# Activity for DNB‭

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Comment Post #283899 "bad English" — This is ungracious. English is not my first language. You can't deny that fluency in English is hard. Or else whole world is fluent! "basic failures of reading comprehension" — Honestly, I have no idea what you mean. Where's evidence? Which posts are you referring? "... really, it te...
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Comment Post #283899 I think you're referring me? No offense, but your post appears inequitable. "[tortured synonym for “know”]" — My teacher used these words. And I don't mean "know". Prognosticating a step isn't the same as KNOWING a step. I know now that GME (GameStop stock price) rocketed to $488, but I didn't progn... (more) about 2 months ago Edit Post #283887 Post edited: about 2 months ago Edit Post #283887 Initial revision about 2 months ago Question Does "both girls, ≥ 1 winter girl" = "both girls, ≥ 1 winter child" let you generalize the problem statement that postulated both children as girls? What does "the fact that "both girls, at least one winter girl" is the same event as both girls, at least one winter child"" imply about the problem statement below that posited merely girls? I feel that this fact lets us breadthen the problem statement. Can it? Adam Bailey's answer details the ... (more) about 2 months ago Edit Post #283886 Post edited: about 2 months ago Edit Post #283886 Initial revision about 2 months ago Question How would you intuit, and soothsay to rewrite, "both girls, ≥ 1 winter girl" as "both girls, ≥ 1 winter child"? 1. Please see the red underline. How can you intuit that "both girls, at least one winter girl" as "both girls, at least one winter child"? I ask this for my 15 y.o. 2. This step feels fey, sibylline! How would you prognosticate to use this fact? We would've never augured to construe "both girls, ... (more) about 2 months ago Edit Post #283388 Post edited: 2 months ago Comment Post #283387 I'd rather not "rephrase the issue you're having", because I want to stick to the text in the book. Have you thought of complaining to the author at blitzstein@stat.harvard.edu? (more) 2 months ago Edit Post #283255 Post edited: 2 months ago Edit Post #283255 Post edited: 2 months ago Edit Post #283255 Post edited: 2 months ago Edit Post #283255 Post edited: 2 months ago Edit Post #283118 Post edited: 2 months ago Edit Post #282942 Post edited: 2 months ago Edit Post #282942 Post edited: 2 months ago Edit Post #282942 Post edited: 2 months ago Edit Post #283388 Post edited: 2 months ago Edit Post #283260 Post edited: 2 months ago Edit Post #283260 Post edited: 2 months ago Edit Post #283260 Post edited: 2 months ago Edit Post #283388 Post edited: 2 months ago Comment Post #283399 "That also means such questions probably aren't appropriate for this site; consider asking a teacher instead." This feels unmannerly? Some of us aren't wealthy enough to afford private tutors. (more) 2 months ago Comment Post #283387 @Moshi I don't think "By definition" answers my question? Why not define these two probabilities the other way around? (more) 2 months ago Edit Post #283388 Post edited: 2 months ago Edit Post #283388 Post edited: 2 months ago Edit Post #283388 Initial revision 2 months ago Question How do I prove Simpson's Paradox, scilicet$P(A|B) > P(A|B^C)$?$\forall a,b,c,d > 0, aP(A|B^C)$! I can multiply the blue and orange inequalities together, but their product is in the opposite direction too! 2. I can't simply multiply the purple inequality by the green inequality, because the green inequality's in the SAME direction as$P(A|B)>P(A|B^C)$. ... (more) 2 months ago Edit Post #283387 Initial revision 2 months ago Question Why "only 1/10,000 men with wives they abuse subsequently murder them" ≠ P(A|G,M) & "50% of husbands who murder their wives abused them” ≠ P(G|A)? All emboldings are mine. See my red side line — the solution identifies "only 1 in 10,000 men with wives they abuse subsequently murder their wives" (in the problem statement) with$\color{red}P(G|A)$. See my green underline — the solution identifies "50% of husbands who murder their wives prev... (more) 2 months ago Edit Post #283371 Initial revision 2 months ago Question How can I exploit symmetry to intuit P(A wins) + P(B wins) = 1, without performing algebra? I'm not asking about algebra that I can execute. How can I intuit P(A wins) + P(B wins) = 1 most quickly, without algebra? Any 15 year old can calculate that for the$p = 1/2$case,$\dfrac{i}{N} + {\color{limegreen}\dfrac{N - i}{N}} = 1$.$p \neq 1/2$case,$\dfrac{1 - (q/p)^i}{1 - (q/p)^N} ...
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Comment Post #283299 Sorry. I think I was editing the old question, but I copypasted my draft for my new question in the wrong box. What would you like me to do? Create a new question for the the Alice question?
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Comment Post #283317 "For this equation to hold (in general) the exponent must be k = 0 and not k + 1 = 1". What do you mean These are equivalent. $k = 0 \iff k + 1 = 1$.
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Comment Post #282945 I don't think you answered by second question above on the "simple geometric interpretation"?
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Question How to vaticinate calculating $w_{k + 1} = w_1(1 + \dfrac{1- p}{p} + \dots + [\dfrac{1- p}{p}]^k)$ separately for $p = 1 - p$ and $p \neq 1 - p$?
Please see the orange, green, red underlines. 1. Why does $\color{limegreen}{w{k + 1} = w1(1 + \dfrac{1- p}{p} + \dots + p\dfrac{1- p}{p}]^k)}$ beget the two cases of $\color{red}p = 1 - p$ and $\color{red}p \neq 1 - p$? 2. Even after reading the solution, I wouldn't have prognosticated to calcul...
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Question In $w_{k + 1} - w_k = (\frac{1 - p}{p})^{exponent}(w_1 - w_0)$, why isn't exponent $k + 1$?
Please see the $r^k$ underlined in red, which is $(\frac{1 - p}{p})^k$ as defined by the green underlines. 1. How do you deduce that the exponent must be $k$? Why isn't the exponent $k + 1$? 2. Is this question related to the Fence Post Error? Have I committed it? Image alt text Tsitsikli...
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Question Why was my question closed: If Alice must've have classes on at least 2 days, why do you need the intersection of 3 's?
1. I don't know why the MathJax isn't processing at https://math.codidact.com/posts/282645. 2. But why was it closed "as not constructive"? >This question cannot be answered in a way that is helpful to anyone. It's not possible to learn something from possible answers, except for the solution...
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Comment Post #282945 1. Thanks for your answer. I agree that P(A) + P(B) can $>1$, but then why can't we change the definition of probability to include all numbers $>1$? 2. "here is a simple geometric interpretation". This is just the area, correct? But how does this answer my question?
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Comment Post #283260 Thanks! Boy am I scatter brained! I already got one downvote here. If I keep getting downvotes, I'll construe them to mean that this question is too half-baked, and I'll delete
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Comment Post #282665 2. Why does "starting with the red equation would only prove that the green equation holds for odd n"? I see no thing in your last summation, or $0 \le k \le 2n + 1$ (the summation bounds) that restricts $k$ to odd integers?
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Comment Post #282665 Thanks. 1. How do I prove that $\{2n + 1 - k \mid k \in \{ 0,\dots,n \}\} \equiv \{n + 1,\dots, 2n + 1\}$? Undeniably, I can see that this is true if I substitute $k = 0,\dots,n$, because $2n + 1 - \color{red}0, 2n + 1 - \color{red}1, \dots, 2n + 1 - \color{red}n = 2n + 1, 2n, \dots, n + 1$. But thi...
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Question How would you vaticinate to $-w_k$ from both sides of $w_{k + 1} = \dfrac{w_k - (1 - p)w_{k - 1}}{p}$?
This question appeared on my pop quiz last week. I got 0%. I achieved everything until the green equation, then I didn't know how to proceed. After reading this solution, I see that you must isolate $pw{k + 1}$ and move $wk$ to the right. $\color{limegreen}{wk = pw{k + 1} + (1- p)w{k - 1}} \if... (more) 3 months ago Edit Post #283255 Post edited: 3 months ago Edit Post #283255 Initial revision 3 months ago Question After$n - 2$unchosen doors are opened, how does the probability of the$n - 2$unchosen doors "shift" or "transfer" to the lone unchosen door? The second quotation below uses the verb "shift" to describe how Monty Hall's opening the 98 unchosen doors (revealing a goat each) ""shifts" [boldening mine] that 99/100 chance to door #100"? The first quotation below uses "transfer". But how can probabilities "shift" or "transfer"? This analog... (more) 3 months ago Edit Post #283254 Post edited: 3 months ago Edit Post #283253 Post edited: 3 months ago Edit Post #283254 Post edited: 3 months ago Edit Post #283254 Initial revision 3 months ago Question How to visualize multiplication in the Odds form of Bayes's Theorem? Here I'm asking solely about the circle pictograms. Please eschew numbers as much as possible. Please explain using solely the circle pictograms. Undeniably, I'm NOT asking about how to multiply numbers. I don't understand Image alt text 1. How do I "visually" multiply Circle 1 (representing ... (more) 3 months ago Edit Post #283253 Post edited: 3 months ago Edit Post #283253 Post edited: 3 months ago Edit Post #283253 Post edited: 3 months ago Edit Post #283253 Initial revision 3 months ago Question How to visualize division in the Odds form of Bayes's Theorem? Here I'm asking solely about the circle pictograms. Please eschew referring to, or using, numbers as much as possible. Please explain using solely the circle pictograms. Undeniably, I'm NOT asking about how to divide numbers. I don't understand Image alt text 1. How do I "visually" divide Cir... (more) 3 months ago Edit Post #283111 Post edited: 3 months ago Edit Post #283252 Initial revision 3 months ago Question Which vertical line signifies "putting the cutoff for a positive result at a very low level"? The author, Karen Stewart MA Natural Sciences (Univ. of Cambridge) PhD Veterinary Microbiology (Cambridge), refers to "a purple dashed line" in her original graph, but I don't see any PURPLE dashed line. Perhaps I need an eye exam! So I re-colored and annotated them. Which of the 3 lines did she mean... (more) 3 months ago Comment Post #282975 Thanks. Can you please recapitulate and underscore my mistake? I read your answer, but I still don't see my mistake? Perhaps you are too diplomatic to chide me, but go ahead! (more) 3 months ago Edit Post #283118 Post edited: 3 months ago Edit Post #283118 Initial revision 3 months ago Question Why would skyrocketing the numbers of doors help laypeople intuit the Monty Hall Problem? Alas, it isn't clear to me that it becomes clear that the probabilities are not 50-50 for the two unopened doors. Had I never seen this exercise or problem, even if there were 1 Billion doors, I would "stubbornly stick with their original choice". What am I misunderstanding? Am I just that witles... (more) 3 months ago Edit Post #283116 Initial revision 3 months ago Question Intuitively, why would organisms — that after one minute, will either die, split into two, or stay the same, with equal probability — all die ultimately? I have no questions on the solution or the algebra, but even after re-reading the solution, I still can't fathom or intuit why$P(D) = 1$from the problem statement. Even now, I couldn't have divined or foretold that$P(D) = 1$! Image alt text >The strategy of first-step analysis works here be... (more) 3 months ago Edit Post #283115 Initial revision 3 months ago Question How does P(Monty opens door 2) = P(Monty opens door 3), and$P(\text{get car}|M_2)P(M_2) = P(\text{get car}|M_3)P(M_3)$? "Monty, who knows where the car is, then opens one of the two remaining doors. The door he opens always has a goat behind it (he never reveals the car!)." So Monty must open ONE of the$Mj (j = 2,3$), the one with the goat! But Monty mustn't and won't open the other$Mj$with the car. So$P(M2) \neq ...
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Edit Post #283113 Initial revision 3 months ago
Question In the Monty Hall problem, why can you just assume the contestant picked door 1? Why are you entitled to relabel the doors, or rewrite this solution with the door numbers permuted?
My bafflement ought be obvious. 1. A contestant could've picked doors 2, 3. So you can't just assume he picked door 1. 2. Correct me if I'm wrong, but the game show didn't authorize contestants "to relabel the doors, or" permute the door numbers. So what permits you to do any of this in this solut...
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Comment Post #282600 That question got removed, and your link no longer works.
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Question If a 2nd test's independent from the 1st test, then why does $\frac{0.95}{0.05}$ figure twice in $\frac{P(D|T_1)}{P(D^C|T_1)}\frac{P(T_2|D,T_1)}{P(T_2|D^C,T_1)}$?
The problem statement postulates that "The new test is independent of the original test (given his disease status)". So where did the two $\frac{0.95}{ 0.05}$, that I underlined in red and purple, stem from? >### Example 2.6.1 (Testing for a rare disease, continued). >Fred, who tested posi...
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Question Why's the true positive rate termed Sensitivity and true negative rate Specificity, not vice versa?
1. To wit, what's "Sensitive" about True Positive Rates, and "Specific" about True Negative Rates? 2. Why weren't these Metaphors) or Imports reversed? Why wasn't "Sensitive" termed to signify True Negative Rates, and "Specific" True Positive Rates instead? I learn best visually. Can these g...
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Question How do these 3 bell curves of Likelihood, Posterior, Prior pictorialize the Odds form of Bayes' rule?
I learn best visually, and I found these graph. 1. Does it furnish intuition on Theorem 2.3.5 below? 2. E.g. Is the Likelihood Ratio always graphically left of Posterior and Prior? If so, why? >This can also be pictorially represented – the graph below shows the new posterior belief for a cer...
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Edit Post #282966 Initial revision 3 months ago
Question What's wrong with evaluating $n(n-1) \dots (n-[k-3])(n-[k-2])\color{red}{(n-[k-1])}$ at $k = 1$?
This snag arose out of this post, and these comments by r. In that post, I couldn't imagine how >By convention, $n(n-1) \dots {\color{red}{(n-k+1)}} = n$ for k = 1. Thus I wrote out the LHS $= n(n-1)(n - 2) \dots (n-[k-3])(n-[k-2])\color{red}{(n-[k-1])}$. Then I substituted $k=1$ into this exp...
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Question Intuitively, why does A, B independent $\iff$ A, $B^C$ independent $\iff A^C, B^C$ independent?
>### Proposition 2.5.3. >If A and B are independent, then A and $B^C$ are independent, $A^C$ and B are independent, and $A^C$ and $B^C$ are independent. Blitzstein, Introduction to Probability (2019 2 ed) p 64. I'm seeking solely intuition. I'm NOT asking about how to prove these independ...
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Question Besides guaranteeing 0 ≤ P(A ∩ B) ≤ 1, why can't Independence be defined as P(A ∩ B) = P(A) +,-, or ÷ P(B)?
I know that $0 \le P(A \cap B) \le 1$ will be violated if $P(A \cap B) = P(A) +,-,\text{or} ÷ P (B)$. I'm not asking about or challenging this reason. But what are the other reasons against defining independence as $P(A \cap B) = P(A)$ 1. $+ P (B)$? 2. or $- P (B)$? 3. or $÷ P (B)$? >#...
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Comment Post #282892 Thanks. Can you pls elaborate why terms like $s/c^n$ will diverge instead of converging to $0$? Can you please respond in, by editing, your answer? Comment chains are cumbersome to read.
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Edit Post #282890 Initial revision 3 months ago
Question How does $\lim\limits_{n \to \infty} \dfrac{p[s + (1 - s)c^n]}{p[ s + (1 - s)c^n] + (1 - p)[s + (1 - s)w^n]} = p?$
To minimize this post's length, I don't repeat the exercise itself. The title refers to the sentence beside my red line below. I divide both the numerator and denominator by $\color{red}{c^n}$. Then $P(G|U) = \dfrac{p[ \dfrac{s}{\color{red}{c^n}} + (1 - s)]}{p[ \dfrac{s}{\color{red}{c^n}} + (1 ... (more) 3 months ago Edit Post #282605 Post edited: 3 months ago Comment Post #282615 Thanks again. 1. Can you please expound why "you can't simply substitute a variable into that kind of informal, descriptive expression"? 2. "It's not an algebraic expression, so the rules of algebra don't apply" How isn't this an algebraic expression? This is a product of variables! (more) 3 months ago Edit Post #282889 Initial revision 3 months ago Question How do you calculate$P(X = n|G), P(X = n|G^C)$by the Law of Total Probability, with extra conditioning? 1. Please see$P(U|G)$and$P(U|G^C)$below, beside my red line. Can you please expound these calculations? 2. How was$s$computed in both equations? >### Example 2.4.5 (Unanimous agreement). >The article "Why too much evidence can be a bad thing" by Lisa Zyga [30] says: >>Under... (more) 3 months ago Edit Post #282888 Initial revision 3 months ago Question How can I visualize the Law of Total Probability with extra conditioning? How can I pictorialized this Theorem 2.4.3? As you can see below, I edited a picture by drawing E inside B.. Is my edit correct? Can my edit be improved? >Theorem 2.4.3 (LOTP with extra conditioning). Let$A{1}, \ldots, A{n}$be a partition of S. Provided that$P\left(A{i} \cap E\right)>0$for a... (more) 3 months ago Comment Post #282873 Your comment doesn't construe my post charitably. "The answer to 1 was given just two paragraphs earlier." I rectified this by coloring that earlier sentence. "2 and 3 are not true" Why not? (more) 3 months ago Edit Post #282873 Post edited: 3 months ago Edit Post #282873 Post edited: 3 months ago Edit Post #282873 Initial revision 3 months ago Question In the Lost Boarding Pass Probability Problem, why couldn't Passengers 2-99 sit in Seats 1 or 100, before Passenger 100 boards? Although Tanae Rao was just a high school graduate when he wrote his solution below, it's the most clear out of the solutions I read. 1. I don't understand the step, that I colored in red. Why must Seats 2-99 be occupied, before Passenger 100 boards the plane? 2. Why couldn't one of Passengers... (more) 3 months ago Edit Post #282771 Initial revision 3 months ago Question How can "information about the birth season" bring "at least one is a girl" closer to "a specific one is a girl"? Please see the sentences beside my red highlighted words. I don't understand how "Conditioning on more and more specific information brings the probability closer and closer to$1/2$"? Example$2.2 .7$(A girl born in winter). A family has two children. Find the probability that both children are... (more) 3 months ago Edit Post #282645 Post edited: 3 months ago Edit Post #282645 Post undeleted 3 months ago Edit Post #282645 Post edited: 3 months ago Edit Post #282642 Post edited: 3 months ago Edit Post #282645 Post deleted 3 months ago Edit Post #282666 Initial revision 4 months ago Question Why isn't the probability of being void in 3 specifi c suits$\frac{1/13}{\dbinom{52}{13}}$? Kindly see the sentence UNDER the red line below. Why isn't the probability being void in 3 specific suits$\frac{1/13}{\dbinom{52}{13}}$? As "the probability of being void in 3 specific suits" means you don't want cards of 3 suits, you desire cards of solely ONE suit. If you remove all cards of any ... (more) 4 months ago Edit Post #282643 Post edited: 4 months ago Edit Post #282643 Post edited: 4 months ago Edit Post #282643 Post undeleted 4 months ago Edit Post #282643 Post edited: 4 months ago Edit Post #282643 Post deleted 4 months ago Edit Post #282645 Post edited: 4 months ago Edit Post #282645 Post edited: 4 months ago Edit Post #282645 Post edited: 4 months ago Edit Post #282645 Post edited: 4 months ago Edit Post #282645 Post edited: 4 months ago Edit Post #282644 Post edited: 4 months ago Edit Post #282645 Post edited: 4 months ago Edit Post #282645 Initial revision 4 months ago Question If Alice must've have classes on at least 2 days, why do you need the intersection of 3$A_i^C$'s? Can someone please rectify my MathJax? Please see the red phrase below. 1. The question itself never touts or postulates outright that Alice "must have classes on at least 2 days", which feels like an esoteric deduction. So why must she have classes on at least 2 days? 2. If Alice must've cl... (more) 4 months ago Edit Post #282644 Post edited: 4 months ago Edit Post #282644 Initial revision 4 months ago Question How does the change of variable$\color{red}{r↦n−r}$transmogrify$\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$into$\sum\limits_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$? I'm unskilled at performing algebra with Capita-sigma notation. This comment by a deleted user alleges that the "change of variables$\color{red}{r↦n−r}$" will transmogrify$\sum\limits{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$into$\sum\limits{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$. I got stuck.... (more) 4 months ago Edit Post #282643 Initial revision 4 months ago Question If A & B are joint, can Arby recoup some of his loss only when$P_{Arby}(A \cup B) < P_{Arby}(A) + P_{Arby}(B)$? 1. Please see the sentence alongside the red line below, but the authors didn't write this sentence for the first case ($P{Arby}(A \cup B) < P{Arby}(A) + P{Arby}(B)$). Thus if A & B ARE joint, can Arby recoup some of his loss in this first class? 2. If I'm correct above, then why do these 2 cases... (more) 4 months ago Edit Post #282642 Post edited: 4 months ago Edit Post #282642 Post edited: 4 months ago Edit Post #282642 Post edited: 4 months ago Edit Post #282642 Initial revision 4 months ago Question$\sum_{k=0}^{n} \binom{n}{k}=2^{n} \overset{?}{\iff} \sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$Jack D'Aurizio narratively proved$\color{red}{\sum\limits{k=0}^{n} \binom{2n+1}{k}=2^{2n}}$. Is this red equation related, and can it be transmogrified, to$\color{limegreen}{\sum\limits{k=0}^{n} \binom{n}{k}=2^{n}}$? I started my attempt by substituting$n = m/2$, because the RHS of the green... (more) 4 months ago Comment Post #282615 You wrote "You started on the right track with$n(n - 1)\ldots(n - [k - 3])(n - [k - 2])(n - [k - 1])$". I just evaluated this expression at$k = 1$. Now do you understand "what the middle expression in that line is supposed to mean"? (more) 4 months ago Comment Post #282616 The issue here appears to be that the English syntax differs from the order of the terms on the RHS? I misconstrued "we could first choose the k team members" as$k$, and "then choose one of them to be captain" as$\dbinom{n}{k}$. (more) 4 months ago Edit Post #282605 Post edited: 4 months ago Comment Post #282614 I saved up these questions over a week. But I'll slow down as you ask. "I'm a bit uncertain about asking tons of questions from other sources" Huh? I see nothing with asking questions from renowned textbooks? (more) 4 months ago Edit Post #282605 Post edited: 4 months ago Comment Post #282611 Absolutely not! Manners please? I couldn't copy and paste from the second website. (more) 4 months ago Edit Post #282606 Post edited: 4 months ago Edit Post #282614 Initial revision 4 months ago Question Why$\color{red}{k\dbinom{k}{1}} \neq$"first choose the k team members and then choose one of time to be captain"? Because you "first choose the k team members and then choose one of time to be captain", shouldn’t the RHS be$\color{red}{k\dbinom{k}{1}}$? The captain is chosen from the$k$team members already chosen.$\color{forestgreen}{k\dbinom{n}{k}}$appears wrong to me, because this means that you're c... (more) 4 months ago Edit Post #282613 Initial revision 4 months ago Question Why shouldn't the Bose-Einstein value be used to calculate birthday probabilities? Can you please expound and simplify the embolden phrase below? >As another example, with n = 365 days in a year and k people, how many possible unordered birthday lists are there? For example, for k = 3, we want to count lists like (May 1, March 31, April 11), where all permutations are consid... (more) 4 months ago Edit Post #282612 Initial revision 4 months ago Question You're sampling k people from a population of size n one at a time, with replacement and with equal probabilities. Order or not? If you're sampling k people from a population of size n one at a time, with replacement and with equal probabilities, then why does it matter whether your samples are ordered? The quotation below doesn't expound the pros and cons of ordering your samples or not. >1.4.23. The Bose-Einstein result ... (more) 4 months ago Edit Post #282611 Initial revision 4 months ago Question What's the bijection between Stars and Bars and Integer Solutions to an Equality? The second quotation below keeps mentioning "bijection", but it never explicitly defines it. So what's the formula for that bijection? A story instead of stars and bars - Making Your Own Sense > On to the third problem. As I said earlier, many people teach students to reduce other problems to ... (more) 4 months ago Comment Post #282602 Thanks. I fixed my typo. (more) 4 months ago Edit Post #282602 Post edited: 4 months ago Edit Post #282609 Initial revision 4 months ago Question Are Stars and Bars in Combinatorics related to the Fence Post Error? The bars in the lower picture look like fences. That's why Stars and Bars reminds me of Fence Post Error? >It is common to replace the balls with “stars”, and to call the separators “bars”, yielding the popular name of the technique. We have 5 stars, and 2 bars in our example: >![](https://www.... (more) 4 months ago Edit Post #282608 Initial revision 4 months ago Question Out of 4 people, why does ways to choose a 2-person committee overcount by 2 the ways to divide the 4 into 2 teams of 2? 1. Please see the sentence alongside my red line below. Why does part (a) overcount part (b) by a factor of c? 2. Scilicet, why aren't the answers to parts (a) and (b) the same? Whenever you choose a 2-person committee #1, the remaining unchosen 2 members automatically can form the 2-person comm... (more) 4 months ago Edit Post #282607 Initial revision 4 months ago Question Explain to a 9 year old — To count each possibility c times, why divide by c? Why not subtract by c? Please see the embolded phrase below. How can you explain to a 9 year old why you 1. must divide by$c$? 2. can't subtract by$c$? >### 1.4.2 Adjusting for overcounting >In many counting problems, it is not easy to directly count each possibility once and only once. If, however, we are... (more) 4 months ago Edit Post #280168 Post edited: 4 months ago Edit Post #280168 Post edited: 4 months ago Edit Post #280168 Post edited: 4 months ago Comment Post #281319 Thanks. Does my edit [to my post] change your answer? (more) 4 months ago Edit Post #280168 Post edited: 4 months ago Edit Post #282606 Initial revision 4 months ago Question If k = 1, why$n(n-1) \dots \color{red}{(n-k+1)} = n$? Please see the boldened sentence below. I write out the LHS$= n(n-1) \dots (n-[k-3])(n-[k-2])\color{red}{(n-[k-1])}$. Then$LHS| {k = 1} = n(n-1) \dots (n+2)(n+1) \neq n$. >### Theorem 1.4.8 (Sampling without replacement). >Consider n objects and making k choices from them, one at a time wi... (more) 4 months ago Edit Post #282605 Post edited: 4 months ago Edit Post #282605 Post edited: 4 months ago Edit Post #282605 Initial revision 4 months ago Question Intuitively, if you pick k out of n objects singly without replacement, why's the number of possible outcomes NOT$n(n-1) \dots [(n-(k - 1)]\color{red}{(n - k)}$? I know that$\color{limegreen}{(n-k+1)} \equiv (n - (k - 1))$. But whenever I contemplate choosing k from n objects singly without replacement, I keep muffing the number of possible outcomes as$n(n-1) \dots \color{limegreen}(n-k+1)\color{red}{(n - k)}$. I bungled by adding the unnecessary and wrong ... (more) 4 months ago Edit Post #282604 Initial revision 4 months ago Question Why aren't the "21 possibilities here" NOT equally likely? Please see the last sentence below, that I highlighted in red. Example$1.4 .5$(Ice cream cones). Suppose you are buying an ice cream cone. You can choose whether to have a cake cone or a waffle cone, and whether to have chocolate, vanilla, or strawberry as your flavor. This decision process ca... (more) 4 months ago Edit Post #282603 Post edited: 4 months ago Edit Post #282603 Initial revision 4 months ago Question Without calculations, how can you visualize "that half the squares are white and half are black"? Please see the 2nd para. below alongside my red highlighted words. I can't "[i]magine rotating the chessboard 90 degrees clockwise." I can't visualize how "all the positions that had a white square now contain a black square, and vice versa". Example 1.4.4 (Chessboard). How many squares are t... (more) 4 months ago Edit Post #282602 Post edited: 4 months ago Edit Post #282602 Initial revision 4 months ago Question "A occurred" vs. "something must happen" 1. Why doesn't "Something must happen" mean$s{actual} \in A$? 2. Scilicet, doesn't "A occurs" mean the same thing as "something must happen"? Something must happen.$\iff$Some event must happen.$\iff$At least one event must happen$\iff$Call this event A. Then A occurred. ![](https://... (more) 4 months ago Edit Post #282134 Initial revision 5 months ago Question What story and TWO-digit Natural Numbers best fit Bayes' Theorem chart? Why did Madam Monica Cellio close What story and TWO-digit Natural Numbers best fit Bayes' Theorem chart? as duplicate of What story and ONE-digit Natural Numbers explain Bayes' Theorem chart most simply?? The difference is blindingly obvious. The first question seeks an example of Bayes' Theorem ... (more) 5 months ago Edit Post #281987 Initial revision 5 months ago Question What story and two-digit Natural Numbers best fit Bayes' Theorem chart? To complete the table below most comfortably for teenagers, 1. what are the simplest stories? 2. what natural numbers$\le 99$contrast the base rate fallacy the most? Please don't repeat a number. I'm trying to improve on this question that uses two-digits just$\le 20$, because 3. the... (more) 5 months ago Edit Post #280741 Post edited: 9 months ago Comment Post #280741 I posted at https://math.codidact.com/posts/280742 about the miffed Mathjax. (more) 9 months ago Edit Post #280742 Initial revision 9 months ago Question Why isn't \hline rendering here, when it does on Stack Exchange? Please see https://math.codidact.com/posts/280741. I just pasted it on Stack Exchange and my MathJax is rendered perfectly. (more) 9 months ago Edit Post #280741 Initial revision 9 months ago Question What story and one-digit Natural Numbers explain Bayes' Theorem chart most simply? Some students have sniveled that most examples of Bayes' Theorem use non-integer numbers. I want to try a Bayes' Theorem chart that uses just single digit Natural Numbers$\le 9\$. To complete the table below most comfortably for teenagers, 1. what are the simplest stories? 2. what natural numbe...
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Edit Post #280168 Initial revision 10 months ago
Question How can I deduce which operation removes redundacies?
Please don't answer by working backwards from the answer, or by appealing to arithmetic. Act as if you're learning this for the first time. 1. How can I deduce which operation ought fill in the red blank beneath? 2. Why can't it be subtraction? I shortened the original explanation: >Quandary: How...
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