Mixing
If $P(A cup B cup C) = 1$, $P(B) = 2P(A) $, $P(C) = 3P(A) $, $P(A cap B) = P(A cap C) = P(B cap C) $, then...
$begingroup$
We have ($P$ is probability):
$P(A cup B cup C) = 1$ ; $P(B) = 2P(A) $ ; $P(C) = 3P(A) $ and $P(A cap B) = P(A cap C) = P(B cap C) $. Prove that $P(A) le frac{1}{4} $.
Well, I tried with the fact that $ 1 = P(A cup B cup C) = 6P(A) - 3P(A cap B) + P(A cap B cap C) $ but I got stuck... Could anyone help me, please?
probability
edited Jan 11 at 22:00
Did
248k23223460
asked Feb 19 '13 at 19:56
AnneAnne
732918
$endgroup$
add a comment |
$begingroup$
We have ($P$ is probability):
$P(A cup B cup C) = 1$ ; $P(B) = 2P(A) $ ; $P(C) = 3P(A) $ and $P(A cap B) = P(A cap C) = P(B cap C) $. Prove that $P(A) le frac{1}{4} $.
Well, I tried with the fact that $ 1 = P(A cup B cup C) = 6P(A) - 3P(A cap B) + P(A cap B cap C) $ but I got stuck... Could anyone help me, please?
probability
edited Jan 11 at 22:00
Did
248k23223460
asked Feb 19 '13 at 19:56
AnneAnne
732918
$endgroup$
add a comment |
$begingroup$
We have ($P$ is probability):
$P(A cup B cup C) = 1$ ; $P(B) = 2P(A) $ ; $P(C) = 3P(A) $ and $P(A cap B) = P(A cap C) = P(B cap C) $. Prove that $P(A) le frac{1}{4} $.
Well, I tried with the fact that $ 1 = P(A cup B cup C) = 6P(A) - 3P(A cap B) + P(A cap B cap C) $ but I got stuck... Could anyone help me, please?
probability
edited Jan 11 at 22:00
Did
248k23223460
asked Feb 19 '13 at 19:56
AnneAnne
732918
$endgroup$
We have ($P$ is probability):
$P(A cup B cup C) = 1$ ; $P(B) = 2P(A) $ ; $P(C) = 3P(A) $ and $P(A cap B) = P(A cap C) = P(B cap C) $. Prove that $P(A) le frac{1}{4} $.
Well, I tried with the fact that $ 1 = P(A cup B cup C) = 6P(A) - 3P(A cap B) + P(A cap B cap C) $ but I got stuck... Could anyone help me, please?
probability
probability
edited Jan 11 at 22:00
Did
248k23223460
asked Feb 19 '13 at 19:56
AnneAnne
732918
edited Jan 11 at 22:00
Did
248k23223460
asked Feb 19 '13 at 19:56
AnneAnne
732918
edited Jan 11 at 22:00
Did
248k23223460
edited Jan 11 at 22:00
Did
248k23223460
edited Jan 11 at 22:00
Did
248k23223460
248k23223460
asked Feb 19 '13 at 19:56
AnneAnne
732918
asked Feb 19 '13 at 19:56
AnneAnne
732918
asked Feb 19 '13 at 19:56
AnneAnne
732918
732918
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $mathbb P(A)=a$, then $mathbb P(B)=2a$ and $mathbb P(C)=3a$. Hence $mathbb P(Bcap C)geqslant mathbb P(B)+mathbb P(C)-1$, that is, $mathbb P(Bcap C)geqslant5a-1$. By hypothesis, $mathbb P(Acap B)=mathbb P(Bcap C)$ hence $mathbb P(Acap B)geqslant5a-1$. But $mathbb P(Acap B)leqslant mathbb P(A)=a$ hence $ageqslant5a-1$, that is, $mathbb P(A)=aleqslantfrac14$.
Note: This does not use the hypotheses on $mathbb P(Acup Bcup C)$ and $mathbb P(Acap C)$.
answered Feb 19 '13 at 21:27
DidDid
248k23223460
$endgroup$
add a comment |
$begingroup$
This solution is quite possibly messier than necessary, but it’s what first occurred to me. For convenience let $x=Pbig((Acap B)setminus Cbig)$ and $y=P(Acap Bcap C)$. Let $a=P(A)-2x-y$, $b=P(B)-2x-y$, and $c=P(C)-2x-y$; these are the probabilities of $Asetminus(Bcup C)$, $Bsetminus(Acup C)$, and $Csetminus(Acup B)$, respectively. (A Venn diagram is helpful here.)
Now $$b+2x+y=2(a+2x+y)=2a+4x+2y;,$$ so $b=2a+2x+y$. Similarly, $$c+2x+y=3(a+2x+y)=3a+6x+3y;,$$ so $c=3a+4x+2y$. Then
$$begin{align*}
1&=P(Acup Bcup C)\
&=a+b+c+3x+y\
&=6a+9x+4y\
&=4(a+2x+y)+2a+x\
&=4P(A)+2a+x;.
end{align*}$$
Since $2a+xge 0$, we must have $P(A)le frac{1}{4}$.
answered Feb 19 '13 at 20:16
Brian M. ScottBrian M. Scott
457k38510909
$endgroup$
$begingroup$
But I guess $a= P(A) -2x+y$ and it changes a lot...
$endgroup$
– Anne
Feb 19 '13 at 20:58
$begingroup$
@Anne: Sorry: I typoed the definition of $x$. You’ll see that everything is correct if $x$ is the probability of $(Acap B)setminus C$, as it was intended to be.
$endgroup$
– Brian M. Scott
Feb 20 '13 at 0:29
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $mathbb P(A)=a$, then $mathbb P(B)=2a$ and $mathbb P(C)=3a$. Hence $mathbb P(Bcap C)geqslant mathbb P(B)+mathbb P(C)-1$, that is, $mathbb P(Bcap C)geqslant5a-1$. By hypothesis, $mathbb P(Acap B)=mathbb P(Bcap C)$ hence $mathbb P(Acap B)geqslant5a-1$. But $mathbb P(Acap B)leqslant mathbb P(A)=a$ hence $ageqslant5a-1$, that is, $mathbb P(A)=aleqslantfrac14$.
Note: This does not use the hypotheses on $mathbb P(Acup Bcup C)$ and $mathbb P(Acap C)$.
answered Feb 19 '13 at 21:27
DidDid
248k23223460
$endgroup$
add a comment |
$begingroup$
Let $mathbb P(A)=a$, then $mathbb P(B)=2a$ and $mathbb P(C)=3a$. Hence $mathbb P(Bcap C)geqslant mathbb P(B)+mathbb P(C)-1$, that is, $mathbb P(Bcap C)geqslant5a-1$. By hypothesis, $mathbb P(Acap B)=mathbb P(Bcap C)$ hence $mathbb P(Acap B)geqslant5a-1$. But $mathbb P(Acap B)leqslant mathbb P(A)=a$ hence $ageqslant5a-1$, that is, $mathbb P(A)=aleqslantfrac14$.
Note: This does not use the hypotheses on $mathbb P(Acup Bcup C)$ and $mathbb P(Acap C)$.
answered Feb 19 '13 at 21:27
DidDid
248k23223460
$endgroup$
add a comment |
$begingroup$
Let $mathbb P(A)=a$, then $mathbb P(B)=2a$ and $mathbb P(C)=3a$. Hence $mathbb P(Bcap C)geqslant mathbb P(B)+mathbb P(C)-1$, that is, $mathbb P(Bcap C)geqslant5a-1$. By hypothesis, $mathbb P(Acap B)=mathbb P(Bcap C)$ hence $mathbb P(Acap B)geqslant5a-1$. But $mathbb P(Acap B)leqslant mathbb P(A)=a$ hence $ageqslant5a-1$, that is, $mathbb P(A)=aleqslantfrac14$.
Note: This does not use the hypotheses on $mathbb P(Acup Bcup C)$ and $mathbb P(Acap C)$.
answered Feb 19 '13 at 21:27
DidDid
248k23223460
$endgroup$
Let $mathbb P(A)=a$, then $mathbb P(B)=2a$ and $mathbb P(C)=3a$. Hence $mathbb P(Bcap C)geqslant mathbb P(B)+mathbb P(C)-1$, that is, $mathbb P(Bcap C)geqslant5a-1$. By hypothesis, $mathbb P(Acap B)=mathbb P(Bcap C)$ hence $mathbb P(Acap B)geqslant5a-1$. But $mathbb P(Acap B)leqslant mathbb P(A)=a$ hence $ageqslant5a-1$, that is, $mathbb P(A)=aleqslantfrac14$.
Note: This does not use the hypotheses on $mathbb P(Acup Bcup C)$ and $mathbb P(Acap C)$.
answered Feb 19 '13 at 21:27
DidDid
248k23223460
answered Feb 19 '13 at 21:27
DidDid
248k23223460
answered Feb 19 '13 at 21:27
DidDid
248k23223460
answered Feb 19 '13 at 21:27
DidDid
248k23223460
248k23223460
add a comment |
add a comment |
$begingroup$
This solution is quite possibly messier than necessary, but it’s what first occurred to me. For convenience let $x=Pbig((Acap B)setminus Cbig)$ and $y=P(Acap Bcap C)$. Let $a=P(A)-2x-y$, $b=P(B)-2x-y$, and $c=P(C)-2x-y$; these are the probabilities of $Asetminus(Bcup C)$, $Bsetminus(Acup C)$, and $Csetminus(Acup B)$, respectively. (A Venn diagram is helpful here.)
Now $$b+2x+y=2(a+2x+y)=2a+4x+2y;,$$ so $b=2a+2x+y$. Similarly, $$c+2x+y=3(a+2x+y)=3a+6x+3y;,$$ so $c=3a+4x+2y$. Then
$$begin{align*}
1&=P(Acup Bcup C)\
&=a+b+c+3x+y\
&=6a+9x+4y\
&=4(a+2x+y)+2a+x\
&=4P(A)+2a+x;.
end{align*}$$
Since $2a+xge 0$, we must have $P(A)le frac{1}{4}$.
answered Feb 19 '13 at 20:16
Brian M. ScottBrian M. Scott
457k38510909
$endgroup$
$begingroup$
But I guess $a= P(A) -2x+y$ and it changes a lot...
$endgroup$
– Anne
Feb 19 '13 at 20:58
$begingroup$
@Anne: Sorry: I typoed the definition of $x$. You’ll see that everything is correct if $x$ is the probability of $(Acap B)setminus C$, as it was intended to be.
$endgroup$
– Brian M. Scott
Feb 20 '13 at 0:29
add a comment |
$begingroup$
This solution is quite possibly messier than necessary, but it’s what first occurred to me. For convenience let $x=Pbig((Acap B)setminus Cbig)$ and $y=P(Acap Bcap C)$. Let $a=P(A)-2x-y$, $b=P(B)-2x-y$, and $c=P(C)-2x-y$; these are the probabilities of $Asetminus(Bcup C)$, $Bsetminus(Acup C)$, and $Csetminus(Acup B)$, respectively. (A Venn diagram is helpful here.)
Now $$b+2x+y=2(a+2x+y)=2a+4x+2y;,$$ so $b=2a+2x+y$. Similarly, $$c+2x+y=3(a+2x+y)=3a+6x+3y;,$$ so $c=3a+4x+2y$. Then
$$begin{align*}
1&=P(Acup Bcup C)\
&=a+b+c+3x+y\
&=6a+9x+4y\
&=4(a+2x+y)+2a+x\
&=4P(A)+2a+x;.
end{align*}$$
Since $2a+xge 0$, we must have $P(A)le frac{1}{4}$.
answered Feb 19 '13 at 20:16
Brian M. ScottBrian M. Scott
457k38510909
$endgroup$
$begingroup$
But I guess $a= P(A) -2x+y$ and it changes a lot...
$endgroup$
– Anne
Feb 19 '13 at 20:58
$begingroup$
@Anne: Sorry: I typoed the definition of $x$. You’ll see that everything is correct if $x$ is the probability of $(Acap B)setminus C$, as it was intended to be.
$endgroup$
– Brian M. Scott
Feb 20 '13 at 0:29
add a comment |
$begingroup$
This solution is quite possibly messier than necessary, but it’s what first occurred to me. For convenience let $x=Pbig((Acap B)setminus Cbig)$ and $y=P(Acap Bcap C)$. Let $a=P(A)-2x-y$, $b=P(B)-2x-y$, and $c=P(C)-2x-y$; these are the probabilities of $Asetminus(Bcup C)$, $Bsetminus(Acup C)$, and $Csetminus(Acup B)$, respectively. (A Venn diagram is helpful here.)
Now $$b+2x+y=2(a+2x+y)=2a+4x+2y;,$$ so $b=2a+2x+y$. Similarly, $$c+2x+y=3(a+2x+y)=3a+6x+3y;,$$ so $c=3a+4x+2y$. Then
$$begin{align*}
1&=P(Acup Bcup C)\
&=a+b+c+3x+y\
&=6a+9x+4y\
&=4(a+2x+y)+2a+x\
&=4P(A)+2a+x;.
end{align*}$$
Since $2a+xge 0$, we must have $P(A)le frac{1}{4}$.
answered Feb 19 '13 at 20:16
Brian M. ScottBrian M. Scott
457k38510909
$endgroup$
This solution is quite possibly messier than necessary, but it’s what first occurred to me. For convenience let $x=Pbig((Acap B)setminus Cbig)$ and $y=P(Acap Bcap C)$. Let $a=P(A)-2x-y$, $b=P(B)-2x-y$, and $c=P(C)-2x-y$; these are the probabilities of $Asetminus(Bcup C)$, $Bsetminus(Acup C)$, and $Csetminus(Acup B)$, respectively. (A Venn diagram is helpful here.)
Now $$b+2x+y=2(a+2x+y)=2a+4x+2y;,$$ so $b=2a+2x+y$. Similarly, $$c+2x+y=3(a+2x+y)=3a+6x+3y;,$$ so $c=3a+4x+2y$. Then
$$begin{align*}
1&=P(Acup Bcup C)\
&=a+b+c+3x+y\
&=6a+9x+4y\
&=4(a+2x+y)+2a+x\
&=4P(A)+2a+x;.
end{align*}$$
Since $2a+xge 0$, we must have $P(A)le frac{1}{4}$.
answered Feb 19 '13 at 20:16
Brian M. ScottBrian M. Scott
457k38510909
edited Feb 20 '13 at 0:30
answered Feb 19 '13 at 20:16
Brian M. ScottBrian M. Scott
457k38510909
answered Feb 19 '13 at 20:16
Brian M. ScottBrian M. Scott
457k38510909
answered Feb 19 '13 at 20:16
Brian M. ScottBrian M. Scott
457k38510909
457k38510909
$begingroup$
But I guess $a= P(A) -2x+y$ and it changes a lot...
$endgroup$
– Anne
Feb 19 '13 at 20:58
$begingroup$
@Anne: Sorry: I typoed the definition of $x$. You’ll see that everything is correct if $x$ is the probability of $(Acap B)setminus C$, as it was intended to be.
$endgroup$
– Brian M. Scott
Feb 20 '13 at 0:29
add a comment |
$begingroup$
But I guess $a= P(A) -2x+y$ and it changes a lot...
$endgroup$
– Anne
Feb 19 '13 at 20:58
$begingroup$
@Anne: Sorry: I typoed the definition of $x$. You’ll see that everything is correct if $x$ is the probability of $(Acap B)setminus C$, as it was intended to be.
$endgroup$
– Brian M. Scott
Feb 20 '13 at 0:29
$begingroup$
But I guess $a= P(A) -2x+y$ and it changes a lot...
$endgroup$
– Anne
Feb 19 '13 at 20:58
$begingroup$
But I guess $a= P(A) -2x+y$ and it changes a lot...
$endgroup$
– Anne
Feb 19 '13 at 20:58
$begingroup$
@Anne: Sorry: I typoed the definition of $x$. You’ll see that everything is correct if $x$ is the probability of $(Acap B)setminus C$, as it was intended to be.
$endgroup$
– Brian M. Scott
Feb 20 '13 at 0:29
$begingroup$
@Anne: Sorry: I typoed the definition of $x$. You’ll see that everything is correct if $x$ is the probability of $(Acap B)setminus C$, as it was intended to be.
$endgroup$
– Brian M. Scott
Feb 20 '13 at 0:29
add a comment |
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