alexTextboxOK, this is totally what I want to hear.

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alexTextboxRight.

alexCallouteach

alexCalloutNote that to calculate this, you'd first reduce 6/36 to 1/6.

alexTextboxI see. This helps with the place of the counting techniques surveyed in Bersektas.

alexTextboxSo, this is obviously very easy.

alexCalloutOK, so this analogy of filling slots, certainly makes sense of/justifies this, as giving the number of permutations.(But review the previous page, if the justification loses its clarity. The justification becomes clear, when one thinks of the tree. First selection defines the number of branches in the first round; then the second is the number of sub-branches on each of these; etc. That's why you MULTIPLY the numbers together).

alexCalloutMmm. Again, this analogy helps a lot.

alexTextboxOne feels it has been treated far more simply, than in Bersektas.

alexTextboxAgain, it feels that the discussion has been infinitely more simple and smooth.

Although possibly we were tackling some more complicated cases?

alexCalloutAs per what has been just worked out.

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alexCalloutThis language is less meaningful to me, but I think it will come up quite a bit.

alexTextboxSo, this is just plugging into the formula.

alexTextboxNote that this certainly works out by the formula, where in the denominator you multiply r! by (n-r)! For then your numbers will just swap places, in terms of the variable that represents them. Instead of 'r' being 4, it will be 13. Then, 'n-r' will be 4, instead of 13.

alexCalloutThis explanation doesn't make much sense to me, although if one refers to the table, the first column of which is the List of Committees, one gets some sense of what is meant.

alexTextboxRecall that 0! = 1.

alexCalloutQuite interesting, to meet a case where we actually see the mathematical arbitrary definition in action.

alexCalloutMmm, yes indeed.

alexCalloutYes, OK.

alexCalloutFirst, note that it is actually OBVIOUS that the answer is 5/52. There is one Qs, so the chances of picking it from the pack are 1/52. But you get 5 picks in a hand.

alexCalloutThat is, combinations of cards, which also include the Qs.

alexTextboxThe key is that they're thinking of the different 5-card combinations that exist, which include a Qs. If you then divide this by the total number of hands (5-card combinations), you've got the probability.

alexCalloutNote that "favorable" here means NOT getting a Qs.

alexTextboxThe manipulations here are pretty easy.

alexTextboxYep

alexCalloutSo, this is the number of possible COMBINATIONS of all-heart hands (or indeed, 5 cards all of any single suit).

alexCalloutRecall (page 19) the analogy of filling 5 slots, where initially we choose from a pool of 13, then from 12, etc.

alexTextboxSo, note this. Although the answer may turn out the same, the fractions in (6) and (7) are not the same. The numbers in (7) are bigger. We are dividing a bigger numerator by a bigger denominator - so, in the end, the ratio is the same.

alexCalloutThat you should multiply these individual combinations together, is not obvious.If thinking of filling slot, the point is that for one of the ways of filling slot 1, you have, then, a whole number of ways that filling slot 2 can be paired with this. Similarly for the next way of filling the slot.Or, on the tree representation - each of the branches, has a number of sub-branches corresponding to, you know, the sumber of possibilities at this second level.

alexTextboxSo, what I said about multiplication in the note, applies identically here.

alexCalloutI guess one way to appreciate this, is to see it as the first splitting of the tree.

alexTextboxSo, there is nothing going on here that is different from the other cases.

alexTextboxNot too sure about this.

alexCalloutThis is the vowel. I suppose choosing 1 out of 5 reduces to this.

alexCalloutOK, so basically, I think this term is maybe indicating that we can arrange each of our chosen elements in a different order. For EACH set of seven elements we choose, you know, we can arrange them in different slots, and there are 7! options each time.

alexCallout26 letters and 10 digits. And (page 19) this is the way you count a permutation of, you know, n elements from among a larger number.

alexTextboxSo, maybe an illuminating summary description of this, is to say that the permutations are occurring WITHN the combination(s).I must say that I feel I'd have no chance of working this out on my own. (See below).

alexTextbox....OK, so this at least gives us some insight into the sort of method we might use to solve such problems. One follows something of a script.

alexTextboxI think that I can appreciate that I might develop an "eye" for spotting things like this more easily.

alexCalloutBoiling things down into elementary situations like this, helps.

alexTextboxYes, clear.

alexCalloutSo, it IS the case that there is a different overall probability, when you are dealing with- and without replacement.

alexTextboxOK.

alexCalloutWhat is a bit difficult to see, is why it is called specifically "symmetry", as opposed to, say, "equivalence". I suppose that maybe because you have a situation of "unchanged/indistinguishable after transformation".

alexTextboxSo, note that this was given in Bersektas (p24) as one of the 'properties' of probability laws that could be derived from the basic axioms.But, so, I remain a little confused as to whether, by contrast, it is simply a definition of exclusive OR.

Basically, the ensuing discussion below makes things clear. Simply - it is surprising.

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alexTextboxSo, what is at first surprising, is that it should apparently ALWAYS work out that, in adding back in the events that are situated where ALL THREE overlap, one will be adding back in ONLY those that have been removed twice.However, note that actually, in the example, p4 is subtracted three times, rather than just twice, so it seems the procedure is not sensitive to this.....I still don't SEE how it works though.

alexCalloutThis feels like a new way of going about things that they have followed here. Perhaps follows from the fact that we are removing (are required to remove) ALL the instances of our required type, from the pool. (Not the case, since the same method is applied next page).

alexCalloutSince it is a case of OR, we are not, like, drawing from the pack, and then drawing a further 4 from the depleted pack. So, you know, the sample space remains.

alexCalloutSo, this is the prob of 8 specified cards

alexCalloutPossibly the same condition as Bersektas calls "disjoint".

alexCalloutIt feels quite instructive to consider the problem "all spades in a hand of 5" without the buildup of the previous pages. You might begin thinking "13 in a suit, out of 52". But then also "But we only need 5 specific cards out of the 52". Etc.

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alexCalloutSo '52 choose 5' is your total sample space, your total number of possible hands. '48 choose 5' is your total number of non-ace hands.So basically we take the COMPLEMENT of this event, to get the prob of the event: 'at least one ace'.

alexCalloutAgain, note how here we are giving the probability of getting a specific card (ace), via giving the probability of getting a hand containing the OTHER components of that hand-which-has-an-ace. What this expression says is that, among the total set of hands (52 choose 5), there are the given number of different hands containing an ace. The number (51 choose 4) is the number of hands that contain 1 ace. We divide that by the number of hands simpliciter. The key is this. Don't try to jump the gun. The numerator term is simply giving the NUMBER OF HANDS containing an ace. It is the expression as a whole that gives the PROB. The numerator should be seen as obviously giving that, because we can choose any 4 cards, from amongst 51.

alexCalloutYes, note that, since we are pairing terms together. So, it is a committee/combination of 2 terms (since order doesn't matter).

alexNoteSo, we have split the procedure up into two stages. First, we choose ace(s) from 4, then the remainder of the hand from the other 48 cards. I don't fully comprehend the rationale. If one reads the next method, it gets a little clearer - the numbers (the 'coefficients') for the number of aces, and the common denominator of '52 choose 5'.

I think maybe the key, as with the one below, is to recognize that in the numerator, we are just giving the NUMBER OF HANDS. It is only when this numerator is over the denominator (giving the total number of possible hands simpliciter) that we get a PROBABILITY.So, with the one-ace case, we are saying the NUMBER OF HANDS that contain one ace.So, we get that number by multiplying together, the number of different ways of getting an ace (namely, '4 choose 1', or just 4), with the number of different hands there are when you choose 4 cards (the remainder of the hand) from among 48 (the remainder of the pack).

And for the rationale behind the multiplication, just think of the tree diagram.

alexTextboxSo note how, because there is a common denominator, and it is fractional addition, you just end up with this denominator.

alexCalloutSo, I think that here, we are dealing with MUTUALLY EXCLUSIVE events, so our 'or's can be turned into simple additions. Regarding the example as a whole, clearly, it makes sense that, to get at "at most", you take the COMPLEMENT of the probability of the events that EXCEED that number.

alexTextboxAgain, the familiar method being employed. Numerators are giving the NUMBER OF HANDS containing required number of (or absence of) spades. Denominator is giving total number of hands.

alexCalloutSo, the first term gives the number of hands, amongst the total, than contain 2 spades. The second term subtracts away, from amongst these hands (which will contain also all those hands containing hearts) the number of hands that contain 2 spades AND which, among the other 3 cards, contain no cards from this other 13-card suit (hearts).

alexTextboxPretty obvious what is happening in this case, I think.

alexCalloutSo, it is the inclusive or. So, we are getting the probability of getting either.

alexCallout...and it is the complement of this that we calculated, above.

alexTextboxOne way to think about this which seems to work, is to consider the MINIMAL way you negate the event.Thus, with at least one of each suit, the condition is broken as soon as you don't have a member of ONE of the suits.

alexTextboxShould be a fair bit steeper.

alexCalloutSo, this method is also used in Bersektas, p23.

alexTextboxSo, one appreciates that this is an extremely useful visualization.

alexTextboxMakes sense to do this. Such an area is basically the size of the sample space.

alexCalloutI think it is a piece of luck that the area here can be computed so easily, i.e. that the triangle happens to have two equal sides.

alexTextboxSo, this is very easy.

alexTextboxThis is a bit abstract to know exactly what is meant.

alexCalloutYeah, fine.

alexCalloutSee Bersektas p22. The reason is that, if there was a positive probability and you add together all such, they would exceed 1, being infinite in number.

alexCalloutInteresting.

alexCalloutThis is the 'universe' specifying possible positions of the pin. All possible combinations of angles and values of 'y'.

alexTextboxThis integral, and re-arrangement, is actually fine.

alexCalloutThis thing of the specification of a new universe, is mentioned also by Bersektas, p29-30.

alexTextboxThe formula given in Bersektas 29-30, where it is derived more formally.

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alexCalloutSo, note that you are getting the double fraction because the numerators, in each case, are giving you numbers of hands, and the denominators are turning the expressions into probabilities. But then, you know, this is the case for each probability there, either side of the fraction in the formula.

alexCalloutSo, note the point that the "at least one picture" in the numerator here, is the conditioning event A, in the numerator from the formula P(A and B)

alexTextboxFine with this.

alexTextboxNot sure how this simplification works out.

alexCalloutQuite an understandable way of putting it, I think.

alexTextboxStated in Bersektas p32.

alexCalloutOh yes, I see this.

alexTextboxSo, this is getting to what Bersektas calls the Multiplication Rule, page 34.

alexCalloutClearly, this is drawing on the combinations formula (page 20). But I don't see why it is the reciprocal, nor how this instantiates the stuff just covered.OK, so, first, it is not supposed to instantiate the new material (shown instead in the next method).Second, there is only one favourable committee. The number of combinations giving the total, is indeed supplied by the combinations rule, and it is the reciprocal, you know, just because that total is on the denominator.

alexCalloutOK, this example makes a lot of intuitive sense of this expansion of the conditioning rule, the Chain Rule or Multiplication Rule from Bersektas.

alexTextboxSimple stuff, obviously.

alexCalloutSo, obviously I'm not totally confident about what is going on with sequences/series, but basically, you have this formula (namely, here, a/1-r) that generates any member of the series, that is appropriate for cases where the series is of the form a+ar_ar^2 etc.So apparently, you can just get the probability from the formula.

alexCalloutInteresting.

alexTextboxI think this may be using the chain rule for AND, from above.

alexTextboxNot sure how this is solved

alexTextboxAn interesting distinction.

alexCalloutMmmm, yes, OK.

alexTextboxOK. Don't think this situation is covered in Bersektas.

alexCalloutI.e. each of the permutations.

alexCalloutOK. I think I fairly follow, although why we use division, you know, as opposed to subtraction or something - I don't know.

alexTextboxThe places you pick are going to end up determining the permutation.

alexTextboxOK yes, I follow this, as a derivation, much better.

alexTextboxFine.

alexTextboxAgain, fine.

alexCalloutQuite an interesting property; worth thinking about.

alexTextboxYes, OK. Very simple.

alexTextboxOK...... So, this is just drawing on the analysis of the previous chapter.

alexTextboxOK, this example seems simple enough.

alexTextboxOne can see it. And its useful!

alexTextboxAh yes, indeed one sees that it is a special case.

alexCalloutThe probability of the outcome 'k', times the number of such trials; same with 'q'.

alexCalloutPermutations.

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alexCalloutOK, interesting.

alexTextboxFine.

alexCalloutYes OK. Just note how it is set up.

alexTextboxMmmm. So, there is no way that I would have been able to know that these two situations should be treated in these very different ways.I suppose that, if you think about it, in the first case, you have the extra "10 choose 8" because you are giving the NUMBER OF PERMUTATIONS that have the requisite number of successes. Whereas in the latter case it is just this single permutation.

alexTextboxOK......

alexCallout'k' successes in 'n' trials.

alexTextboxOK, yes I see.

alexTextboxCertainly, this makes a lot of sense.

alexTextboxOK.

alexTextboxThis formula seems good.

alexTextboxThis all, is the one-head case.

alexTextboxOK, yes. Though notice that there is nothing in this formula that is trying to represent that, you know, the first 9 failures come specifically BEFORE the success.

alexTextboxNot sure about this.

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alexCalloutSo, everything here except the final 'p', is - is it not? - just instantiating the Binomial formula - k successes in n trials.

alexTextboxSo, this is the formula for combinations/committees.

alexCalloutThis is surely a typing error

alexTextboxCertainly doesn't look too hard to do these simulations.

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alexTextboxBecause these are, so t speak, self-sufficient experiments.

alexCalloutCan't fully recall the AND rule - is it just adding probabilities?

alexTextboxOK.

alexTextboxBersektas page 38.

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alexCalloutSo, this echoes the talk about 'partitioning' in Bersektas.

alexTextboxYes, recall this.

alexTextboxOne can wrap oneself into knots, but really it is not so difficult.

alexTextboxOK fine. Obviously, very similar example examined in Bersektas.

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alexTextboxImportant philosophical stuff.

alexTextboxSo, I am not presently "up" on series...

alexTextboxOK, interpreting this way is fine.

alexCalloutIn the circumstances given by the conditions.

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alexCalloutOK, so this is just a re-writing of the 'binomial coefficient', where you have n! / k! * (n-k)! See page 61 above.

alexCalloutNote that the denominator in this term and the "main term" have swapped. In other words, where, for the p^k term, substitution would have given you lambda^k / n^k, we have put that n^k denominator under the main term, and put the k! under this term.

alexTextboxAh, we hit a wall of my lack of knowledge of series.So basically, this is going to mean that I don't understand the DERIVATION, or rather, the equivalence of the Poisson to the Bernouille.

alexCalloutOK, so we see why there is this need to be able to alternatively use the Poisson.

alexCalloutSo, note that here, we HAVE just written out the Poisson formula with each of the 4 cases and added them. It is simply that we factored out the common factor of e^-30

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alexCalloutAverage number of successes.

alexTextboxOK fine.

alexTextboxObviously, just plugging in.

alexTextboxYeah fine - a copy of the one on the previous page.

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alexTextboxClever setup.

alexCalloutMmm, I see.

alexTextboxVery int.

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alexTextboxWell, note this point.

alexCalloutI think the fraction here comes from having created a common denominator.

alexCalloutSo, this is the weighting part.

alexCalloutSo: formerly, 'k'.

alexTextboxOK yes. We recall this from the previous chapter.

alexTextboxSee page 62, for where the idea of the 'geometric distribution' is first introduced. Basically, you are finding the probability of the bunch of unsuccessful trials (and it is construed as all the trials up to the trial before you succeed).

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alexTextboxPretty simple.

alexCalloutSo, if I recall properly, this is the average value.

alexTextboxI don't follow this proof. (And this is laughably unfriendly).

alexTextboxFocus on what it is we're talking about. It is the VALUE OF SOME PROPERTY (called the 'random variable'). The 'expected value' is the value this property IS DISCOVERED TO HAVE when we've applied the appropriate probability formula.

alexTextboxAgain, it just is not really very clear to me what is going on here.

alexCalloutI.e. 'n' trials, and our prob of occurrence is 'p'.

alexTextboxIndeed, obviously the expansion of the sum

alexCallout?

alexCalloutThe number of trials.

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alexCalloutThat is to say, we want the expected values of these.

alexCalloutSo, this is the value that can be taken on, multiplied by the probability of occurrence of that value.

alexTextboxMust say that the presentation has been too abstract/sketchy for me to follow properly, really.

alexCalloutYes, certainly this is intuitive.

alexTextboxFairly helpful summary discussion.

alexCalloutOur "random variable".

alexTextboxOK fine. Though, obviously, don't feel I understand more.

alexCalloutIs this supposed to tell us that the probability would be the same for any?

alexCalloutBernoulli trials introduced page 61. Basically, when there are two outcomes.

alexCalloutThe 'random variable'.

alexCalloutThe value we expect 'X' to have.

alexCalloutSeems we choose '1' for F (failure) since we're interested in 'number of tries before success'. So our 'hits' are 'F's.

alexCalloutSo, it doesn't count 'F's, but rather, the number of trials it takes to get an S.

alexCallout...see

alexCalloutSo, this is the success...

alexCalloutI.e. their probability, times the number of these, i.e. the number of events having this probability.

alexCalloutSo this (for example) is the probability of getting two failures, I think (i.e. the first two as failures).

alexTextboxSo here, we have recognized the series, and given its formula. See page 47.

alexTextboxMmm yes. I see this reasonableness.

alexCalloutYes, OK, note that point.

alexTextboxIs there, you know, something I should be aware of about this?

alexCalloutSo, this is giving the 'expected value' (and drawing upon the result from the discussion just above, to get it).

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alexTextboxOK, this is pretty informative.

alexCallout1/p. Then: 1/(10/24), which would be 1/1 * 24/10, etc. And we apply this, because it is a case of "expected number of Bernouilli trials till success, as covered by (4) above.

alexTextboxSo, we are focusing on the black balls, since we want to find the number of balls before a white. (That is our chosen random variable X).

alexCalloutThe success.

alexCalloutTo see this, just focus on the fact that, here, we are basically treating every B as totally distinct - such that indeed it would be confusing to call them all 'B'. For we are here treating the prob of a particular B.

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alexCalloutThe value of the card, times its probability of occurring.

alexTextboxThis is pretty much fine.

alexTextboxYup, OK. Just remember what these numbers are - an average value of the number of reds from each urn.

alexCalloutBecause P(heads) = 0.4

alexTextboxOK.

alexTextboxYep fine. Refer back to the previous example if it stops making sense.

alexTextboxAgain, don't really follow. Not really too sure what

alexCalloutFractions give probabilities. E.g. 3 ways of getting 4.

alexCalloutThe number of rolls, GIVEN that it is one of these results?

alexCalloutI.e. this "player's point" business.

alexCalloutBecause there are 6 ways of getting a 7.

alexTextboxOK, makes sense.

alexCalloutExpectation of the number of rolls, GIVEN first roll is 4.

alexCalloutSo, this is 1/9 plus 36/9, i.e. 45/9, i.e.5.

alexTextboxSo, to solidify this, the point is that, if a 4 comes up, then you are going to keep rolling till either a 4 or a 7 turns up. So, naturally, the further number of rolls is going to be determined by the probabilities of those two coming up. Of course, we have also to factor in the probability of getting a result on the first throw.

alexTextboxSo, note how it indeed matches the form of the 'theorem of total expectation', above.

alexTextboxI don't know how we would solve this.

alexCalloutOther name for expected value.

alexTextboxOK.

alexTextboxI seem to get pretty much nothing in this review.

alexTextboxPretty understandable so far, surely.

alexTextboxThis seems a usefully picturesque analogy.

alexCalloutOK, so note that that is what f(x) signifies - a function yielding the density, at a point 'x'.

alexCalloutProducing probabilities, as a function of points.

alexTextboxAgain, what a lovely image!

alexCalloutNote this detail.

alexCalloutYes, I see. Integrating over the relevant limits of integration (corresponding to the interval).

alexTextboxAll seems OK.

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alexTextboxComfortable with this.

alexTextboxOne can appreciate how they 'work'.

alexCalloutHappy with this integration.

alexCalloutNot too sure about this business.

alexTextboxThe probabilities over these various intervals. Feel I understand the idea.

alexTextboxMakes sense.

alexLine

alexTextbox...because the EXACT points 'a' and 'b' have zero prob.

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alexTextboxQuite interesting.

alexPolyline

alexTextboxSeems an important point

alexCalloutSo, this is getting you the probability, not the prob density.

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alexTextboxPicturing the height as giving the density, seems like a useful analogy.

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alexCalloutProbability per unit length.

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alexTextboxPretty comprehensible, it would seem.

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alexTextboxvery comprehensible

So, just bear in mind that the PDF here is giving CUMULATIVE probs.

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alexTextboxNot clear what is gained by this.

alexCalloutMakes sense.

alexCalloutPortion of distance covered, divided by total distance.

Remember that what we are giving is a breakdown of f(x), which is probability density, or probability per unit length.

alexCalloutUseful visualization.

alexCallouti.e. dollars, winnings.

alexTextboxAll seems fine.

alexTextboxYes, fine.

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alexTextboxVery comprehensible

alexTextboxA little bit of a problem properly interpreting.

alexTextboxDefinitely need a diagram to properly interpret.

alexTextboxRight, yes, fine.

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alexTextboxThe nub of it.

alexTextboxClear.

alexTextboxOK note this.

alexTextboxThe height after, and before, 'the chunk of prob is fed in'.

alexTextboxSo, you slice off that bit more probability (equal to the prob fed in at the jump), if you get the cumulative prob greater-than OR EQUAL TO the value at which there is a jump.

alexCalloutI.e. here, whether less-than, or less-than-or-equal makes no difference since there is no jump. And, the function at this point churns out 1/8.

alexTextboxYeah, fine.

alexTextboxNote the diff.

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alexTextboxNote this distinction.

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alexTextboxOK, yes, so this gives the relation

alexTextboxSeems fine.

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alexCalloutNot totally sure why the integral works out as that.

alexCalloutThe cumulative probability up to the point 0. (Rather than, the cumulative prob up to the point x. And yes, the function supplying this prob, ends its realm of application at 0.

alexCalloutNot totally sure how the integral works out as that.

But still, one gets the idea of what happens in this case, where you add the contribution of the two continuous distributions.

alexTextboxOK.

alexTextboxI understand that there are no jumps, but don't feel I 100% get the notation.

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alexTextboxNote.

alexCalloutI.e. the derivative.

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alexCalloutSo, with this differentiation, you are using the power rule. Which means you would have 2 * 1/8, which is going to give you 1/4.

alexTextboxSo basically, you differentiate each sub-component of F(x).

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alexTextboxUseful for clarity, clearly.

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alexTextboxNote this distinction, for understanding.

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alexTextboxAll usefully clarifying.

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alexCalloutSo, does this mean that it is going to give the distribution of probabilities of waiting times. E.g. a certain probability that the waiting time will be.....

alexCalloutOur 'random variable'; our property of interest.

alexTextboxI THINK I get the sense of this.

alexCalloutDo not really follow where this comes from. So the derivation as a whole is not very informative to me.

alexCalloutCan't recall the exact differentiation rule being used here. Could look it up...

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alexTextboxNote the result, anyway.

alexCalloutSo, not totally confident how to carry out this integration.

alexCalloutI'm not totally confident how you get rid of the minus.

alexTextboxSo, this example at least tells us what is going on; concretizes. You can find a numerical probability, for a particular value of a random variable.

alexCalloutSo, you know, the exponential is the function that models the probability distribution for this sort of situation.

alexCalloutSo, this expresses the point about independence.

alexCalloutSo, the parameter '2' has been put into the function.

alexTextboxRight, so, again, this is just the point about independence.

alexTextboxHave not followed this.

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alexTextboxNote the distinction.

alexCalloutYes I think I can see that the gamma collapses to the exponential if 'x' is set to 1.

alexTextboxHave not followed this derivation.

alexTextboxThe graphs help remind what the density is all about. The probability of values occurring in those regions.

alexTextboxThink I understand pretty well what is going on here.

alexCalloutSo, half of the probability has accumulated at 0 - what you would obviously expect.

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alexTextboxHave not followed this proof.

alexTextboxI get this, although it is not easy to keep the point in mind.

alexCalloutOK yes, note that that is equivalent to the previous.

alexCalloutOK, so (3) works with actual values.

alexCalloutThe location of the mean

alexTextboxSo, you know, I follow all the steps, but I am not really sure of, like, what it is exactly that we're doing here.

alexCalloutYes, fine.

alexTextboxSort of, a bit unsure what is being said.

alexTextboxVery much kind of losing track of exactly what is being said.

alexCalloutnumber of successes.

alexTextboxSo, we'd have to remember how to actually calculate this.

alexTextboxYou know, I am sort of following.

alexTextboxOK. I vaguely recall that this is the method.

alexCalloutI think this is from lookin up 6/9 in the table...

alexTextboxAgain, a sort of iffy understanding.

alexPencil