mm-394 mm-394 Subject: Re: Uniqueness of gcd(a,b) = as + btSuppose a, b are positive integers with gcd(a,b) = 1. Then there exists > positive integers s and t, with 0 < s < b, such that as - bt = 1, and > in particular (as) mod b = 1. Show that s is unique.Any hints on this one?It's easy: s = 1/a (mod b). But recall inverses are unique: 1 = as = aS => s = s(aS) = (sa)S = SA slightly slicker unqueness proof uses two-way evaluation:Inverses Unique: --- s a S since over/underlined = 1 --- => s = S Law of Signs: ----------- a b + a(-b) + (-a)(-b) since over/underlined = 0 ---------------- by the distributive law => a b equals (-a)(-b)as excerpted from one of my (many) prior posts on this topichttp://google.com/groups?selm=y8ziss57jx5.fsf% 40nestle.ai.mit.edu===Subject: Re: Daryl Shawn Kabatoff - 100 Stupid Messages> Unfortunately, OE doesn't allow you to block threads initiated by specific> senders. So whilst I have DAR and James Harris blocked, I still get the> responses to their idiot posts. Which is even more reason not to feed the> trolls - I can block trolls, but not troll-feeders.> One of the best features of the Outlook Express or the Netscape 7.1> Newsgroup Readers -- is their ability to block out specific senders such> as:> DAR (aka Probabb...), even James Harris.> There is no way to stop idiots from posting - but rather than get> flustered> by it - just put the offenders on the BLOCKED list in you newsgroup> reader> !It is possible to block threads that contain certain combinations of words.Usually if crazy trolls like Mr. K are posting in a thread, the whole threadis dominated by their madness and negative responses to the same.You don't lose much by plonking a thread of this type. Pease===Subject: Re: Daryl Shawn Kabatoff - 100 Stupid Messages> It is possible to block threads that contain certain combinations of words.> Usually if crazy trolls like Mr. K are posting in a thread, the whole thread> is dominated by their madness and negative responses to the same.> You don't lose much by plonking a thread of this type.> PeaseGood advice. But it would have been lost to those who plonked this thread. ----http://www.crbond.com=== ===Subject: summation using matriciesI'm trying to sum from k=1:10, A(k)=k and t=0:3 with 1000 spacesA(k)sin(2*pi*k*t+pi/k) using matricies. I tried to do t as a column vectorwith length 1000 then multiplied k as row vector with a height of 10 so iget a 10*1000 matrix. Then I add pi/k which is a 10x1 matrix to each row sorow 1 of matrix pi/k is added to all of row one in the 10*1000 matrix row 2of pi/k is added to all of row 2 in the 10*1000 matrix etc. Then Imultiplied A(k) as a column vector with the 10*1000 matrix resulting in a1*1000 matrix. This doesn't seem to give the correct answer though. Anysuggestions on how to set up the matrix. I'm sorry if this is a littleconfusing.===Subject: Re: summation using matriciesContent-transfer-encoding: 8bitI'm trying to sum from k=1:10, A(k)=k and t=0:3 with 1000 spaces> A(k)sin(2*pi*k*t+pi/k) using matricies. I tried to do t as a column vector> with length 1000 then multiplied k as row vector with a height of 10 so i> get a 10*1000 matrix. Then I add pi/k which is a 10x1 matrix to each row so> row 1 of matrix pi/k is added to all of row one in the 10*1000 matrix row 2> of pi/k is added to all of row 2 in the 10*1000 matrix etc. Then I> multiplied A(k) as a column vector with the 10*1000 matrix resulting in a> 1*1000 matrix. This doesn't seem to give the correct answer though. Any> suggestions on how to set up the matrix. I'm sorry if this is a little> confusing.Since you seem happy with the size of your 1 x 1000 matrix, I guess youare asking for the 1000 sums: sum(k*sin(6*pi*k*t/1000 + pi/k),k = 1..10) for t = 1..1000.In any case, you seem to be trying sin(2*pi*k*t) + pi/k = sin(2*pi*k*t + pi/k) which is, of course, nottrue.-- ===Subject: Re: summation using matriciesWow I cant believe I missed that. Thank you very much for the help> I'm trying to sum from k=1:10, A(k)=k and t=0:3 with 1000 spaces> A(k)sin(2*pi*k*t+pi/k) using matricies. I tried to do t as a columnvector> with length 1000 then multiplied k as row vector with a height of 10 soi> get a 10*1000 matrix. Then I add pi/k which is a 10x1 matrix to each rowso> row 1 of matrix pi/k is added to all of row one in the 10*1000 matrixrow 2> of pi/k is added to all of row 2 in the 10*1000 matrix etc. Then I> multiplied A(k) as a column vector with the 10*1000 matrix resulting ina> 1*1000 matrix. This doesn't seem to give the correct answer though. Any> suggestions on how to set up the matrix. I'm sorry if this is a little> confusing.> Since you seem happy with the size of your 1 x 1000 matrix, I guess you> are asking for the 1000 sums:> sum(k*sin(6*pi*k*t/1000 + pi/k),k = 1..10)> for t = 1..1000.> In any case, you seem to be trying> sin(2*pi*k*t) + pi/k = sin(2*pi*k*t + pi/k) which is, of course, not> true.> --===Subject: Re: Map Colouring in RP2http://mathworld.wolfram.com/ProjectivePlane.hAt the bottom there is an example of a graph which requires six colours> It's a long time since I did this so I may have some of these details> wrong but here goes. Part of a course we did at college was> generalisations of the 4 colour map theorem. IIRC in the real> projective plane the fewest colours required to colour a map so that> no two countries which share a border are the same colour is six. We> were shown that six is enough and then we were set the problem of> finding a map that required six colours. I found a solution I liked. I> wondered if there were resources on this problem on the web? Is there> anywhere I can see maps in RP2 that require six colours?> >===Subject: Re: Planning to buy a calculus textbook!>I'm a senior in high school, and I'm taking AP calculus. The book I'm using>is by Larson, ISBN 0669327093, which is the 5th edition, and it sucks! I'm>looking for a better book (to buy for my own) to use, which would clearly>explains much wider types of problems. The one I'm using only shows about 3>examples; on the excercise, I have to sit there for hours trying to even>begin understanding how they even got the answer (answer to odd questions on>back). Usually the next day my teacher show us how that certain problem was>done AFTER he assigns the homework.Why not go over to the college bookstore, which is almost withinwalking distance for you. Look at the book they use. You can decidewhether you like it.===Subject: Re: Planning to buy a calculus textbook!I used james stuart's fourth edition book for calc I II and III. It's agreat textbook.message> I'm a senior in high school, and I'm taking AP calculus. The book I'm> using> is by Larson, ISBN 0669327093, which is the 5th edition, and it sucks!I'm> looking for a better book (to buy for my own) to use, which wouldclearly> explains much wider types of problems. The one I'm using only showsabout> 3> examples; on the excercise, I have to sit there for hours trying to even> begin understanding how they even got the answer (answer to oddquestions> on> back). Usually the next day my teacher show us how that certain problem> was> done AFTER he assigns the homework.> I have 2 editions of a book by James Stewart and it is one of the bestbooks> I've seen.> ===Subject: Probability Question?I'm having a hard time understanding something that undoubtedly has a simplesolution and am hoping someone here can point out the obvious to me. I assureyou this is not a homework problem (I'm 57 years old ;-)If the probability of one event (P1) is say 0.5, and the probability of anotherevent (P2) is 0.25, what is the probability of *either* event happening?According to the way I misunderstand the Countable Additivity ProbabilityAxiom at Wolfram's Mathworld(http://mathworld.wolfram.com/ CountableAdditivityProbabilityAxiom.h), theprobabilities are strictly additive, but this obviously cannot be true becauseif we had the probabilities P1=0.6 and P2=0.6, then P1+P2=1.2 which is greaterthan 1.Obviously the correct answer to my original question P1=0.5 and P2=0.25, liessomewhere between 0.5 and 1.0, but I am unsure how to compute it.===Subject: Re: Probability Question?> Obviously the correct answer to my original question P1=0.5 and P2=0.25,lies> somewhere between 0.5 and 1.0, but I am unsure how to compute it.If they are mutually exclusive, then there are three possibilities: P1happens, P2 happens, or NeitherOne happens (which in this case has aprobability of 1 - .5 - .25 = .25). If they are not mutually exclusive--ifthere is a chance that both can happen--then more info is needed. One pieceof information missing from the puzzle: How much do P1 and P2 overlap?Check out this venn diagram model athttp://stat-www.berkeley.edu/users/stark/Java/Venn.htmYou can get a good feel for the interplay of probabilities bymoving/resizing the parts.===Subject: Re: Probability Question?> ... If they are not mutually exclusive--if> there is a chance that both can happen--then more info is needed.> One piece of information missing from the puzzle:> How much do P1 and P2 overlap?> Check out this venn diagram model at> http://stat-www.berkeley.edu/users/stark/Java/Venn.htm> You can get a good feel for the interplay of probabilities by> moving/resizing the parts.diagram and did find it to be very helpful, but I am puzzled about one thing.You say that If they are not mutually exclusive--if there is a chance thatboth can happen--then more info is needed, and yet your venn diagram exampledoes indeed give the probability of two independent events happening.For example, if I set the probability of P(A) to 20% and P(B) to 20%, andarrange the two rectangles so they do not overlap (ie; so that they areindependent and either could happen with equal probability), your Java programtells me that there is a 40% chance of either of these two events happening.That seems to fly in the face of the other's assertions that the probability oftwo independent (NOT mutually exclusive) events happening cannot becalculated(?) Unfortunately the state space in the diagram is too small toallow me to set the probabilities of events A and B so that each is > 50%,because that is truly what I would like to see. I assume you did that purposelyso that the combined probability can never exceed 100%, but I think you willagree that it's certainly possible for two independent (not mutually exclusive)events to be possible in real life. For example, let's say there is a 80%chance that I will eat lunch today and there is a 90% chance that I will drinksomething today (two mutually exclusive events in which both can happen); BUT,it is also possible that I might fast today and neither eat nor drink, or doone but not the other. Surely it must be possible to calculate the odds of atleast one of these events happening?===Subject: Re: Probability Question?> ... If they are not mutually exclusive--if> there is a chance that both can happen--then more info is needed.> One piece of information missing from the puzzle:> How much do P1 and P2 overlap?> Check out this venn diagram model at> http://stat-www.berkeley.edu/users/stark/Java/Venn.htm> You can get a good feel for the interplay of probabilities by> moving/resizing the parts.> diagram and did find it to be very helpful, but I am puzzled about onething.> You say that If they are not mutually exclusive--if there is a chancethat> both can happen--then more info is needed, and yet your venn diagramexample> does indeed give the probability of two independent events happening.Maybe you are confusing two different concepts - independent events andmutually exclusive events?Events A and B are mutually exclusive if (A intersect B) = empty set. (I.e.the events cannot both happen)Events A and B are independant if P(A intersect B) = P(A) * P(B). This isharder to grasp, but here is an example: I roll two dice (a red and a greenone). The events (Red die shows 6) and (Green die shows 6) are independent: P(Red=6 intersect Green=6) = P(Red=6) * P(Green=6) = 1/6 * 1/6 = 1/36> For example, if I set the probability of P(A) to 20% and P(B) to 20%, and> arrange the two rectangles so they do not overlap (ie; so that they are> independent and either could happen with equal probability),> your Java programIf the rectangles do not overlap, the events are mutually exclusive. Theevents are certainly *not* independent.> tells me that there is a 40% chance of either of these two eventshappening.> That seems to fly in the face of the other's assertions that theprobability of> two independent (NOT mutually exclusive) events happening cannot be> calculated(?)I think others have commented on the probability of disjoint (mutuallyexclusive) events happening. (Not sure if anyone has discussedindependance)> Unfortunately the state space in the diagram is too small to> allow me to set the probabilities of events A and B so that each is > 50%,> because that is truly what I would like to see. I assume you did thatpurposely> so that the combined probability can never exceed 100%, but I think youwill> agree that it's certainly possible for two independent (not mutuallyexclusive)> events to be possible in real life. For example, let's say there is a 80%> chance that I will eat lunch today and there is a 90% chance that I willdrink> something today (two mutually exclusive events in which both can happen);BUT,> it is also possible that I might fast today and neither eat nor drink, ordo> one but not the other. Surely it must be possible to calculate the odds ofat> least one of these events happening?Maybe this is a poor example, because in real life these events are notreally independent, e.g. if you're struck by lightening you probably won'tbe doing much driving after that to end up in an accident!But anyway, I see what you're getting at. If we know two events A and B areindependent, then we can indeed work out the probability of either eventhappening (including the case of both occuring). With your example, A =eat lunch today and B = drink today, and you want P(A union B) = P(A) + P(B) - P(A intersect B)this is just the formula you have seen already, and applies for *any* eventsA,B. Additionally, if A and B are independent, then we know the last termand the equation becomes P(A union B) = P(A) + P(B) - P(A) * P(B) = 0.8 + 0.9 - (0.8 * 0.9) = 0.98--===Subject: Re: Probability Question?> Maybe you are confusing two different concepts - independent events and> mutually exclusive events?No , I assure you that the difference between the two has been pounded intomy head quite throughly by everyone who has participated in this thread. Infact I would be quite willing to shout from the rooftops that I am interestedonly in NON MUTUALLY EXCLUSIVE EVENTS - NO OTHERS NEED APPLY! ;-) In fact atthis point I should probably just abandon this thread and start another oneelsewhere that is worded more precisely. I suppose it's probably due either tothe fact that I ERRONOUSLY linked to a Mathworld page pertaining to mutuallyexclusive events that got this thread off on the wrong foot, or perhaps becausethere is no way to deal mathematically with NON Mutually Exclusive Events thatno one wishes to discuss NON Mutually Exclusive Events.> If the rectangles do not overlap, the events are mutually exclusive. The> events are certainly *not* independent.Ok, if that's the case then I misinterpreted AngleWrym's venn diagrams and theydo not apply to the case in which I'm interested.> But anyway, I see what you're getting at. If we know two events A and B are> independent, then we can indeed work out the probability of either event> happening (including the case of both occuring). With your example, A => eat lunch today and B = drink today, and you want> P(A union B) = P(A) + P(B) - P(A intersect B)AH yes, and that last term is the crux of my dilemma, because for events thatare NOT mutually exclusive there is NO WAY of knowing the combined probabilityof both occurring simultaneously because the events do not necessarilyintersect.At least that explains why I was unable to find any discussion of these typesof events in any of my probability or statistics books.===Subject: Re: Probability Question?>I'm having a hard time understanding something that undoubtedly has a simple>solution and am hoping someone here can point out the obvious to me. I assure>you this is not a homework problem (I'm 57 years old ;-)I don't know what your age has to do with it. People any age can be students. But I'll accept your stipulation that this isn't homework.>If the probability of one event (P1) is say 0.5, and the probability of another>event (P2) is 0.25, what is the probability of *either* event happening?The question as stated cannot be answered, because you don't know the probability of P1 and P2 both happening. If they are mutually exclusive, i.e. if they can't both happen, then the probability of P1 or P2 is 0.5+0.25 = 0.75.But if P1 and P2 could happen together, then P(P1 or P2) = P(P1) + P(P2) - p(P1 and P2)You might find my Probability Summary at http://www.acad.sunytccc.edu/instruct/sbrown/stat/ probsumm.htmhelpful, particularly under mutually exclusive events.>According to the way I misunderstand the Countable Additivity Probability>Axiom at Wolfram's Mathworld>(http://mathworld.wolfram.com/ CountableAdditivityProbabilityAxiom.h), the>probabilities are strictly additive, but this obviously cannot be true because>if we had the probabilities P1=0.6 and P2=0.6, then P1+P2=1.2 which is greater>than 1.Mathworld is correct, _if_ the events are mutually exclusive. A synonym for ME events is disjoint events, and now that you know to look for it you see that the Mathworld page states the formula is for disjoint events.>Obviously the correct answer to my original question P1=0.5 and P2=0.25, lies>somewhere between 0.5 and 1.0, but I am unsure how to compute it.Without more information about the probability of P1 and P2 happening simultaneously, the question can't be answered.-- Address munging may or may not reduce the spam you get; it surelyreduces the number of useful answers you get. http://www.cs.tut.fi/~jkorpela/usenet/laws.h===Subject: Re: Probability Question?Oops, I spoke too soon because I'm confused again. I must have misunderstoodsomething. To take another example, if I had three coins, c1, c2, and c3, thenthe probability of tossing a Head on each of the three coins would be:P(c1) = 0.5P(c2) = 0.5P(c3) = 0.5According to your previous equation (if I understand it correctly), theprobability of independently getting Heads on ANY of the three coins would be:P(c1 or c2 or c3) = P(c1)+P(c2)+P(c3) - p(c1 and c2 and c3)P(c1)+P(c2)+P(c3) = 0.5+0.5+0.5 = 1.5p(c1 and c2 and c3) = 0.5*0.5*0.5 = 0.125P(c1 or c2 or c3) = 1.5 - 0.125 = 1.375What have I done wrong? That implies that the probability of getting Heads onat least one of the coins is greater than an absolute certainty, and yet it'sobviously possible to get tails on all three coins. In fact (as I mentioned inmy previous reply to ), it seems to me the correct answer would be 0.875because the probability of getting tails on all three coins is (0.5)^3, so theprobability of NOT getting tails on all three coins (ie; of getting at leastone Head on one of them) would be 1.0 - (0.5)^3 = 0.875 ???===Subject: Re: Probability Question? > Oops, I spoke too soon because I'm confused again. I must have misunderstood> something. To take another example, if I had three coins, c1, c2, and c3, then> the probability of tossing a Head on each of the three coins would be:P(c1) = 0.5> P(c2) = 0.5> P(c3) = 0.5Here is where you are going astray. There are not just two possible outcomes but 8: HHH, HHT, HTT, etc.It is true that P(Hxx) = P(xHx) = P(xxH) = 1/2.Using your notation: C1 is the event that coin 1 comes up H, C2 and C3siimilarly.P(C1 or C2 or C3) =P(C1) + P(C2) + P(C3) - P(C1 and C2) - P(C1 and C3) - P(C2 and C3) + P(C1 and C2 and C3) =1/2 + 1/2 + 1/2 - 1/4 - 1/4 - 1/4 + 1/8 = 7/8.You can verify that this is correct by noting that all but one of the 8possible outcomes (TTT) has at least one H.This all follows from the Inclusion/Exclusion principle: For a set A,let |A| be the number of elements in A, A / B are the elements ineither A or B or both; A / B is the elements in both A and B.Then |A / B| = |A| + |B| - |A / B|The things in A / B get counted twice, once in |A| and once in |B|.Likewise, for a third set C,|A / B / C| = |A| + |B| + |C| - |A / B| - |A / C| - |B / C| + |A/ B / C|For your coin problem, you have your outcome space:O = {HHH, HHT, HTH, HTT, THT, TTH, THH, TTT}C1 = {HHH, HHT, HTH, HTT} and so on. So P(C1) = |C1|/|O| = 1/2.C1 / C2 = {HHT, HHH} so P(C1 and C2) = |C1 / C2|/|O| = 1/4 etc.HTH. [...]-- ===Subject: Re: Probability Question?Ah, I see now how I generalized Stan's equation incorrectly. As I nowunderstand it, the point behind the equation you and Stan posted is to subtractout the intersection of each pair of possible outcomes that are encompassedwithin the additive sum (the inclusive OR) of the three possible events P(C1)+ P(C2) + P(C3) in order to get the contribution of each event to the totalprobability(?)For example, I assume that the probability contributed by event C1 alone is notsimply P(C1) but is actually = P(C1) - P(C2 and C3). Or in other words:P(C1) alone = P(C1) - P(C2 and C3) = 1/2 - 1/4 = 1/4P(C2) alone = P(C2) - P(C1 and C3) = 1/2 - 1/4 = 1/4P(C3) alone = P(C3) - P(C1 and C2) = 1/2 - 1/4 = 1/4P(C1/C2/C3) together = P(C1 and C2 and C3) = (1/2)^3 = 1/8Add these up and you get the probability of ANY or ALL of the three events(C1,C2,C3) occurring, and this probability will always be less than or equal to1.That's wonderful - thank you , I think(?) I now understand the concept Stanand (and you of course) were getting at. Unfortunately, I think I alsounderstand why Stan said that my original problem statement had no solution.I seem to recall reading somewhere in one of my old college physics books thatall physical problems have real physical solutions (although those solutionsmay not necessarily be easy to determine of course). That's why I have a realproblem accepting the concept that the probability of either or both of twoindependent events occurring is unknowable.Let's say for example that I attract lightening like a lightening rod, and I aman *extremely* poor driver, so that the probability that I will get struck bylightening is 0.6 and the probability of my being involved in an automobilegather, since these are independent and NOT mutually exclusive events (thepossibility of getting struck by lightening while being involved in a caraccident is admittedly pretty remote, but is nonetheless possible), it isimpossible to know what the combined probability is of having either one orboth of those events happen to me.Is that correct? Is such a probability truly unknowable?P.S.Sorry for using an X in my username in my initial posting. It was somethingleft over from when I was fiddling around with my account settings awhile back.===Subject: Re: Probability Question?> But if P1 and P2 could happen together, then> P(P1 or P2) = P(P1) + P(P2) - p(P1 and P2)So for example I had a two sided coin and a six sided die, and P1= probabilityof tossing a Head on the coin, and P2 = probability of rolling say a 6 on thedie, then the probability of either tossing a Head on the coin OR rolling a sixon the die would be:P1= 1/2P2= 1/6P(P1) + P(P2)= 1/2 + 1/6 = 2/3P1 and P2= (1/2)*(1/6)= 1/12P(P1 OR P2) = 2/3 - 1/12 = 7/12Is that correct?If so, that's exactly the type of answer I was looking for.===Subject: Re: Probability Question?> I'm having a hard time understanding something that undoubtedly has asimple> solution and am hoping someone here can point out the obvious to me. Iassure> you this is not a homework problem (I'm 57 years old ;-)> If the probability of one event (P1) is say 0.5, and the probability ofanother> event (P2) is 0.25, what is the probability of *either* event happening?> According to the way I misunderstand the Countable Additivity Probability> Axiom at Wolfram's Mathworld> (http://mathworld.wolfram.com/ CountableAdditivityProbabilityAxiom.h),the> probabilities are strictly additive, but this obviously cannot be truebecause> if we had the probabilities P1=0.6 and P2=0.6, then P1+P2=1.2 which isgreater> than 1.> Obviously the correct answer to my original question P1=0.5 and P2=0.25,lies> somewhere between 0.5 and 1.0, but I am unsure how to compute it.I am no expert in probability, but I believe I understand the jist of yourconcern and hopefully can address it to your satisfaction. If anything hereis not correct, surely someone will correct me.Apparently you are wondering how this can make sense if the problem wasdifferent, as in P(E)=.6 and P(F)=.6. We would have P(E/F)=.6+.6=1.2. Thekey is understanding that MUTUALLY EXCLUSIVE events E and F, taken from thesame sample space, can't sum to >1, however events that are not mutuallyexclusive can. The formula you refer to is for mutually exclusive eventsonly. If P(E)=.6, then P(not E)=.4, of which all other mutually excusiveevents *sum to*. If P(F) is >.4, then E and F can't be mutually exclusiveevents, and the formula simply does not apply.By mutually exclusive we mean the intersection of E and F (written E/F)is the null set, ie they share no common element. In the mathworlddefinition you cited, the key word there is disjoint. Perhaps a simpleexample will clarify. The sample space for rolling a die is:S = {1,2,3,4,5,6}let E be the event that a 4 is rolled.let F be the event that a 7 is rolled.E = {4}F = {6}P(E) = 1/6P(F) = 1/6{4} / {6} = null.The events are mutually exclusive, and so the probability of E OR Foccurring is given by:P(E / F) = P(E) + P(F) = 1/6 + 1/6 = 1/3.Let E be the event that a 5 is rolled.Let F be the event that an odd number is rolled.E = {5}F = {1,3,5}P(E) = 1/6P(F) = 3/6 = 1/2{5} / {1,3,5} = {5}We can't say the probability of E or F occurring is simply 1/6+1/2, becauseE and F are not mutually exclusive events. In this case, since 5 isincluded in the set of odd numbers, then P(E/F) is the greater of P(E) andP(F). 1/2 in this case. it becomes quite clear when you think about it;the probability of rolling a 5 OR an odd number, is the same as rolling anodd number, in this case 1/2.Of course, neither are E and F mutually exclusive is P(E)=.6 and P(F)=.6.Can't be; these probabilities sum to >1. The sets E and F are therefore notdisjoint (they share at least one common element).-- ===Subject: Re: Probability Question?> let E be the event that a 4 is rolled.> let F be the event that a 7 is rolled.oops, that should have been let F be the event that a 6 is rolled.-- ===Subject: Re: Probability Question?Thank you . I now understand why I misunderstood the formula atMathworld, but I don't think I've made myself clear about the type of problemI'm trying to solve. I'm looking for a formula that gives me the probability ofmutually *INDEPENDENT* and *inclusive* events happening.If I understand you correctly, you said that events that are NOT mutuallyexclusive can have a probability > 1. I'm sorry , but I have a realhard time believing that anything can have a probability greater than certainty(ie; probability = 1). I have no idea what that would mean in a physical sense.For example, lets say I have three independent coins. It's clear that theprobability that at least one of them will be tails is 0.875 because it's thesame as the probability that not all of them are heads (ie; 1.0 - (0.5)^3 =0.875). If one were to assume that events that are NOT mutually exclusive (suchas the tossing of these three coins) can be > 1, one might be led to believethat it could be computed as: 0.5 + 0.5 + 0.5 = 1.5, which is clearlyincorrect.Let me use some simple sample data from the actual problem I'm working on. I'mtrying to compute the probability that the carries in some boolean (modulo 2)equations will be = 1. The first 2 of the 30,624 equations are:c1:=a2*b2 mod 2;c2:=((a3+a2+b3)*c1+1+(a3*a2+1)*(a3*b3+1)*(a2*b3+1)) mod 2;The probability that any of the a's or b's = 1, is 0.5, so I can simplifythusly:c1:=0.5*0.5 mod 2;c1:= 0.25; (ie; The probability that c1=1 is 0.25)And for the second equation:c2:=((0.5+0.5+0.5)*0.25 + 1 + (0.5*0.5+1)*(0.5*0.5+1)*(0.5*0.5+1)) mod 2;c2:=(0.375 + 1 + 1.953125) mod 2;c2:=(0.375 + 2 + 0.953125) mod 2;Since (2 mod 2 = 0), I can say:c2:= 0.375 + 0.953125;But since 0.375 and 0.953125 are independent and NOT mutually exclusive*probabilities* rather than just plain numbers, it is my understanding that Icannot simply add them, and that the true probability that c2=1 is somewherebetween 0.953125 and 1.0, right?===Subject: Re: Probability Question?<...If I understand you correctly, you said that events that are NOT mutually> exclusive can have a probability > 1.No, I said that the *sum* of the (probabilities of) events that are notmutually exclusive can be >1, not the individual probabilities themselves.Example (back to our standard die):E = 2,3,4,5, or 6 is rolledF = 2 or 3 is rolledP(E) = 5/6P(F) = 1/3P(E) + P(F) = 7/6The point was, 7/6 is NOT P(E/F), ie the probability of either E or F asyou thought the formula was trying to tell you despite your common sense tothe contrary. Events E and F are not mutually exclusive, thus the formulain queston simply does not apply. This directly addressed your concern, orso I hoped.... The fact remains, however, that P(E)+P(F)>1, as useless afact as that may or may not be it's still a demonstrable fact....<...Let me use some simple sample data from the actual problem I'm working on.I'm> trying to compute the probability that the carries in some boolean (modulo2)> equations will be = 1. The first 2 of the 30,624 equations are:<...>I'll leave further comments for those more comfortable with your veryspecific question. I hope I have at least satisifed your original question;that of the axiom you cited. Like I said I'm no probability expert so I'llleave the boolean stuff to others...-- > c1:=a2*b2 mod 2;> c2:=((a3+a2+b3)*c1+1+(a3*a2+1)*(a3*b3+1)*(a2*b3+1)) mod 2;> The probability that any of the a's or b's = 1, is 0.5, so I can simplify> thusly:> c1:=0.5*0.5 mod 2;> c1:= 0.25; (ie; The probability that c1=1 is 0.25)> And for the second equation:> c2:=((0.5+0.5+0.5)*0.25 + 1 + (0.5*0.5+1)*(0.5*0.5+1)*(0.5*0.5+1)) mod 2;> c2:=(0.375 + 1 + 1.953125) mod 2;> c2:=(0.375 + 2 + 0.953125) mod 2;> Since (2 mod 2 = 0), I can say:> c2:= 0.375 + 0.953125;> But since 0.375 and 0.953125 are independent and NOT mutually exclusive> *probabilities* rather than just plain numbers, it is my understandingthat I> cannot simply add them, and that the true probability that c2=1 issomewhere> between 0.953125 and 1.0, right?===Subject: Re: Probability Question?> ... The fact remains, however, that P(E)+P(F)>1, as useless a> fact as that may or may not be it's still a demonstrable fact....Ok, I'm not arguing with you , it's just that it appeared to be aphysical impossibility. I now realize that you were pointing it out as amathematical artifact without trying to attribute any physical meaning to it.> I'll leave further comments for those more comfortable with your very> specific question. I hope I have at least satisifed your original question;> that of the axiom you cited.Yes you did , and thanks very much for your help in clearing up mymisunderstanding due to my own failure to notice the disjoint events sentencein the page I cited. :-)===Subject: Linear Algebra questionI have a question I hope someone can help me with. How do you show that theset of 2 x 2 matrices with real coefficients form a linear space over R withdimension 4?===Subject: Re: Linear Algebra questionContent-transfer-encoding: 8bitI have a question I hope someone can help me with. How do you show that the> set of 2 x 2 matrices with real coefficients form a linear space over R with> dimension 4?> Tom Risager's response is certainly correct and probably what you wereintended to do.However, there is a shortcut if you have enough background.There is a natural one-to-one onto function, f, from R^4 to the 2 x 2'ssuch that f(u + v) = f(u) + f(v) and f(a*v) = a*f(v). It followsimmediately that, since R^4 is a v.s. of dim 4, so are the 2 x 2's.-- ===Subject: Re: Linear Algebra questionNeat. Is it always true that if a linear transform is onto and one-to-onethen it maps between spaces of the same dimension?Tom> I have a question I hope someone can help me with. How do you show thatthe> set of 2 x 2 matrices with real coefficients form a linear space over Rwith> dimension 4?> Tom Risager's response is certainly correct and probably what you were> intended to do.> However, there is a shortcut if you have enough background.> There is a natural one-to-one onto function, f, from R^4 to the 2 x 2's> such that f(u + v) = f(u) + f(v) and f(a*v) = a*f(v). It follows> immediately that, since R^4 is a v.s. of dim 4, so are the 2 x 2's.> -- ===Subject: Re: Linear Algebra questionContent-transfer-encoding: 8bit> I have a question I hope someone can help me with. How do you show that> the> set of 2 x 2 matrices with real coefficients form a linear space over R> with> dimension 4?>> Tom Risager's response is certainly correct and probably what you were> intended to do.> However, there is a shortcut if you have enough background.> There is a natural one-to-one onto function, f, from R^4 to the 2 x 2's> such that f(u + v) = f(u) + f(v) and f(a*v) = a*f(v). It follows> immediately that, since R^4 is a v.s. of dim 4, so are the 2 x 2's. > Neat. Is it always true that if a linear transform is onto and one-to-one> then it maps between spaces of the same dimension?> Tom> Yes. The image of a basis is a basis.But note that you can't have a linear transformation until you haveboth the domain and codomain vector spaces. The real thrust of mysuggestion was that you can avoid laboriously (and trivially) verifyingall those axioms for the 2 x 2's by using f and the fact that R^4satisfies them.-- ===Subject: Re: Linear Algebra questionI'm trying to learn this stuff too - so I'm clearly not an expert - but myunderstanding is you need to show that all of the following hold for all 2x2matrices a and b with real coefficients:1) if u and v are 2x2 matrices, then so is u + v2) u + v = v + u3) (u + v) + w = u + (v + w)4) there is a 2x2 matrix 0 such that u + 0 = u5) there is a 2x2 matrix -u such that u + (-u) = 06) ku is also a 2x2 matrix if k is a scalar7) k(u + v) = ku + kv8) (k + l)u = ku + lu, k and l are scalars9) k(lu) = (kl)u10 1u = uThat establishes the set of 2x2 matrices as a linear space. To show thatthey have dimension 4 you need to find a basis for your space and count theelements. In this case you could take {[1, 0; 0 , 0], [0, 1; 0, 0], [0, 0;1, 0], [0, 0; 0, 1]} as your basis.Tom> I have a question I hope someone can help me with. How do you show thatthe> set of 2 x 2 matrices with real coefficients form a linear space over Rwith> dimension 4?> ===Subject: Re: Linear Algebra questionthats to show for ALL 2 x 2 matrices.===Subject: Diff of a Quot: HELP!!!!!!!Goin MAD!!!!!!!!hi guys,how do i diff this?(ln2t)/(sqrt(t))to this.....(1-(1/2)ln2t)/(sqrt(t^3))done previous 7 examples not too many problems, just seem to be missingsomething on this. i've used the quotient rule and the standard derivativelnax= 1/xbest ===Subject: Re: Diff of a Quot: HELP!!!!!!!Goin MAD!!!!!!!!thanx guys, sussed it now, can go to bed!!!!===Subject: Re: Diff of a Quot: HELP!!!!!!!Goin MAD!!!!!!!!hi guys,thought i sussed it, but can't get from sqrt(t)/t - ln(2t)/(2*sqrt(t)).to [1 - ln(2t)/2]/sqrt(t), or [1 - (1/2)*ln(2t)]/sqrt(t).i know that sqrt(t)/t=1/sqrt(t).....maybe i'm tired.....but i just don't see it....the rest i'm positive i know butcan't suss this step===Subject: Re: Diff of a Quot: HELP!!!!!!!Goin MAD!!!!!!!!> hi guys,> how do i diff this?> (ln2t)/(sqrt(t))> to this.....> (1-(1/2)ln2t)/(sqrt(t^3))> done previous 7 examples not too many problems, just seem to be missing> something on this. i've used the quotient rule and the standard derivative> lnax= 1/x> best > I think for this, I'd use logarithmic differentiation (I hate the quotientrule!). If you don't know how to do this, it works like this...Take the log of both sidesln y=ln(ln2t)-ln(t^1/2))(1/y)(dy/dx)=(1/t)((1/t)/ln(2t))-1/2lnt(1 /y)(dy/dx)=(1/t^2)/ln2t-1/(2t)dy/dx=(ln2t/sqrt(t))((1/t^2)/( ln2t)-1/(2t)))Please correct me if my differentiation is wrong.===Subject: Re: Diff of a Quot: HELP!!!!!!!Goin MAD!!!!!!!!alt.math.undergrad: > hi guys,> how do i diff this?> (ln2t)/(sqrt(t))> to this.....> (1-(1/2)ln2t)/(sqrt(t^3))Just plugging into the quotient rule gives: [sqrt(t)*(ln(2t))' - ln(2t)*(sqrt(t))']/(sqrt(t))^2The denominator simplifies to t, so we'll worry about the numerator. (ln(2t))' = 1/t, so the first term is sqrt(t)/t.(sqrt(t))' = (t^(1/2))' = (1/2)t^(-1/2) = 1/(2*sqrt(t)), so the whole numerator is sqrt(t)/t - ln(2t)/(2*sqrt(t)).But sqrt(t)/t = 1/sqrt(t), so the numerator can be simplified to [1 - ln(2t)/2]/sqrt(t), or [1 - (1/2)*ln(2t)]/sqrt(t).Put the denominator back in to get [1 - (1/2)*ln(2t)]/[t*sqrt(t)].Finally, t*sqrt(t) = t * t^(1/2) = t^(3/2) = sqrt(t^3), and you're done.===Subject: Tipped ellipse equationG'day all, I NEED a generic equation that will describe the following:Given Point A on Y axis at (0,6) for exampleGiven Point B on X axis at (14,0) for exampleand Point C on Y axis at (0,3) Points A and B are foci and C is a point on the ellipseWhat is the equation for that ellipse?I have come at this problem from two ways : each leaving me stuck1. using x^2/a^2 + y^2/b^2 = 1 , solving for y, and then literally manually tweaking the result to create the curve that is close. My equation for this direction is:y=9-sqrt((1-(x-7.2)^2/(17.317821063/2)^2)*(8.2405659017/2)^ 2)-((6/14)*x)but this is only gotten by MUCH tweaking. But it is REAL Close! :)2. using good olde pythagorus, I determine what the length of the 'string' is an call it k, then for a point P(x,y) :k = sqrt( (x-A1)^2 + (y-A2)^2 ) + sqrt( (x-B1)^2 + (y-B2)^2 )where point A( A1, A2 ) and point B( B1, B2 ). with this I visited my local community college and begged a few moments of Maple time to solve this equation for y, and got 1 full page if we use k as is, and 12 pages if we use the equation to get k in the first place. Clearly not the direction I wanted to go.This is from a real world question that involves a string that runs from A to B where an ellipse is drawn that includes c. We need the equation to specifically get the Y value for all points of X from 0 to just past (14,0).Al 'Z'===Subject: Re: Tipped ellipse equation> Given Point A on Y axis at (0,6) for example> Given Point B on X axis at (14,0) for example> and Point C on Y axis at (0,3) > Points A and B are foci and C is a point on the ellipse> What is the equation for that ellipse?if P is on the ellipse thend(A, P) + d(B, P) = d(A, C) + d(B, C)so your equation is sqrt(x^2 + (y-6)^2) + sqrt((x-14)^2 + y^2) = 3 + sqrt(205)which is not much fun and can't be simplified a great deal afaik, but you are welcome to try.-- If this message helped you, consider buying an itemfrom my wish list: ===Subject: HELPPPP!!!!!!!!!!!!how to get from this to thishi guys,someone helped me last night with how to solve a diff quot problem, it reallyhelped but this part i don't get: sqrt(t)/t - ln(2t)/(2*sqrt(t)).to this.. i know that sqrt(t)/t=1/sqrt(t) [1 - (1/2)*ln(2t)]/sqrt(t).best ===Subject: Re: HELPPPP!!!!!!!!!!!!how to get from this to this> hi guys,> someone helped me last night with how to solve a diff quot problem, it really> helped but this part i don't get: sqrt(t)/t - ln(2t)/(2*sqrt(t)).to this.. i know that sqrt(t)/t=1/sqrt(t) sqrt(t) ln(2t) -------- - ----------- t 2 sqrt(t) 1 ln(2t) -------- - --------- sqrt(t) 2 sqrt(t) ^^^^you said you know thatNow just factor out the common bit 1 / 1 ------- | 1 - --- ln(2t) | sqrt(t) 2 /> [1 - (1/2)*ln(2t)]/sqrt(t).-- ===Subject: Re: Probability Question?----- Original Message ----- X I'm having a hard time understanding something that undoubtedly has asimple> solution and am hoping someone here can point out the obvious to me. Iassure> you this is not a homework problem (I'm 57 years old ;-)> If the probability of one event (P1) is say 0.5, and the probability ofanother> event (P2) is 0.25, what is the probability of *either* event happening?> According to the way I misunderstand the Countable Additivity Probability> Axiom at Wolfram's Mathworld> (http://mathworld.wolfram.com/ CountableAdditivityProbabilityAxiom.h),the> probabilities are strictly additive, but this obviously cannot be truebecause> if we had the probabilities P1=0.6 and P2=0.6, then P1+P2=1.2 which isgreater> than 1.> Obviously the correct answer to my original question P1=0.5 and P2=0.25,lies> somewhere between 0.5 and 1.0, but I am unsure how to compute it.> You must know the probability that both events occur simultaneously. Theprobability( that either occurs) is the probability( that first occurs)+probability( that second occurs) - probability( that both occursimultaneously)The website you referenced appears to apply only to disjoint events( i.e.eventswhich cannot occur simultaneously.)Hope this helps.===Subject: Re: Probability Question?> You must know the probability that both events occur simultaneously. The> probability( that either occurs) is the probability( that first occurs)+> probability( that second occurs) - probability( that both occur> simultaneously)> Hope this helps.> Yes, thank you Rob. Actually though, I'm interested in calculating theprobability that EITHER or BOTH events occur. The website I cited turns out tobe irrelevant to my question, and I should have omitted it.===Subject: Subject: Probability Question?----- Original Message ----- X I'm having a hard time understanding something that undoubtedly has asimple> solution and am hoping someone here can point out the obvious to me. Iassure> you this is not a homework problem (I'm 57 years old ;-)> If the probability of one event (P1) is say 0.5, and the probability ofanother> event (P2) is 0.25, what is the probability of *either* event happening?> According to the way I misunderstand the Countable Additivity Probability> Axiom at Wolfram's Mathworld> (http://mathworld.wolfram.com/ CountableAdditivityProbabilityAxiom.h),the> probabilities are strictly additive, but this obviously cannot be truebecause> if we had the probabilities P1=0.6 and P2=0.6, then P1+P2=1.2 which isgreater> than 1.> Obviously the correct answer to my original question P1=0.5 and P2=0.25,lies> somewhere between 0.5 and 1.0, but I am unsure how to compute it.> You must know the probability that both events occur simultaneously. Theprobability( that either occurs) is the probability( that first occurs)+probability( that second occurs) - probability( that both occursimultaneously)The website you referenced appears to apply only to disjoint events( i.e.eventswhich cannot occur simultaneously.)Hope this helps.===Subject: Surface area I have tried to calculate the surface area of the curve defined by thefollowing parametric equations: x = 5 cos ( t ) andy = 3 sin ( t ). I am not looking for the answer, just a hint. Every timeI try to reduce the equation to something that can be integrated, I end upat the same place I started.Once again, any hints would be appreciated.Gord.===Subject: Re: Surface area> I have tried to calculate the surface area of the curve defined by the> following parametric equations: x = 5 cos ( t ) and> y = 3 sin ( t ). A curve doesn't have a surface area. A surface has a surface area, a curve has length. What are you really asking?-- If this message helped you, consider buying an itemfrom my wish list: ===Subject: Re: Surface area> I have tried to calculate the surface area of the curve defined by the> following parametric equations: x = 5 cos ( t ) and> y = 3 sin ( t ).> A curve doesn't have a surface area. A surface has a surface area, a> curve has length. What are you really asking?Maybe he wants the area of the (planar) region bounded by the curve. If so:The curve is an ellipse with semiaxes a = 5 and b = 3. The area would thenbe Pi*a*b = 15*Pi.David===Subject: using diff func of a func..get from this to this???!!!!hi guys,came across this one in book i'm using, its in the diff func of a func sectionso no cheating using the standard derivative e^ax=ae^ax. how do i wtite this instandard form so i can use func of a func?? knowing my luck will come up onexam, so a solution would be appreciated.5e^2t+1 to 10e^2t+1best ===Subject: Re: using diff func of a func..get from this to this???!!!!> hi guys,> came across this one in book i'm using, its in the diff func of a funcsection> so no cheating using the standard derivative e^ax=ae^ax. how do i wtitethis in> standard form so i can use func of a func?? knowing my luck will come upon> exam, so a solution would be appreciated.> 5e^2t+1 to 10e^2t+1> best > I'll rewrite it as y=(5e^2t+1)^(10e^2t+1)First take the log of both sides and use the Power Rule.ln y=(10e^2t+1)ln(5e^2t+1)Now differentiate implicitly(1/y)(dy/dx)=(10e^2t+1)(10e^2t/(5e^2t+1))+(10e^2t)ln (5e^2t+1)dy/dx=(5e^2t+1)^(10e^2t+1)((10e^2t+1)(10e^2t/(5e^2t+1 ))+(10e^2t)ln(5e^2t+1)