Probability and condition
up vote
1
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favorite
I have a simple problem where I am getting 2 different results.
Its given that
$P(A) = 0.22, ;P(B) = 0.35,; P(A cup B) = 0.54$
Now using below method
$P(A) + P(B) - P(A cap B) = P(A cup B)$
$0.22 + 0.35 - P(A cap B) = 0.54$
$P(A cap B) = 0.03$
Now using a simple method
$P(A) cdot P(B) = P(A cap B)$
$P(A cap B) = 0.077$
Can someone explain where my understanding is incorrect?
probability
edited Nov 17 at 13:22
amWhy
191k27223438
asked Nov 17 at 10:23
LoveWithMaths
267
add a comment |
up vote
1
down vote
favorite
I have a simple problem where I am getting 2 different results.
Its given that
$P(A) = 0.22, ;P(B) = 0.35,; P(A cup B) = 0.54$
Now using below method
$P(A) + P(B) - P(A cap B) = P(A cup B)$
$0.22 + 0.35 - P(A cap B) = 0.54$
$P(A cap B) = 0.03$
Now using a simple method
$P(A) cdot P(B) = P(A cap B)$
$P(A cap B) = 0.077$
Can someone explain where my understanding is incorrect?
probability
edited Nov 17 at 13:22
amWhy
191k27223438
asked Nov 17 at 10:23
LoveWithMaths
267
3
the second equality $P(A)P(B)=P(Acap B) $ holds iff $A$ and $B$ are independenly. You just showed thati this is not the case
– ALG
Nov 17 at 10:28
If e.g. $Asubseteq B$ then $P(Acap B)=P(A)=0.22$. As @ALG says: the equality $P(Acap B)=P(A)P(B)$ states that $A$ and $B$ are independent events, which is not always the case.
– drhab
Nov 17 at 10:32
@ALG can you please explain how do did you infer from above that 2 events are not independent.
– LoveWithMaths
Nov 17 at 12:10
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a simple problem where I am getting 2 different results.
Its given that
$P(A) = 0.22, ;P(B) = 0.35,; P(A cup B) = 0.54$
Now using below method
$P(A) + P(B) - P(A cap B) = P(A cup B)$
$0.22 + 0.35 - P(A cap B) = 0.54$
$P(A cap B) = 0.03$
Now using a simple method
$P(A) cdot P(B) = P(A cap B)$
$P(A cap B) = 0.077$
Can someone explain where my understanding is incorrect?
probability
edited Nov 17 at 13:22
amWhy
191k27223438
asked Nov 17 at 10:23
LoveWithMaths
267
I have a simple problem where I am getting 2 different results.
Its given that
$P(A) = 0.22, ;P(B) = 0.35,; P(A cup B) = 0.54$
Now using below method
$P(A) + P(B) - P(A cap B) = P(A cup B)$
$0.22 + 0.35 - P(A cap B) = 0.54$
$P(A cap B) = 0.03$
Now using a simple method
$P(A) cdot P(B) = P(A cap B)$
$P(A cap B) = 0.077$
Can someone explain where my understanding is incorrect?
probability
probability
edited Nov 17 at 13:22
amWhy
191k27223438
asked Nov 17 at 10:23
LoveWithMaths
267
edited Nov 17 at 13:22
amWhy
191k27223438
asked Nov 17 at 10:23
LoveWithMaths
267
edited Nov 17 at 13:22
amWhy
191k27223438
edited Nov 17 at 13:22
amWhy
191k27223438
edited Nov 17 at 13:22
amWhy
191k27223438
191k27223438
asked Nov 17 at 10:23
LoveWithMaths
267
asked Nov 17 at 10:23
LoveWithMaths
267
asked Nov 17 at 10:23
LoveWithMaths
267
267
3
the second equality $P(A)P(B)=P(Acap B) $ holds iff $A$ and $B$ are independenly. You just showed thati this is not the case
– ALG
Nov 17 at 10:28
If e.g. $Asubseteq B$ then $P(Acap B)=P(A)=0.22$. As @ALG says: the equality $P(Acap B)=P(A)P(B)$ states that $A$ and $B$ are independent events, which is not always the case.
– drhab
Nov 17 at 10:32
@ALG can you please explain how do did you infer from above that 2 events are not independent.
– LoveWithMaths
Nov 17 at 12:10
add a comment |
3
the second equality $P(A)P(B)=P(Acap B) $ holds iff $A$ and $B$ are independenly. You just showed thati this is not the case
– ALG
Nov 17 at 10:28
If e.g. $Asubseteq B$ then $P(Acap B)=P(A)=0.22$. As @ALG says: the equality $P(Acap B)=P(A)P(B)$ states that $A$ and $B$ are independent events, which is not always the case.
– drhab
Nov 17 at 10:32
@ALG can you please explain how do did you infer from above that 2 events are not independent.
– LoveWithMaths
Nov 17 at 12:10
3
3
the second equality $P(A)P(B)=P(Acap B) $ holds iff $A$ and $B$ are independenly. You just showed thati this is not the case
– ALG
Nov 17 at 10:28
the second equality $P(A)P(B)=P(Acap B) $ holds iff $A$ and $B$ are independenly. You just showed thati this is not the case
– ALG
Nov 17 at 10:28
If e.g. $Asubseteq B$ then $P(Acap B)=P(A)=0.22$. As @ALG says: the equality $P(Acap B)=P(A)P(B)$ states that $A$ and $B$ are independent events, which is not always the case.
– drhab
Nov 17 at 10:32
If e.g. $Asubseteq B$ then $P(Acap B)=P(A)=0.22$. As @ALG says: the equality $P(Acap B)=P(A)P(B)$ states that $A$ and $B$ are independent events, which is not always the case.
– drhab
Nov 17 at 10:32
@ALG can you please explain how do did you infer from above that 2 events are not independent.
– LoveWithMaths
Nov 17 at 12:10
@ALG can you please explain how do did you infer from above that 2 events are not independent.
– LoveWithMaths
Nov 17 at 12:10
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
The "simple method" is clearly incorrect, as the independence of $A$ and $B$, an assumption the method makes, has not been established beforehand. Therefore it cannot be applied at all, and the Venn diagram approach has to be employed instead.
By deriving $P(Acap B)$ through this correct approach, we do show that $A$ and $B$ are not independent, but that is beyond the scope of the question.
answered Nov 17 at 10:30
Parcly Taxel
41k137198
Its not clear how the probabilities given shows that they are not independent event. Can you please explain ? how its intuitive to you ?
– LoveWithMaths
Nov 17 at 11:54
can you please elaborate as i do not understand.
– LoveWithMaths
Nov 17 at 13:03
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The "simple method" is clearly incorrect, as the independence of $A$ and $B$, an assumption the method makes, has not been established beforehand. Therefore it cannot be applied at all, and the Venn diagram approach has to be employed instead.
By deriving $P(Acap B)$ through this correct approach, we do show that $A$ and $B$ are not independent, but that is beyond the scope of the question.
answered Nov 17 at 10:30
Parcly Taxel
41k137198
Its not clear how the probabilities given shows that they are not independent event. Can you please explain ? how its intuitive to you ?
– LoveWithMaths
Nov 17 at 11:54
can you please elaborate as i do not understand.
– LoveWithMaths
Nov 17 at 13:03
add a comment |
up vote
1
down vote
The "simple method" is clearly incorrect, as the independence of $A$ and $B$, an assumption the method makes, has not been established beforehand. Therefore it cannot be applied at all, and the Venn diagram approach has to be employed instead.
By deriving $P(Acap B)$ through this correct approach, we do show that $A$ and $B$ are not independent, but that is beyond the scope of the question.
answered Nov 17 at 10:30
Parcly Taxel
41k137198
Its not clear how the probabilities given shows that they are not independent event. Can you please explain ? how its intuitive to you ?
– LoveWithMaths
Nov 17 at 11:54
can you please elaborate as i do not understand.
– LoveWithMaths
Nov 17 at 13:03
add a comment |
up vote
1
down vote
up vote
1
down vote
The "simple method" is clearly incorrect, as the independence of $A$ and $B$, an assumption the method makes, has not been established beforehand. Therefore it cannot be applied at all, and the Venn diagram approach has to be employed instead.
By deriving $P(Acap B)$ through this correct approach, we do show that $A$ and $B$ are not independent, but that is beyond the scope of the question.
answered Nov 17 at 10:30
Parcly Taxel
41k137198
The "simple method" is clearly incorrect, as the independence of $A$ and $B$, an assumption the method makes, has not been established beforehand. Therefore it cannot be applied at all, and the Venn diagram approach has to be employed instead.
By deriving $P(Acap B)$ through this correct approach, we do show that $A$ and $B$ are not independent, but that is beyond the scope of the question.
answered Nov 17 at 10:30
Parcly Taxel
41k137198
edited Nov 17 at 13:09
answered Nov 17 at 10:30
Parcly Taxel
41k137198
answered Nov 17 at 10:30
Parcly Taxel
41k137198
answered Nov 17 at 10:30
Parcly Taxel
41k137198
41k137198
Its not clear how the probabilities given shows that they are not independent event. Can you please explain ? how its intuitive to you ?
– LoveWithMaths
Nov 17 at 11:54
can you please elaborate as i do not understand.
– LoveWithMaths
Nov 17 at 13:03
add a comment |
Its not clear how the probabilities given shows that they are not independent event. Can you please explain ? how its intuitive to you ?
– LoveWithMaths
Nov 17 at 11:54
can you please elaborate as i do not understand.
– LoveWithMaths
Nov 17 at 13:03
Its not clear how the probabilities given shows that they are not independent event. Can you please explain ? how its intuitive to you ?
– LoveWithMaths
Nov 17 at 11:54
Its not clear how the probabilities given shows that they are not independent event. Can you please explain ? how its intuitive to you ?
– LoveWithMaths
Nov 17 at 11:54
can you please elaborate as i do not understand.
– LoveWithMaths
Nov 17 at 13:03
can you please elaborate as i do not understand.
– LoveWithMaths
Nov 17 at 13:03
add a comment |
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3
the second equality $P(A)P(B)=P(Acap B) $ holds iff $A$ and $B$ are independenly. You just showed thati this is not the case
– ALG
Nov 17 at 10:28
If e.g. $Asubseteq B$ then $P(Acap B)=P(A)=0.22$. As @ALG says: the equality $P(Acap B)=P(A)P(B)$ states that $A$ and $B$ are independent events, which is not always the case.
– drhab
Nov 17 at 10:32
@ALG can you please explain how do did you infer from above that 2 events are not independent.
– LoveWithMaths
Nov 17 at 12:10