Find the probability of $ A, B, C, D $
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Suppose I have four events and I'll call them $ A,B,C,D $ .
This is the information I'm given:
$$ D subseteq C subseteq B ,P(A) = 0.59 , P(A cap B cap C cap D) = 0.10 \ P(A | D) = 0.50, P(A cup B cup C cup D) =0.8\ P(A cap B^c cap C^c cap D^c) = 0.2 , P(A^c cap B cap C^c cap D^c) = 0.1$$
$ 1/3 $ of those that are part of three events ( $ A, B,C,D $ ) do not belong to $ D $ .
I need to find $ P(A), P(B), P(C), P(D) $ . so far I got that because $ D subseteq C subseteq B $ then $ P(A cap B cap C cap D) = P(A cap D) = 0.10 \ P(A | D) = frac{P(Acap D)}{P(D)} = 0.10 Rightarrow P(D) = 0.2$
But couldn't find a way to find $ P(B), P(C) $
conditional-probability
asked Nov 17 at 18:46
bm1125
55116
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1
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Suppose I have four events and I'll call them $ A,B,C,D $ .
This is the information I'm given:
$$ D subseteq C subseteq B ,P(A) = 0.59 , P(A cap B cap C cap D) = 0.10 \ P(A | D) = 0.50, P(A cup B cup C cup D) =0.8\ P(A cap B^c cap C^c cap D^c) = 0.2 , P(A^c cap B cap C^c cap D^c) = 0.1$$
$ 1/3 $ of those that are part of three events ( $ A, B,C,D $ ) do not belong to $ D $ .
I need to find $ P(A), P(B), P(C), P(D) $ . so far I got that because $ D subseteq C subseteq B $ then $ P(A cap B cap C cap D) = P(A cap D) = 0.10 \ P(A | D) = frac{P(Acap D)}{P(D)} = 0.10 Rightarrow P(D) = 0.2$
But couldn't find a way to find $ P(B), P(C) $
conditional-probability
asked Nov 17 at 18:46
bm1125
55116
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose I have four events and I'll call them $ A,B,C,D $ .
This is the information I'm given:
$$ D subseteq C subseteq B ,P(A) = 0.59 , P(A cap B cap C cap D) = 0.10 \ P(A | D) = 0.50, P(A cup B cup C cup D) =0.8\ P(A cap B^c cap C^c cap D^c) = 0.2 , P(A^c cap B cap C^c cap D^c) = 0.1$$
$ 1/3 $ of those that are part of three events ( $ A, B,C,D $ ) do not belong to $ D $ .
I need to find $ P(A), P(B), P(C), P(D) $ . so far I got that because $ D subseteq C subseteq B $ then $ P(A cap B cap C cap D) = P(A cap D) = 0.10 \ P(A | D) = frac{P(Acap D)}{P(D)} = 0.10 Rightarrow P(D) = 0.2$
But couldn't find a way to find $ P(B), P(C) $
conditional-probability
asked Nov 17 at 18:46
bm1125
55116
Suppose I have four events and I'll call them $ A,B,C,D $ .
This is the information I'm given:
$$ D subseteq C subseteq B ,P(A) = 0.59 , P(A cap B cap C cap D) = 0.10 \ P(A | D) = 0.50, P(A cup B cup C cup D) =0.8\ P(A cap B^c cap C^c cap D^c) = 0.2 , P(A^c cap B cap C^c cap D^c) = 0.1$$
$ 1/3 $ of those that are part of three events ( $ A, B,C,D $ ) do not belong to $ D $ .
I need to find $ P(A), P(B), P(C), P(D) $ . so far I got that because $ D subseteq C subseteq B $ then $ P(A cap B cap C cap D) = P(A cap D) = 0.10 \ P(A | D) = frac{P(Acap D)}{P(D)} = 0.10 Rightarrow P(D) = 0.2$
But couldn't find a way to find $ P(B), P(C) $
conditional-probability
conditional-probability
asked Nov 17 at 18:46
bm1125
55116
asked Nov 17 at 18:46
bm1125
55116
asked Nov 17 at 18:46
bm1125
55116
asked Nov 17 at 18:46
bm1125
55116
asked Nov 17 at 18:46
bm1125
55116
55116
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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1
down vote
accepted
Partial solution:
Since $B^csubseteq C^csubseteq D^c$ we have:
$$P(A cap B^c cap C^c cap D^c) = 0.2implies P(Acap B^c)=0.2$$
so $$boxed{P(A)-P(Acap B)=0.2}$$
and since $Bcup Ccup D=B$ we have $$P(Acup B) =0.8$$
so $$boxed{P(A)+P(B)-P(Acap B)=0.8}$$
so we get $P(B)=0.6$
From here $$ P(A^c cap B cap C^c) = 0.1$$ we should somehow find $P(C)$...
answered Nov 17 at 19:21
greedoid
35.1k114489
Could you just expalin why $ P(A cup B) = 0.8 $ that is not clear to me?
– bm1125
Nov 18 at 13:41
1
I made an edit...
– greedoid
Nov 18 at 13:52
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
accepted
Partial solution:
Since $B^csubseteq C^csubseteq D^c$ we have:
$$P(A cap B^c cap C^c cap D^c) = 0.2implies P(Acap B^c)=0.2$$
so $$boxed{P(A)-P(Acap B)=0.2}$$
and since $Bcup Ccup D=B$ we have $$P(Acup B) =0.8$$
so $$boxed{P(A)+P(B)-P(Acap B)=0.8}$$
so we get $P(B)=0.6$
From here $$ P(A^c cap B cap C^c) = 0.1$$ we should somehow find $P(C)$...
answered Nov 17 at 19:21
greedoid
35.1k114489
Could you just expalin why $ P(A cup B) = 0.8 $ that is not clear to me?
– bm1125
Nov 18 at 13:41
1
I made an edit...
– greedoid
Nov 18 at 13:52
add a comment |
up vote
1
down vote
accepted
Partial solution:
Since $B^csubseteq C^csubseteq D^c$ we have:
$$P(A cap B^c cap C^c cap D^c) = 0.2implies P(Acap B^c)=0.2$$
so $$boxed{P(A)-P(Acap B)=0.2}$$
and since $Bcup Ccup D=B$ we have $$P(Acup B) =0.8$$
so $$boxed{P(A)+P(B)-P(Acap B)=0.8}$$
so we get $P(B)=0.6$
From here $$ P(A^c cap B cap C^c) = 0.1$$ we should somehow find $P(C)$...
answered Nov 17 at 19:21
greedoid
35.1k114489
Could you just expalin why $ P(A cup B) = 0.8 $ that is not clear to me?
– bm1125
Nov 18 at 13:41
1
I made an edit...
– greedoid
Nov 18 at 13:52
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Partial solution:
Since $B^csubseteq C^csubseteq D^c$ we have:
$$P(A cap B^c cap C^c cap D^c) = 0.2implies P(Acap B^c)=0.2$$
so $$boxed{P(A)-P(Acap B)=0.2}$$
and since $Bcup Ccup D=B$ we have $$P(Acup B) =0.8$$
so $$boxed{P(A)+P(B)-P(Acap B)=0.8}$$
so we get $P(B)=0.6$
From here $$ P(A^c cap B cap C^c) = 0.1$$ we should somehow find $P(C)$...
answered Nov 17 at 19:21
greedoid
35.1k114489
Partial solution:
Since $B^csubseteq C^csubseteq D^c$ we have:
$$P(A cap B^c cap C^c cap D^c) = 0.2implies P(Acap B^c)=0.2$$
so $$boxed{P(A)-P(Acap B)=0.2}$$
and since $Bcup Ccup D=B$ we have $$P(Acup B) =0.8$$
so $$boxed{P(A)+P(B)-P(Acap B)=0.8}$$
so we get $P(B)=0.6$
From here $$ P(A^c cap B cap C^c) = 0.1$$ we should somehow find $P(C)$...
answered Nov 17 at 19:21
greedoid
35.1k114489
edited Nov 18 at 13:51
answered Nov 17 at 19:21
greedoid
35.1k114489
answered Nov 17 at 19:21
greedoid
35.1k114489
answered Nov 17 at 19:21
greedoid
35.1k114489
35.1k114489
Could you just expalin why $ P(A cup B) = 0.8 $ that is not clear to me?
– bm1125
Nov 18 at 13:41
1
I made an edit...
– greedoid
Nov 18 at 13:52
add a comment |
Could you just expalin why $ P(A cup B) = 0.8 $ that is not clear to me?
– bm1125
Nov 18 at 13:41
1
I made an edit...
– greedoid
Nov 18 at 13:52
Could you just expalin why $ P(A cup B) = 0.8 $ that is not clear to me?
– bm1125
Nov 18 at 13:41
Could you just expalin why $ P(A cup B) = 0.8 $ that is not clear to me?
– bm1125
Nov 18 at 13:41
1
1
I made an edit...
– greedoid
Nov 18 at 13:52
I made an edit...
– greedoid
Nov 18 at 13:52
add a comment |
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