If $P(A) = frac13, P(B) = frac12$, and $P(A cup B) = frac34$ find…
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If $P(A) = frac13, P(B) = frac12$, and $P(A cup B) = frac34$ Find:
$P(A cap B),\
P(A^complement cup B^complement) \
P(A^complement cap B)$
Here is what I did:
$P(A cap B) = P(A) + P(B) - P(A cup B) = frac112$
$P(A^complement cup B^complement) = P( [A cap B]^complement ) = 1 - frac112 = frac1112$ (Not sure if this is correct)
$P(A^complement cap B) = ...$ (I'm not quite sure how to approach this)
probability elementary-set-theory
edited Sep 6 at 4:14
spaceisdarkgreen
29.1k21549
asked Sep 6 at 3:43
AznBoyStride
324
add a comment |Â
up vote
2
down vote
favorite
If $P(A) = frac13, P(B) = frac12$, and $P(A cup B) = frac34$ Find:
$P(A cap B),\
P(A^complement cup B^complement) \
P(A^complement cap B)$
Here is what I did:
$P(A cap B) = P(A) + P(B) - P(A cup B) = frac112$
$P(A^complement cup B^complement) = P( [A cap B]^complement ) = 1 - frac112 = frac1112$ (Not sure if this is correct)
$P(A^complement cap B) = ...$ (I'm not quite sure how to approach this)
probability elementary-set-theory
edited Sep 6 at 4:14
spaceisdarkgreen
29.1k21549
asked Sep 6 at 3:43
AznBoyStride
324
This is really a problem in set theory - how can you rewrite $A^complement cap B$ in terms of $A$, $B$, $A cap B$ etc. As $A cup A^complement$ is the whole space, and the union is disjoint, you also have for every subset $B$: $B = (B cap A) cup (B cap A^complement)$ and the union is disjoint. This will lead you to Ahmad Bazzi's answer.
– mathguy
Sep 6 at 3:55
Try drawing a venn diagram with 2 circles, one for A and one for B.
– DanielV
Sep 6 at 4:01
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If $P(A) = frac13, P(B) = frac12$, and $P(A cup B) = frac34$ Find:
$P(A cap B),\
P(A^complement cup B^complement) \
P(A^complement cap B)$
Here is what I did:
$P(A cap B) = P(A) + P(B) - P(A cup B) = frac112$
$P(A^complement cup B^complement) = P( [A cap B]^complement ) = 1 - frac112 = frac1112$ (Not sure if this is correct)
$P(A^complement cap B) = ...$ (I'm not quite sure how to approach this)
probability elementary-set-theory
edited Sep 6 at 4:14
spaceisdarkgreen
29.1k21549
asked Sep 6 at 3:43
AznBoyStride
324
If $P(A) = frac13, P(B) = frac12$, and $P(A cup B) = frac34$ Find:
$P(A cap B),\
P(A^complement cup B^complement) \
P(A^complement cap B)$
Here is what I did:
$P(A cap B) = P(A) + P(B) - P(A cup B) = frac112$
$P(A^complement cup B^complement) = P( [A cap B]^complement ) = 1 - frac112 = frac1112$ (Not sure if this is correct)
$P(A^complement cap B) = ...$ (I'm not quite sure how to approach this)
probability elementary-set-theory
probability elementary-set-theory
edited Sep 6 at 4:14
spaceisdarkgreen
29.1k21549
asked Sep 6 at 3:43
AznBoyStride
324
edited Sep 6 at 4:14
spaceisdarkgreen
29.1k21549
asked Sep 6 at 3:43
AznBoyStride
324
edited Sep 6 at 4:14
spaceisdarkgreen
29.1k21549
edited Sep 6 at 4:14
spaceisdarkgreen
29.1k21549
edited Sep 6 at 4:14
spaceisdarkgreen
29.1k21549
29.1k21549
asked Sep 6 at 3:43
AznBoyStride
324
asked Sep 6 at 3:43
AznBoyStride
324
asked Sep 6 at 3:43
AznBoyStride
324
324
This is really a problem in set theory - how can you rewrite $A^complement cap B$ in terms of $A$, $B$, $A cap B$ etc. As $A cup A^complement$ is the whole space, and the union is disjoint, you also have for every subset $B$: $B = (B cap A) cup (B cap A^complement)$ and the union is disjoint. This will lead you to Ahmad Bazzi's answer.
– mathguy
Sep 6 at 3:55
Try drawing a venn diagram with 2 circles, one for A and one for B.
– DanielV
Sep 6 at 4:01
add a comment |Â
This is really a problem in set theory - how can you rewrite $A^complement cap B$ in terms of $A$, $B$, $A cap B$ etc. As $A cup A^complement$ is the whole space, and the union is disjoint, you also have for every subset $B$: $B = (B cap A) cup (B cap A^complement)$ and the union is disjoint. This will lead you to Ahmad Bazzi's answer.
– mathguy
Sep 6 at 3:55
Try drawing a venn diagram with 2 circles, one for A and one for B.
– DanielV
Sep 6 at 4:01
This is really a problem in set theory - how can you rewrite $A^complement cap B$ in terms of $A$, $B$, $A cap B$ etc. As $A cup A^complement$ is the whole space, and the union is disjoint, you also have for every subset $B$: $B = (B cap A) cup (B cap A^complement)$ and the union is disjoint. This will lead you to Ahmad Bazzi's answer.
– mathguy
Sep 6 at 3:55
This is really a problem in set theory - how can you rewrite $A^complement cap B$ in terms of $A$, $B$, $A cap B$ etc. As $A cup A^complement$ is the whole space, and the union is disjoint, you also have for every subset $B$: $B = (B cap A) cup (B cap A^complement)$ and the union is disjoint. This will lead you to Ahmad Bazzi's answer.
– mathguy
Sep 6 at 3:55
Try drawing a venn diagram with 2 circles, one for A and one for B.
– DanielV
Sep 6 at 4:01
Try drawing a venn diagram with 2 circles, one for A and one for B.
– DanielV
Sep 6 at 4:01
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Ponder the following diagram, where $()$ denotes $A$ and $$ denotes $B$:
$$( *** [ * ) ***** ] ***$$
$A$ has four stars, out of twelve, i.e. $1/3$.
$B$ has six stars, out of twelve, i.e. $1/2$.
$Acup B$ has nine stars, out of twelve, i.e. $3/4$.
You now have the complete Venn diagram, and can answer all the questions directly. $Bcap A^c$ has five stars.
answered Sep 6 at 3:50
vadim123
74.6k896186
add a comment |Â
up vote
3
down vote
All is correct. For the last one you'd need:
$$P(A^complement cap B) = P(B) - P(A cap B) = frac12 - frac112 = frac512$$
Image to help visualize stuff
answered Sep 6 at 3:49
Ahmad Bazzi
6,1991624
How did you arrive to $P(A^complement cap B)$ = $P(B) - P(A cap B)$. Is this a general rule? Or something that should be intuitive?
– AznBoyStride
Sep 6 at 3:55
@AznBoyStride Take a look at the image that i just added, can you see why ?
– Ahmad Bazzi
Sep 6 at 4:01
Yes makes perfect sense now!
– AznBoyStride
Sep 6 at 4:10
1
Its just a rearrangement of the Law of Total Probability. $mathsf P(B)=mathsf P(Acap B)+mathsf P(A^complementcap B)$
– Graham Kemp
Sep 6 at 4:11
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Ponder the following diagram, where $()$ denotes $A$ and $$ denotes $B$:
$$( *** [ * ) ***** ] ***$$
$A$ has four stars, out of twelve, i.e. $1/3$.
$B$ has six stars, out of twelve, i.e. $1/2$.
$Acup B$ has nine stars, out of twelve, i.e. $3/4$.
You now have the complete Venn diagram, and can answer all the questions directly. $Bcap A^c$ has five stars.
answered Sep 6 at 3:50
vadim123
74.6k896186
add a comment |Â
up vote
2
down vote
accepted
Ponder the following diagram, where $()$ denotes $A$ and $$ denotes $B$:
$$( *** [ * ) ***** ] ***$$
$A$ has four stars, out of twelve, i.e. $1/3$.
$B$ has six stars, out of twelve, i.e. $1/2$.
$Acup B$ has nine stars, out of twelve, i.e. $3/4$.
You now have the complete Venn diagram, and can answer all the questions directly. $Bcap A^c$ has five stars.
answered Sep 6 at 3:50
vadim123
74.6k896186
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Ponder the following diagram, where $()$ denotes $A$ and $$ denotes $B$:
$$( *** [ * ) ***** ] ***$$
$A$ has four stars, out of twelve, i.e. $1/3$.
$B$ has six stars, out of twelve, i.e. $1/2$.
$Acup B$ has nine stars, out of twelve, i.e. $3/4$.
You now have the complete Venn diagram, and can answer all the questions directly. $Bcap A^c$ has five stars.
answered Sep 6 at 3:50
vadim123
74.6k896186
Ponder the following diagram, where $()$ denotes $A$ and $$ denotes $B$:
$$( *** [ * ) ***** ] ***$$
$A$ has four stars, out of twelve, i.e. $1/3$.
$B$ has six stars, out of twelve, i.e. $1/2$.
$Acup B$ has nine stars, out of twelve, i.e. $3/4$.
You now have the complete Venn diagram, and can answer all the questions directly. $Bcap A^c$ has five stars.
answered Sep 6 at 3:50
vadim123
74.6k896186
answered Sep 6 at 3:50
vadim123
74.6k896186
answered Sep 6 at 3:50
vadim123
74.6k896186
answered Sep 6 at 3:50
vadim123
74.6k896186
74.6k896186
add a comment |Â
add a comment |Â
up vote
3
down vote
All is correct. For the last one you'd need:
$$P(A^complement cap B) = P(B) - P(A cap B) = frac12 - frac112 = frac512$$
Image to help visualize stuff
answered Sep 6 at 3:49
Ahmad Bazzi
6,1991624
How did you arrive to $P(A^complement cap B)$ = $P(B) - P(A cap B)$. Is this a general rule? Or something that should be intuitive?
– AznBoyStride
Sep 6 at 3:55
@AznBoyStride Take a look at the image that i just added, can you see why ?
– Ahmad Bazzi
Sep 6 at 4:01
Yes makes perfect sense now!
– AznBoyStride
Sep 6 at 4:10
1
Its just a rearrangement of the Law of Total Probability. $mathsf P(B)=mathsf P(Acap B)+mathsf P(A^complementcap B)$
– Graham Kemp
Sep 6 at 4:11
add a comment |Â
up vote
3
down vote
All is correct. For the last one you'd need:
$$P(A^complement cap B) = P(B) - P(A cap B) = frac12 - frac112 = frac512$$
Image to help visualize stuff
answered Sep 6 at 3:49
Ahmad Bazzi
6,1991624
How did you arrive to $P(A^complement cap B)$ = $P(B) - P(A cap B)$. Is this a general rule? Or something that should be intuitive?
– AznBoyStride
Sep 6 at 3:55
@AznBoyStride Take a look at the image that i just added, can you see why ?
– Ahmad Bazzi
Sep 6 at 4:01
Yes makes perfect sense now!
– AznBoyStride
Sep 6 at 4:10
1
Its just a rearrangement of the Law of Total Probability. $mathsf P(B)=mathsf P(Acap B)+mathsf P(A^complementcap B)$
– Graham Kemp
Sep 6 at 4:11
add a comment |Â
up vote
3
down vote
up vote
3
down vote
All is correct. For the last one you'd need:
$$P(A^complement cap B) = P(B) - P(A cap B) = frac12 - frac112 = frac512$$
Image to help visualize stuff
answered Sep 6 at 3:49
Ahmad Bazzi
6,1991624
All is correct. For the last one you'd need:
$$P(A^complement cap B) = P(B) - P(A cap B) = frac12 - frac112 = frac512$$
Image to help visualize stuff
answered Sep 6 at 3:49
Ahmad Bazzi
6,1991624
edited Sep 6 at 4:01
answered Sep 6 at 3:49
Ahmad Bazzi
6,1991624
answered Sep 6 at 3:49
Ahmad Bazzi
6,1991624
answered Sep 6 at 3:49
Ahmad Bazzi
6,1991624
6,1991624
How did you arrive to $P(A^complement cap B)$ = $P(B) - P(A cap B)$. Is this a general rule? Or something that should be intuitive?
– AznBoyStride
Sep 6 at 3:55
@AznBoyStride Take a look at the image that i just added, can you see why ?
– Ahmad Bazzi
Sep 6 at 4:01
Yes makes perfect sense now!
– AznBoyStride
Sep 6 at 4:10
1
Its just a rearrangement of the Law of Total Probability. $mathsf P(B)=mathsf P(Acap B)+mathsf P(A^complementcap B)$
– Graham Kemp
Sep 6 at 4:11
add a comment |Â
How did you arrive to $P(A^complement cap B)$ = $P(B) - P(A cap B)$. Is this a general rule? Or something that should be intuitive?
– AznBoyStride
Sep 6 at 3:55
@AznBoyStride Take a look at the image that i just added, can you see why ?
– Ahmad Bazzi
Sep 6 at 4:01
Yes makes perfect sense now!
– AznBoyStride
Sep 6 at 4:10
1
Its just a rearrangement of the Law of Total Probability. $mathsf P(B)=mathsf P(Acap B)+mathsf P(A^complementcap B)$
– Graham Kemp
Sep 6 at 4:11
How did you arrive to $P(A^complement cap B)$ = $P(B) - P(A cap B)$. Is this a general rule? Or something that should be intuitive?
– AznBoyStride
Sep 6 at 3:55
How did you arrive to $P(A^complement cap B)$ = $P(B) - P(A cap B)$. Is this a general rule? Or something that should be intuitive?
– AznBoyStride
Sep 6 at 3:55
@AznBoyStride Take a look at the image that i just added, can you see why ?
– Ahmad Bazzi
Sep 6 at 4:01
@AznBoyStride Take a look at the image that i just added, can you see why ?
– Ahmad Bazzi
Sep 6 at 4:01
Yes makes perfect sense now!
– AznBoyStride
Sep 6 at 4:10
Yes makes perfect sense now!
– AznBoyStride
Sep 6 at 4:10
1
1
Its just a rearrangement of the Law of Total Probability. $mathsf P(B)=mathsf P(Acap B)+mathsf P(A^complementcap B)$
– Graham Kemp
Sep 6 at 4:11
Its just a rearrangement of the Law of Total Probability. $mathsf P(B)=mathsf P(Acap B)+mathsf P(A^complementcap B)$
– Graham Kemp
Sep 6 at 4:11
add a comment |Â
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This is really a problem in set theory - how can you rewrite $A^complement cap B$ in terms of $A$, $B$, $A cap B$ etc. As $A cup A^complement$ is the whole space, and the union is disjoint, you also have for every subset $B$: $B = (B cap A) cup (B cap A^complement)$ and the union is disjoint. This will lead you to Ahmad Bazzi's answer.
– mathguy
Sep 6 at 3:55
Try drawing a venn diagram with 2 circles, one for A and one for B.
– DanielV
Sep 6 at 4:01