#1
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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? |
#2
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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. |
#3
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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 |
#4
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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. |
#5
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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 |
#6
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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 |
#7
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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 |
#8
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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. |
#9
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thank you
outstanding proof. I appreciate that you took the time to write it out for me.
thanks. eric |
#10
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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 |
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