probability question
 

Imagine 5 events, a, b, c, d and e. They may be mutually exclusive, or they may not be. All you know about them is the following:

P(a U c) = .4
P(a U b U d U e) = .8
P(c U d) = .4
P(a U b U c U d U e) = 1

What is P(a)? Is there a unique solution?

Re: probability question
 

Suppose that c and b U e are mutually exclusive, with P(c) = 0.4 and P(b U e) = 0.6. a and d only happen if c happens and P(a U d) = 0.2. That satisfies all four expressions and allows Pa(a) to take any value between 0 and 0.2.

We know P(a) can't be greater than 0.2 because then the first and last expressions conflict. So 0 <= P(a) <= 0.2.

Re: probability question
 

[ QUOTE ]
Imagine 5 events, a, b, c, d and e. They may be mutually exclusive, or they may not be. All you know about them is the following:

P(a U c) = .4
P(a U b U d U e) = .8
P(c U d) = .4
P(a U b U c U d U e) = 1

What is P(a)? Is there a unique solution?

[/ QUOTE ]


[ QUOTE ]
Suppose that c and b U e are mutually exclusive, with P(c) = 0.4 and P(b U e) = 0.6. a and d only happen if c happens and P(a U d) = 0.2. That satisfies all four expressions and allows Pa(a) to take any value between 0 and 0.2.

We know P(a) can't be greater than 0.2 because then the first and last expressions conflict. So 0 <= P(a) <= 0.2.

[/ QUOTE ]

As it relates to the problem in the other thread that's gotten so much discussion, notice also that under Aaron's conditions , if P(a) is strictly less than 0.2 then

(a U b U d U e) intersect (a U c) is not equal to a.

In fact, under aaron's conditions they are equal iff P(a)=.2

Regardless, your original conditions still imply that

P( (a U b U d U e) intersect (a U c) ) = .2

PairTheBoard

Re: probability question
 

[ QUOTE ]
Imagine 5 events, a, b, c, d and e. They may be mutually exclusive, or they may not be. All you know about them is the following:

P(a U c) = .4
P(a U b U d U e) = .8
P(c U d) = .4
P(a U b U c U d U e) = 1

What is P(a)? Is there a unique solution?

[/ QUOTE ]

There is no unique solution.

thank you
 

ahhh.... Outstanding counter-example, thanks. I'm convinced there is no unique solution.


I'd like to hear more about your claim that a must not be greater than .2... can you explain this further? I don't see the conflict in the equations you mentioned.

edit: n/m. I'm still not sure about the equations you mentioned, but I have found a proof that a <= .2 (see response to PairTheBoard for this). Thanks! /edit

thanks,
eric

Re: probability question
 

Hi PairTheBoard,

[ QUOTE ]
(a U b U d U e) intersect (a U c) is not equal to a.

In fact, under aaron's conditions they are equal iff P(a)=.2

[/ QUOTE ]

A very insightful point. Thank you. I'll fill the proof you left out by mentioning that under Aaron's conditions, aUc = c, so

(a U b U d U e) intersect (a U c) = aUd as c contains all points in a and d, and is ME with bUd.

We know that aUd = .2 from equation 2) aUd U bUe = .8 with Aaron assuming bUe is ME with aUd and equal to .6. Therefore, this intersection = a only when p(d) = 0 and p(a) = .2.



[ QUOTE ]
Regardless, your original conditions still imply that

P( (a U b U d U e) intersect (a U c) ) = .2

[/ QUOTE ]

Frankly, I'm getting rather bored with your simply stating that this is the case. You may be right, but "PairTheBoard said so" does not constitute a proof in my mind.

There is obviously confusion over this point, so if you can prove what you claim, why don't you just do it?

-eric

Re: probability question
 

[ QUOTE ]
Quote:
--------------------------------------------------------------------------------

Regardless, your original conditions still imply that

P( (a U b U d U e) intersect (a U c) ) = .2


--------------------------------------------------------------------------------



Frankly, I'm getting rather bored with your simply stating that this is the case. You may be right, but "PairTheBoard said so" does not constitute a proof in my mind.

There is obviously confusion over this point, so if you can prove what you claim, why don't you just do it?


[/ QUOTE ]

Given any two events E,F of a Sample Space, it's well known that,

* P(E union F) = P(E)+P(F)-P(E intersect F)

Letting E=(a U b U d U e)
and F=(a U c)
the proof follows from the fact that
(a U b U d U e) U (a U c) = (a U b U c U d U e)
and your assumption that
P(a U b U c U d U e) = 1

ie. P(E union F) = P(E) + P(F) - P(E intersect F)
so, 1 = .8 + .4 - P(E intersect F)
so, P(E intersect F) = .8 + .4 - 1 = .2

To see the * equation holds for any two Events E,F, using the set notation,
A-B = {x: x is in A and x is not in B)
we have the following partitions into mutually exlusive events,

E = (E-F) union (E intersect F)
F = ((F-E) union (E intersect F)
(E U F) = (E-F) union (F-E) union (E intersect F)
so that,

P(E-F) = P(E) - P(E intersect F)
P(F-E) = P(F) - P(E intersect F)
and
P(E union F) = P(E-F) + P(F-E) + P(E intersect F) =
= P(E)-P(E intersect F) + P(F) - P(E intersect F) + P(E intersect F) =
= P(E) + P(F) - P(E intersect F)

PairTheBoard

Re: thank you
 

[ QUOTE ]
ahhh.... Outstanding counter-example, thanks. I'm convinced there is no unique solution.

I'd like to hear more about your claim that a must not be greater than .2... can you explain this further? I don't see the conflict in the equations you mentioned.


[/ QUOTE ]
Sorry, I misstated it in my answer. You need the second equation as well. The second and fourth together tell us that the probability of c U (not (a U b U d U e)) is 0.2 (since taking c out reduces the probability from 1 to 0.8). Therefore P(c U (not a)) >= 0.2. P(a U c) = P(a) + P(not a U c) = 0.4, so P(a) <= 0.2.

If you only want to prove there's no unique solution, it's easy to see that:

P(a) = P(b) = P(c) = P(d) = P(e) = 0.2 is a solution with everything mutually exclusive and,

P(a) = P(d) = 0, P(b) = P(c) = P(e) = 0.4 is another solution with b and c mutually exclusive and P(not b U c U not e) = 0.2.

thank you
 

outstanding proof. I appreciate that you took the time to write it out for me.

thanks.
eric

Re: thank you
 

[ QUOTE ]
Sorry, I misstated it in my answer. You need the second equation as well. The second and fourth together tell us that the probability of c U (not (a U b U d U e)) is 0.2 (since taking c out reduces the probability from 1 to 0.8). Therefore P(c U (not a)) >= 0.2. P(a U c) = P(a) + P(not a U c) = 0.4, so P(a) <= 0.2.


[/ QUOTE ]

I think you mean,

Sorry, I misstated it in my answer. You need the second equation as well. The second and fourth together tell us that the probability of c intersect (not (a U b U d U e)) is 0.2 (since taking c out reduces the probability from 1 to 0.8). Therefore P(c intersect (not a)) >= 0.2. P(a U c) = P(a) + P( (not a) intersect c) = 0.4, so P(a) <= 0.2.

PairTheBoard