Conditional probability - unify 2 independent conditional probabilities
$begingroup$
I have a ball which can be black or white. There are two methods which can independently predict if the ball is black, each with a probability of success of $90%$. Assuming both methods predict the ball is black, what is the chance the ball is actually black?
In other words, $P(text{Black} mid A) = 0.9$, $P(text{Black} mid B) = 0.9$, $P(text{Black} mid A ∩ B) = $?
probability
edited Dec 19 '18 at 7:33
Saad
20.3k92352
asked Dec 19 '18 at 6:36
user1028741user1028741
1032
$endgroup$
add a comment |
$begingroup$
I have a ball which can be black or white. There are two methods which can independently predict if the ball is black, each with a probability of success of $90%$. Assuming both methods predict the ball is black, what is the chance the ball is actually black?
In other words, $P(text{Black} mid A) = 0.9$, $P(text{Black} mid B) = 0.9$, $P(text{Black} mid A ∩ B) = $?
probability
edited Dec 19 '18 at 7:33
Saad
20.3k92352
asked Dec 19 '18 at 6:36
user1028741user1028741
1032
$endgroup$
add a comment |
$begingroup$
I have a ball which can be black or white. There are two methods which can independently predict if the ball is black, each with a probability of success of $90%$. Assuming both methods predict the ball is black, what is the chance the ball is actually black?
In other words, $P(text{Black} mid A) = 0.9$, $P(text{Black} mid B) = 0.9$, $P(text{Black} mid A ∩ B) = $?
probability
edited Dec 19 '18 at 7:33
Saad
20.3k92352
asked Dec 19 '18 at 6:36
user1028741user1028741
1032
$endgroup$
I have a ball which can be black or white. There are two methods which can independently predict if the ball is black, each with a probability of success of $90%$. Assuming both methods predict the ball is black, what is the chance the ball is actually black?
In other words, $P(text{Black} mid A) = 0.9$, $P(text{Black} mid B) = 0.9$, $P(text{Black} mid A ∩ B) = $?
probability
probability
edited Dec 19 '18 at 7:33
Saad
20.3k92352
asked Dec 19 '18 at 6:36
user1028741user1028741
1032
edited Dec 19 '18 at 7:33
Saad
20.3k92352
asked Dec 19 '18 at 6:36
user1028741user1028741
1032
edited Dec 19 '18 at 7:33
Saad
20.3k92352
edited Dec 19 '18 at 7:33
Saad
20.3k92352
edited Dec 19 '18 at 7:33
Saad
20.3k92352
20.3k92352
asked Dec 19 '18 at 6:36
user1028741user1028741
1032
asked Dec 19 '18 at 6:36
user1028741user1028741
1032
asked Dec 19 '18 at 6:36
user1028741user1028741
1032
1032
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $C$ denote the event that the ball is black. Let $p$ denote the probability that the ball is black.
$P(C|AB)$
$=frac{P(CAB)}{P(AB)}$
$=frac{P(AB|C)P(C)}{P(AB)}$
$=frac{P(A|C)P(B|C)P(C)}{P(AB)}$
Note that $A$ and $B$ are conditionally independent given $C$, but not otherwise.
The numerator is given by $0.9 times 0.9 times p = 0.81p$.
The denominator can be written as $P(AB) = P(AB|C)P(C) + P(AB|bar{C})P(bar{C})$ = $P(A|C)P(B|C)P(C) + P(A|bar{C})P(B|bar{C})P(bar{C})$ = $0.81p + 0.01(1-p)$.
Putting it together:
$P(C|AB) = frac{0.81p}{0.81p + 0.01(1-p)}$.
If the ball is equally likely to be black or white, you can plug in $p=0.5$ and get $P(C|AB) = 81/82$.
answered Dec 19 '18 at 7:12
Aditya DuaAditya Dua
1,15418
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $C$ denote the event that the ball is black. Let $p$ denote the probability that the ball is black.
$P(C|AB)$
$=frac{P(CAB)}{P(AB)}$
$=frac{P(AB|C)P(C)}{P(AB)}$
$=frac{P(A|C)P(B|C)P(C)}{P(AB)}$
Note that $A$ and $B$ are conditionally independent given $C$, but not otherwise.
The numerator is given by $0.9 times 0.9 times p = 0.81p$.
The denominator can be written as $P(AB) = P(AB|C)P(C) + P(AB|bar{C})P(bar{C})$ = $P(A|C)P(B|C)P(C) + P(A|bar{C})P(B|bar{C})P(bar{C})$ = $0.81p + 0.01(1-p)$.
Putting it together:
$P(C|AB) = frac{0.81p}{0.81p + 0.01(1-p)}$.
If the ball is equally likely to be black or white, you can plug in $p=0.5$ and get $P(C|AB) = 81/82$.
answered Dec 19 '18 at 7:12
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
$begingroup$
Let $C$ denote the event that the ball is black. Let $p$ denote the probability that the ball is black.
$P(C|AB)$
$=frac{P(CAB)}{P(AB)}$
$=frac{P(AB|C)P(C)}{P(AB)}$
$=frac{P(A|C)P(B|C)P(C)}{P(AB)}$
Note that $A$ and $B$ are conditionally independent given $C$, but not otherwise.
The numerator is given by $0.9 times 0.9 times p = 0.81p$.
The denominator can be written as $P(AB) = P(AB|C)P(C) + P(AB|bar{C})P(bar{C})$ = $P(A|C)P(B|C)P(C) + P(A|bar{C})P(B|bar{C})P(bar{C})$ = $0.81p + 0.01(1-p)$.
Putting it together:
$P(C|AB) = frac{0.81p}{0.81p + 0.01(1-p)}$.
If the ball is equally likely to be black or white, you can plug in $p=0.5$ and get $P(C|AB) = 81/82$.
answered Dec 19 '18 at 7:12
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
$begingroup$
Let $C$ denote the event that the ball is black. Let $p$ denote the probability that the ball is black.
$P(C|AB)$
$=frac{P(CAB)}{P(AB)}$
$=frac{P(AB|C)P(C)}{P(AB)}$
$=frac{P(A|C)P(B|C)P(C)}{P(AB)}$
Note that $A$ and $B$ are conditionally independent given $C$, but not otherwise.
The numerator is given by $0.9 times 0.9 times p = 0.81p$.
The denominator can be written as $P(AB) = P(AB|C)P(C) + P(AB|bar{C})P(bar{C})$ = $P(A|C)P(B|C)P(C) + P(A|bar{C})P(B|bar{C})P(bar{C})$ = $0.81p + 0.01(1-p)$.
Putting it together:
$P(C|AB) = frac{0.81p}{0.81p + 0.01(1-p)}$.
If the ball is equally likely to be black or white, you can plug in $p=0.5$ and get $P(C|AB) = 81/82$.
answered Dec 19 '18 at 7:12
Aditya DuaAditya Dua
1,15418
$endgroup$
Let $C$ denote the event that the ball is black. Let $p$ denote the probability that the ball is black.
$P(C|AB)$
$=frac{P(CAB)}{P(AB)}$
$=frac{P(AB|C)P(C)}{P(AB)}$
$=frac{P(A|C)P(B|C)P(C)}{P(AB)}$
Note that $A$ and $B$ are conditionally independent given $C$, but not otherwise.
The numerator is given by $0.9 times 0.9 times p = 0.81p$.
The denominator can be written as $P(AB) = P(AB|C)P(C) + P(AB|bar{C})P(bar{C})$ = $P(A|C)P(B|C)P(C) + P(A|bar{C})P(B|bar{C})P(bar{C})$ = $0.81p + 0.01(1-p)$.
Putting it together:
$P(C|AB) = frac{0.81p}{0.81p + 0.01(1-p)}$.
If the ball is equally likely to be black or white, you can plug in $p=0.5$ and get $P(C|AB) = 81/82$.
answered Dec 19 '18 at 7:12
Aditya DuaAditya Dua
1,15418
answered Dec 19 '18 at 7:12
Aditya DuaAditya Dua
1,15418
answered Dec 19 '18 at 7:12
Aditya DuaAditya Dua
1,15418
answered Dec 19 '18 at 7:12
Aditya DuaAditya Dua
1,15418
1,15418
add a comment |
add a comment |
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