:22
Kanwaljit SinghKanwaljit Singh
8,5201516
$endgroup$
add a comment |
$begingroup$
Hint -
In this question intersection of two events not given. You have to find that first.
For example -
$P(A|F) = frac{P(A cap F)}{P(F)}$
answered Jun 2 '17 at 2:22
Kanwaljit SinghKanwaljit Singh
8,5201516
$endgroup$
Hint -
In this question intersection of two events not given. You have to find that first.
For example -
$P(A|F) = frac{P(A cap F)}{P(F)}$
answered Jun 2 '17 at 2:22
Kanwaljit SinghKanwaljit Singh
8,5201516
answered Jun 2 '17 at 2:22
Kanwaljit SinghKanwaljit Singh
8,5201516
answered Jun 2 '17 at 2:22
Kanwaljit SinghKanwaljit Singh
8,5201516
answered Jun 2 '17 at 2:22
Kanwaljit SinghKanwaljit Singh
8,5201516
8,5201516
add a comment |
add a comment |
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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