Finding Probability Using Sets
Clash Royale CLAN TAG#URR8PPP
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How do I go about solving the following probability problem.
$P(Acup(B^complementcup C^complement)^complement)$.
Events $A$, $B$ and $C$ are disjoint and $P(A)=2/5$
Using De Morgan's for
$B^complementcup C^complement=(Bcap C)^complement$
and the compliment of this gives
$Bcap C$.
So do I end up with
$Acup (Bcap C)$
If $P(A)=2/5$. How do I find the propability of
$Acup(B^complementcup C^complement)^complement$
probability probability-theory
edited Sep 10 at 8:15
drhab
89.4k541123
asked Sep 10 at 8:00
rizzo
12
add a comment |Â
up vote
0
down vote
favorite
How do I go about solving the following probability problem.
$P(Acup(B^complementcup C^complement)^complement)$.
Events $A$, $B$ and $C$ are disjoint and $P(A)=2/5$
Using De Morgan's for
$B^complementcup C^complement=(Bcap C)^complement$
and the compliment of this gives
$Bcap C$.
So do I end up with
$Acup (Bcap C)$
If $P(A)=2/5$. How do I find the propability of
$Acup(B^complementcup C^complement)^complement$
probability probability-theory
edited Sep 10 at 8:15
drhab
89.4k541123
asked Sep 10 at 8:00
rizzo
12
Note $cup$ turns to $cap$ in De Morgan: $ B^complement cup C^complement = (B colorredcap C)^complement $
– Ben
Sep 10 at 8:06
Yes. Thank you. I corrected the question.
– rizzo
Sep 10 at 8:13
If $B$ and $C$ are disjoint, then $B cap C $ is empty, and hence $Acup (Bcap C) = A$.
– Hayk
Sep 10 at 8:19
@rizzo Before asking any new questions first take a look at MathJax.
– drhab
Sep 10 at 8:25
Thank you Hayk. And thanks for the link drhab.
– rizzo
Sep 10 at 8:36
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do I go about solving the following probability problem.
$P(Acup(B^complementcup C^complement)^complement)$.
Events $A$, $B$ and $C$ are disjoint and $P(A)=2/5$
Using De Morgan's for
$B^complementcup C^complement=(Bcap C)^complement$
and the compliment of this gives
$Bcap C$.
So do I end up with
$Acup (Bcap C)$
If $P(A)=2/5$. How do I find the propability of
$Acup(B^complementcup C^complement)^complement$
probability probability-theory
edited Sep 10 at 8:15
drhab
89.4k541123
asked Sep 10 at 8:00
rizzo
12
How do I go about solving the following probability problem.
$P(Acup(B^complementcup C^complement)^complement)$.
Events $A$, $B$ and $C$ are disjoint and $P(A)=2/5$
Using De Morgan's for
$B^complementcup C^complement=(Bcap C)^complement$
and the compliment of this gives
$Bcap C$.
So do I end up with
$Acup (Bcap C)$
If $P(A)=2/5$. How do I find the propability of
$Acup(B^complementcup C^complement)^complement$
probability probability-theory
probability probability-theory
edited Sep 10 at 8:15
drhab
89.4k541123
asked Sep 10 at 8:00
rizzo
12
edited Sep 10 at 8:15
drhab
89.4k541123
asked Sep 10 at 8:00
rizzo
12
edited Sep 10 at 8:15
drhab
89.4k541123
edited Sep 10 at 8:15
drhab
89.4k541123
edited Sep 10 at 8:15
drhab
89.4k541123
89.4k541123
asked Sep 10 at 8:00
rizzo
12
asked Sep 10 at 8:00
rizzo
12
asked Sep 10 at 8:00
rizzo
12
12
Note $cup$ turns to $cap$ in De Morgan: $ B^complement cup C^complement = (B colorredcap C)^complement $
– Ben
Sep 10 at 8:06
Yes. Thank you. I corrected the question.
– rizzo
Sep 10 at 8:13
If $B$ and $C$ are disjoint, then $B cap C $ is empty, and hence $Acup (Bcap C) = A$.
– Hayk
Sep 10 at 8:19
@rizzo Before asking any new questions first take a look at MathJax.
– drhab
Sep 10 at 8:25
Thank you Hayk. And thanks for the link drhab.
– rizzo
Sep 10 at 8:36
add a comment |Â
Note $cup$ turns to $cap$ in De Morgan: $ B^complement cup C^complement = (B colorredcap C)^complement $
– Ben
Sep 10 at 8:06
Yes. Thank you. I corrected the question.
– rizzo
Sep 10 at 8:13
If $B$ and $C$ are disjoint, then $B cap C $ is empty, and hence $Acup (Bcap C) = A$.
– Hayk
Sep 10 at 8:19
@rizzo Before asking any new questions first take a look at MathJax.
– drhab
Sep 10 at 8:25
Thank you Hayk. And thanks for the link drhab.
– rizzo
Sep 10 at 8:36
Note $cup$ turns to $cap$ in De Morgan: $ B^complement cup C^complement = (B colorredcap C)^complement $
– Ben
Sep 10 at 8:06
Note $cup$ turns to $cap$ in De Morgan: $ B^complement cup C^complement = (B colorredcap C)^complement $
– Ben
Sep 10 at 8:06
Yes. Thank you. I corrected the question.
– rizzo
Sep 10 at 8:13
Yes. Thank you. I corrected the question.
– rizzo
Sep 10 at 8:13
If $B$ and $C$ are disjoint, then $B cap C $ is empty, and hence $Acup (Bcap C) = A$.
– Hayk
Sep 10 at 8:19
If $B$ and $C$ are disjoint, then $B cap C $ is empty, and hence $Acup (Bcap C) = A$.
– Hayk
Sep 10 at 8:19
@rizzo Before asking any new questions first take a look at MathJax.
– drhab
Sep 10 at 8:25
@rizzo Before asking any new questions first take a look at MathJax.
– drhab
Sep 10 at 8:25
Thank you Hayk. And thanks for the link drhab.
– rizzo
Sep 10 at 8:36
Thank you Hayk. And thanks for the link drhab.
– rizzo
Sep 10 at 8:36
add a comment |Â
1 Answer
1
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0
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Since the sets $A,B,C$ are disjoint we have:$$P(Acup (B^complementcup C^complement)^complement)=P(Acup (Bcap C))=P(A)+P(Bcap C)=P(A)+P(varnothing)=frac25+0=frac25$$
answered Sep 10 at 8:19
drhab
89.4k541123
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Since the sets $A,B,C$ are disjoint we have:$$P(Acup (B^complementcup C^complement)^complement)=P(Acup (Bcap C))=P(A)+P(Bcap C)=P(A)+P(varnothing)=frac25+0=frac25$$
answered Sep 10 at 8:19
drhab
89.4k541123
add a comment |Â
up vote
0
down vote
accepted
Since the sets $A,B,C$ are disjoint we have:$$P(Acup (B^complementcup C^complement)^complement)=P(Acup (Bcap C))=P(A)+P(Bcap C)=P(A)+P(varnothing)=frac25+0=frac25$$
answered Sep 10 at 8:19
drhab
89.4k541123
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Since the sets $A,B,C$ are disjoint we have:$$P(Acup (B^complementcup C^complement)^complement)=P(Acup (Bcap C))=P(A)+P(Bcap C)=P(A)+P(varnothing)=frac25+0=frac25$$
answered Sep 10 at 8:19
drhab
89.4k541123
Since the sets $A,B,C$ are disjoint we have:$$P(Acup (B^complementcup C^complement)^complement)=P(Acup (Bcap C))=P(A)+P(Bcap C)=P(A)+P(varnothing)=frac25+0=frac25$$
answered Sep 10 at 8:19
drhab
89.4k541123
answered Sep 10 at 8:19
drhab
89.4k541123
answered Sep 10 at 8:19
drhab
89.4k541123
answered Sep 10 at 8:19
drhab
89.4k541123
89.4k541123
add a comment |Â
add a comment |Â
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Note $cup$ turns to $cap$ in De Morgan: $ B^complement cup C^complement = (B colorredcap C)^complement $
– Ben
Sep 10 at 8:06
Yes. Thank you. I corrected the question.
– rizzo
Sep 10 at 8:13
If $B$ and $C$ are disjoint, then $B cap C $ is empty, and hence $Acup (Bcap C) = A$.
– Hayk
Sep 10 at 8:19
@rizzo Before asking any new questions first take a look at MathJax.
– drhab
Sep 10 at 8:25
Thank you Hayk. And thanks for the link drhab.
– rizzo
Sep 10 at 8:36