adding probabilities
in [Math]
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From: SomeOne on 7 Mar 2005 13:45 Hi, I'm kinda stuck with this problem I'm having. Maybe you can help me. Let's say we have an event A. This event can occur through two mechanisms. The probability that event A occurs through mechanism 1 is P1=0.9. The probability that event A occurs through mechanism 2 is P2=0.8. The question I'm struggling with is what is the probability that event A will occur? Now, just adding probabilities doesn't seem like a good idea because then we end up with a probability that is greater than 1, and that would be ridiculous. Also, multiplying probabilities seems silly because then you end up with a probability that is less than the original two. So, after long pondering this problem, I figured that the probability must be something like: P1+(1-P1)*P2 This looks right to me. Even more so, because if I turn it around, thus: P2+(1-P2)*P2, it yields the same result. And the result sounds about right. Now, the obvious question is of course: is this correct? And if it's not: what am I doing wrong? It seems to me that I'm trying to re-invent the wheel and that there is some general rule for complementary probabilities and how to derive the ultimate probability. But, it escapes me. What am I overlooking? -- Maurice http://get-me.to/hendrix
From: clemenr on 7 Mar 2005 14:07 SomeOne wrote: > Hi, > > I'm kinda stuck with this problem I'm having. Maybe you can help me. > > Let's say we have an event A. This event can occur through two > mechanisms. The probability that event A occurs through mechanism 1 is > P1=0.9. The probability that event A occurs through mechanism 2 is P2=0.8. > > The question I'm struggling with is what is the probability that event A > will occur? Disclaimer: I'm no expert statistician, but that rarely shuts me up. Actually, without further details, your question appears to me to be more difficult than it looks. It depends a lot on the real world meaning of event A, and the mechanisms. First I'll assume that the mechanism occurring does not definitely result in A being true: Can A occur with both mechanism 1 and mechanism 2 being true? If not, then you'd have: p( a ) = p( a | m1 ) * p( m1 ) + p( a | m2 ) * p( m2 ) However, if both m1 and m2 can occur at the same time, then you'd have: p( a ) = p( a | m1, ¬m2 ) + p( a | ~m1, m2 ) + p( a | m1, m2 ) The first term summed is for "A occuring solely because of m1 with m2 not occurring", the second is for "A occurring solely because of m2 with m1 not occurring", and the third is for "A occurring due to both m1 and m2". The probabilities of these three cases (I'm assuming that A never occurs without either m1 or m2) are going to depend on the real world situation. Secondly: The case where a mechanism occurring means that A is definitely true. That's simpler. if AM1 is the probability that A occurs by mechanism 1, AM2 is the probability that A occurs by mechanism 2, and AM1M2 is A occurring with both mechanisms present, then: p( A ) = P( AM1 ) + P( AM2 ) - P( AM1M2 ) The reason for minusing the third term is because otherwise when both occur, the event is counted twice by both the previous terms. Cheers, Ross-c
From: Bruce Weaver on 7 Mar 2005 14:10 SomeOne wrote: > Hi, > > I'm kinda stuck with this problem I'm having. Maybe you can help me. > > Let's say we have an event A. This event can occur through two > mechanisms. The probability that event A occurs through mechanism 1 is > P1=0.9. The probability that event A occurs through mechanism 2 is P2=0.8. > > The question I'm struggling with is what is the probability that event A > will occur? > > Now, just adding probabilities doesn't seem like a good idea because > then we end up with a probability that is greater than 1, and that would > be ridiculous. Also, multiplying probabilities seems silly because then > you end up with a probability that is less than the original two. > > So, after long pondering this problem, I figured that the probability > must be something like: > > P1+(1-P1)*P2 > > This looks right to me. Even more so, because if I turn it around, thus: > P2+(1-P2)*P2, it yields the same result. And the result sounds about right. > > Now, the obvious question is of course: is this correct? And if it's > not: what am I doing wrong? > It seems to me that I'm trying to re-invent the wheel and that there is > some general rule for complementary probabilities and how to derive the > ultimate probability. But, it escapes me. What am I overlooking? > Look up the "general addition rule". p(A1 or A2) = p(A1) + p(A2) - p(A1 and A2). If you don't understand why you need to subtract p(A1 and A2), draw a Venn diagram, with one circle representing p(A1), the other representing p(A2), and the overlap of the two representing p(A1 and A2). Notice that the intersection gets counted twice when you add p(A1) and p(A2), so you need to subtract it once. -- Bruce Weaver bweaver(a)lakeheadu.ca www.angelfire.com/wv/bwhomedir
From: Matt Gutting on 7 Mar 2005 14:09 SomeOne wrote: > Hi, > > I'm kinda stuck with this problem I'm having. Maybe you can help me. > > Let's say we have an event A. This event can occur through two > mechanisms. The probability that event A occurs through mechanism 1 is > P1=0.9. The probability that event A occurs through mechanism 2 is P2=0.8. > > The question I'm struggling with is what is the probability that event A > will occur? > > Now, just adding probabilities doesn't seem like a good idea because > then we end up with a probability that is greater than 1, and that would > be ridiculous. Also, multiplying probabilities seems silly because then > you end up with a probability that is less than the original two. > > So, after long pondering this problem, I figured that the probability > must be something like: > > P1+(1-P1)*P2 > > This looks right to me. Even more so, because if I turn it around, thus: > P2+(1-P2)*P2, it yields the same result. And the result sounds about right. > > Now, the obvious question is of course: is this correct? And if it's > not: what am I doing wrong? > It seems to me that I'm trying to re-invent the wheel and that there is > some general rule for complementary probabilities and how to derive the > ultimate probability. But, it escapes me. What am I overlooking? > What you've got is a formula that calculates the probability that either A occurs through mechanism 1 (P1), or else A occurs through mechanism 2 (P2) *and* it does not occur through mechanism 1 (1 - P1). This will work as long as there are only two mechanisms to produce the event, and as long as the two mechanisms are mutually exclusive. Matt
From: Randy Poe on 7 Mar 2005 14:15
SomeOne wrote: > Hi, > > I'm kinda stuck with this problem I'm having. Maybe you can help me. > > Let's say we have an event A. This event can occur through two > mechanisms. The probability that event A occurs through mechanism 1 is > P1=0.9. The probability that event A occurs through mechanism 2 is P2=0.8. What do these statements mean? Probability as a fraction of what? It's not at all clear what probability you're defining here. Also, calling it "event A" in both cases is confusing, since in the language of probability, an "event" gets assigned a single probability (the fraction of total outcomes in which that event occurs). So I can call your events A1 and A2, but I'm not sure their relation. > The question I'm struggling with is what is the probability that event A > will occur? > > Now, just adding probabilities doesn't seem like a good idea because > then we end up with a probability that is greater than 1, and that would > be ridiculous. Also, multiplying probabilities seems silly because then > you end up with a probability that is less than the original two. Those aren't good reasons. You add probabilities if the two events are mutually exclusive. If it was caused by 1, is it impossible that it was also caused by 2? If so, then the probabilities you've assiged make no sense. > So, after long pondering this problem, I figured that the probability > must be something like: > > P1+(1-P1)*P2 > > This looks right to me. Maybe it is, maybe it isn't. This is appropriate if the causes are independent: If the probability that it is caused by both 1 and 2 is P1*P2. The actual expression you are looking for is: P1 + P2 - P(1 & 2) where P(1 & 2) = probability of being caused by both 1 and 2. If they are independent, this is P1*P2. If they are mutually exclusive, it's 0. If they are dependent, it might be something else. - Randy |