Determine P(M) using given statements
$begingroup$
So I have an assignment in probability.
I have the events M, A and B.
After some calculations I found that P(MAB)=0.064, P(MAB/)=0.192 and P(MA/B)=0.084.
(B/ is B complement and A/ is A complement).
My question is is there any way to determine P(M) using the given statements?
probability
asked Dec 18 '18 at 23:44
David DanielsDavid Daniels
132
$endgroup$
add a comment |
$begingroup$
So I have an assignment in probability.
I have the events M, A and B.
After some calculations I found that P(MAB)=0.064, P(MAB/)=0.192 and P(MA/B)=0.084.
(B/ is B complement and A/ is A complement).
My question is is there any way to determine P(M) using the given statements?
probability
asked Dec 18 '18 at 23:44
David DanielsDavid Daniels
132
$endgroup$
add a comment |
$begingroup$
So I have an assignment in probability.
I have the events M, A and B.
After some calculations I found that P(MAB)=0.064, P(MAB/)=0.192 and P(MA/B)=0.084.
(B/ is B complement and A/ is A complement).
My question is is there any way to determine P(M) using the given statements?
probability
asked Dec 18 '18 at 23:44
David DanielsDavid Daniels
132
$endgroup$
So I have an assignment in probability.
I have the events M, A and B.
After some calculations I found that P(MAB)=0.064, P(MAB/)=0.192 and P(MA/B)=0.084.
(B/ is B complement and A/ is A complement).
My question is is there any way to determine P(M) using the given statements?
probability
probability
asked Dec 18 '18 at 23:44
David DanielsDavid Daniels
132
asked Dec 18 '18 at 23:44
David DanielsDavid Daniels
132
asked Dec 18 '18 at 23:44
David DanielsDavid Daniels
132
asked Dec 18 '18 at 23:44
David DanielsDavid Daniels
132
asked Dec 18 '18 at 23:44
David DanielsDavid Daniels
132
132
add a comment |
add a comment |
1 Answer
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$begingroup$
Not just with that information, since $P(Mcap A^c cap B^c)$ could be anything from $0$ to $0.66$
and so $P(Mcap A cap B)+P(Mcap A cap B^c)+ P(Mcap A^c cap B)+ P(Mcap A^c cap B^c)$ could be anything from $0.34$ to $1$
answered Dec 18 '18 at 23:55
HenryHenry
100k480166
$endgroup$
$begingroup$
Thanks for the info. What about P(Mc∩Ac∩Bc)?
$endgroup$
– David Daniels
Dec 19 '18 at 21:06
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@DavidDaniels $1-left(P(M^ccap A cap B)+P(M^ccap A cap B^c)+ P(M^ccap A^c cap B)+ P(M^ccap A^c cap B^c)right)$ would work too, but you do not know any of those
$endgroup$
– Henry
Dec 19 '18 at 22:15
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
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votes
$begingroup$
Not just with that information, since $P(Mcap A^c cap B^c)$ could be anything from $0$ to $0.66$
and so $P(Mcap A cap B)+P(Mcap A cap B^c)+ P(Mcap A^c cap B)+ P(Mcap A^c cap B^c)$ could be anything from $0.34$ to $1$
answered Dec 18 '18 at 23:55
HenryHenry
100k480166
$endgroup$
$begingroup$
Thanks for the info. What about P(Mc∩Ac∩Bc)?
$endgroup$
– David Daniels
Dec 19 '18 at 21:06
$begingroup$
@DavidDaniels $1-left(P(M^ccap A cap B)+P(M^ccap A cap B^c)+ P(M^ccap A^c cap B)+ P(M^ccap A^c cap B^c)right)$ would work too, but you do not know any of those
$endgroup$
– Henry
Dec 19 '18 at 22:15
add a comment |
$begingroup$
Not just with that information, since $P(Mcap A^c cap B^c)$ could be anything from $0$ to $0.66$
and so $P(Mcap A cap B)+P(Mcap A cap B^c)+ P(Mcap A^c cap B)+ P(Mcap A^c cap B^c)$ could be anything from $0.34$ to $1$
answered Dec 18 '18 at 23:55
HenryHenry
100k480166
$endgroup$
$begingroup$
Thanks for the info. What about P(Mc∩Ac∩Bc)?
$endgroup$
– David Daniels
Dec 19 '18 at 21:06
$begingroup$
@DavidDaniels $1-left(P(M^ccap A cap B)+P(M^ccap A cap B^c)+ P(M^ccap A^c cap B)+ P(M^ccap A^c cap B^c)right)$ would work too, but you do not know any of those
$endgroup$
– Henry
Dec 19 '18 at 22:15
add a comment |
$begingroup$
Not just with that information, since $P(Mcap A^c cap B^c)$ could be anything from $0$ to $0.66$
and so $P(Mcap A cap B)+P(Mcap A cap B^c)+ P(Mcap A^c cap B)+ P(Mcap A^c cap B^c)$ could be anything from $0.34$ to $1$
answered Dec 18 '18 at 23:55
HenryHenry
100k480166
$endgroup$
Not just with that information, since $P(Mcap A^c cap B^c)$ could be anything from $0$ to $0.66$
and so $P(Mcap A cap B)+P(Mcap A cap B^c)+ P(Mcap A^c cap B)+ P(Mcap A^c cap B^c)$ could be anything from $0.34$ to $1$
answered Dec 18 '18 at 23:55
HenryHenry
100k480166
answered Dec 18 '18 at 23:55
HenryHenry
100k480166
answered Dec 18 '18 at 23:55
HenryHenry
100k480166
answered Dec 18 '18 at 23:55
HenryHenry
100k480166
100k480166
$begingroup$
Thanks for the info. What about P(Mc∩Ac∩Bc)?
$endgroup$
– David Daniels
Dec 19 '18 at 21:06
$begingroup$
@DavidDaniels $1-left(P(M^ccap A cap B)+P(M^ccap A cap B^c)+ P(M^ccap A^c cap B)+ P(M^ccap A^c cap B^c)right)$ would work too, but you do not know any of those
$endgroup$
– Henry
Dec 19 '18 at 22:15
add a comment |
$begingroup$
Thanks for the info. What about P(Mc∩Ac∩Bc)?
$endgroup$
– David Daniels
Dec 19 '18 at 21:06
$begingroup$
@DavidDaniels $1-left(P(M^ccap A cap B)+P(M^ccap A cap B^c)+ P(M^ccap A^c cap B)+ P(M^ccap A^c cap B^c)right)$ would work too, but you do not know any of those
$endgroup$
– Henry
Dec 19 '18 at 22:15
$begingroup$
Thanks for the info. What about P(Mc∩Ac∩Bc)?
$endgroup$
– David Daniels
Dec 19 '18 at 21:06
$begingroup$
Thanks for the info. What about P(Mc∩Ac∩Bc)?
$endgroup$
– David Daniels
Dec 19 '18 at 21:06
$begingroup$
@DavidDaniels $1-left(P(M^ccap A cap B)+P(M^ccap A cap B^c)+ P(M^ccap A^c cap B)+ P(M^ccap A^c cap B^c)right)$ would work too, but you do not know any of those
$endgroup$
– Henry
Dec 19 '18 at 22:15
$begingroup$
@DavidDaniels $1-left(P(M^ccap A cap B)+P(M^ccap A cap B^c)+ P(M^ccap A^c cap B)+ P(M^ccap A^c cap B^c)right)$ would work too, but you do not know any of those
$endgroup$
– Henry
Dec 19 '18 at 22:15
add a comment |
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