# Activity for DNB‭

Type On... Excerpt Status Date
Edit Post #286453 Initial revision 5 months ago
Answer A: $g(x)\xrightarrow{x\to\infty}\infty$ Implies $g'(x)\leq g^{1+\varepsilon}(x)$
Consider $g^{-\epsilon}$. Then its first derivative is $Dx \\; g^{-\epsilon} = \epsilon g^{-1-\epsilon} g'$. Then $g^{-\epsilon} > 0$ and tends to 0, and ${Dx \\; g^{-\epsilon}} < 0$. If $Dx \\; g^{-\epsilon} < -\epsilon$ on a set of infinite measure, then $\int^{x}0 Dx \, g^{-\epsilon} \quad dx ... (more) 5 months ago Edit Post #285679 Post edited: 8 months ago Edit Post #285679 Post edited: 8 months ago Edit Post #285679 Post edited: 8 months ago Edit Post #285679 Post edited: 8 months ago Edit Post #285679 Post edited: 8 months ago Edit Post #285679 Post edited: 8 months ago Edit Post #285679 Post edited: 8 months ago Edit Post #285680 Post edited: 8 months ago Edit Post #285680 Post edited: 8 months ago Edit Post #285680 Post edited: 8 months ago Edit Post #285680 Post edited: 8 months ago Comment Post #285681 Thanks. How can you rewrite the Generalized Vandermonde's Identity with a Summation Index, and Upper and Limits of Summation like Rothe-Hagen Identity? (more) 8 months ago Edit Post #285691 Initial revision 8 months ago Question Why ways to pick a 2-person committee from 4 people$3!$? Why aren't ways to form a 3-member committee from 8 people$8 \times 7 \times 6$? Why was https://math.codidact.com/posts/285679 closed as duplicate of https://math.codidact.com/posts/280168? They may involve the same concept, but the questions AND ANSWERS differ! (more) 8 months ago Edit Post #285680 Post edited: 8 months ago Edit Post #285680 Post edited: 8 months ago Edit Post #285680 Initial revision 8 months ago Question How can a 15 year old construe the LHS of Generalized Vandermonde's Identity, when it lacks summation limits and a summation index? Paradoxically, though Rothe-Hagen Identity (henceforth RHI)$\sum\limits{k=0}^n\frac{x}{x+kz}{x+kz \choose k}\frac{y}{y+(n-k)z}{y+(n-k)z \choose n-k}=\frac{x+y}{x+y+nz}{x+y+nz \choose n}$generalizes Generalized Vandermonde's Identity (henceforth GVI),$\sum\limits{k1+\cdots +kp = m} {n1\c...
(more)
8 months ago
Edit Post #285679 Post edited:
8 months ago
Edit Post #285679 Initial revision 8 months ago
Question How to intuit, construe multiplicands and multiplicators $\le 10$ resulting from $\dbinom pc$, WITHOUT division or factorials?
I grok, am NOT asking about, the answers below. Rather — how can I deduce and intuit the multiplicands and multiplicators $\le 10$, resulting from simplifying $\dbinom {p \text{ people}}{c\text{-person committee}}$ DIRECTLY? WITHOUT division or factorials! Orange underline 1. Unquestionably...
(more)
8 months ago
Comment Post #285670 I have, but I haven't found an expatiation of multiplication vs. addition. Most books merely present the rule and expect you to memorize it.
(more)
8 months ago
Edit Post #285673 Post edited:
8 months ago
Edit Post #285673 Post edited:
8 months ago
Edit Post #285673 Initial revision 8 months ago
Question How can you prognosticate to rewrite the same sum backwards, then add the same sum (twice)?
This comment doesn't fulfill me, because it doesn't demystify this trick of writing $Sn$ forward, then backwards, then adding. What would spur you to action these unnatural, unpredictable steps? How would you vaticinate this trick? I'm hankering after an answer other than "You just have to memorize t...
(more)
8 months ago
Edit Post #285672 Post edited:
8 months ago
Edit Post #285671 Post edited:
8 months ago
Edit Post #285672 Post edited:
8 months ago
Edit Post #285672 Initial revision 8 months ago
Question To correct for overcounting, why can't we divide by the sum of the over-counts?
The book doesn't expatiate the sentence underlined in red below. 1. Why "it isn't 2 + 2"? 2. Note that $2 \times 2 = 2 + 2 = 4$! Is this a coincidence? >Problem 3.4: How many distinct arrangements are there of PAPA? > >Image alt text David Patrick, BS Math & Computer Science, MS Math (C...
(more)
8 months ago
Edit Post #285671 Initial revision 8 months ago
Question How can you "easily see that such squares [of side length $\sqrt{13}$ and $\sqrt{18}$] will not fit into the [4 × 4] grid"?
>Problem 2.4: How many squares of any size can be formed by connecting dots in the grid shown in Figure 2.2. I skip p 31, but apprise me if you want me to include it. 1. Side lengths of squares must be equal. Thus how can $m \neq n$ below? 2. How do you most "easily see that such squares wil...
(more)
8 months ago
Edit Post #285670 Post edited:
8 months ago
Edit Post #285670 Initial revision 8 months ago
Question Why 1. multiply the number of independent options? 2. add the number of exclusive options?
Ironically, this textbook highlights understanding over memorization, but it doesn't expatiate the two WHY's in the question title! >When faced with a series of independent choices, one after the other, we multiply the number of options at each step. When faced with exclusive options (meaning ...
(more)
8 months ago
Edit Post #285596 Post edited:
8 months ago
Edit Post #285443 Post edited:
8 months ago
Edit Post #285613 Initial revision 8 months ago
Question Why isn't MathJax table appearing?
$\begin{array} {|r|r|}\hline \text{If I already earned a} & \text{then I must roll another} \\ \hline 9 & 1 \\ \hline 8 & 2 \\ \hline 7 & 3 \\ \hline 6 & 4 \\ \hline 5 & 5 \\ \hline 4 & 6 \\ \hline 3 & \text{Impossible: I need$\ge 2$more rolls to gain a 10.} \\ \hline 2 & Merge with cell above. \\... (more) 8 months ago Edit Post #285432 Post edited: 8 months ago Edit Post #285433 Post edited: 8 months ago Edit Post #285594 Post edited: 8 months ago Comment Post #285596 Thanks. Can you pls elaborate why this hitch doesn't happen on Stack Exchange, as far as I know? (more) 8 months ago Suggested Edit Post #285596 Suggested edit: (more) helpful 8 months ago Edit Post #285594 Post edited: 8 months ago Edit Post #285594 Post edited: 9 months ago Edit Post #285594 Initial revision 9 months ago Question Why doesn't \$ work?
Why isn't $1 rendering? Both \$1 and $\$$1 failed! (more) 9 months ago Edit Post #285524 Initial revision 9 months ago Question How's it possible to arrange 0 objects? How can 0! = 1? 1. I don't understand the following explanation. Isn't it physically impossible to "arrange 0 objects", or nothing? It's senseless to raise the concept of arrangement when you have nothing, like in a vacuum! 2. Division by zero is undefined because you can't divide something (e.g. cookies) by not... (more) 9 months ago Edit Post #285431 Post edited: 9 months ago Comment Post #285443 Completed. I just edited my post. (more) 9 months ago Edit Post #285443 Post edited: 9 months ago Edit Post #285443 Initial revision 9 months ago Question Why's Pr(the running total of a fair dice rolled repeatedly = n) =$\frac16 (p_{n-1} + p_{n-2} + p_{n-3} + p_{n-4} + p_{n-5} + p_{n-6})$? Where did$\color{red}{pn = \frac16 (p{n-1} + p{n-2} + p{n-3} + p{n-4} + p{n-5} + p{n-6})}$spring from? The solution doesn't expatiate, and makes it appear out of the blue? Can you please familiarize and naturalize me with this? I've tried to contemplate the ways that rolling can sum to$n$. For ... (more) 9 months ago Edit Post #285434 Post edited: 9 months ago Edit Post #285434 Initial revision 9 months ago Question "the people who move have a percentage of Democrats which is between these two values" — How does this furnish intuition for Simpson's Paradox? Why does the phrase colored in red below ($\color{red}{\text{the people who move have a percentage of Democrats which is between these two values}}$) matter? How does it assist with intuiting the peaceful coexistence of$P{new}(D|B) > P{old}(D|B)$and$P{new}(D|B^C) > P{old}(D|B^C)$? >59. The ... (more) 9 months ago Edit Post #285433 Initial revision 9 months ago Question Why's a 2-player game with "win by two" rule = Gambler's Ruin where each player starts with$2? Why N = 4?
1. Why can this problem "be thought of as a gambler's ruin where each player starts out with $2"? 2. Please see my red arrow. Why is this exponent 4? I quote op. cit. p 73. >Example 2.7.3 (Gambler's ruin). Two gamblers, A and B, make a sequence of \\$1 bets. In each bet, gambler A has probabi...
(more)
9 months ago
Edit Post #285429 Post edited:
9 months ago
Edit Post #285432 Post edited:
9 months ago
Edit Post #285432 Initial revision 9 months ago
Question How to intuit p = Calvin's probability of winning each game independently = $1/2 \implies$ P(Calvin wins the match) = 1/2?
Please see the sentence beside my red line. The notion of a "sanity check" suggests that these resultant integers should be obvious, without calculation or contemplation. But why's it plain and intuitive that $p = 1/2 \implies P(C) = 1/2$? Indubitably, a game differs from a match. Just because ...
(more)
9 months ago
Edit Post #285431 Initial revision 9 months ago
Question If C = Calvin wins the match, and $X \thicksim Bin(2, p) =$ how many of the first 2 games he wins — then why P(C|X = 1) = P(C)?
The author's solution doesn't expatiate why $\color{red}{P(C|X = 1) = P(C)}$? This similar question on Math Stack Exchange has 0 answers, as at 4 January 2022. >50. Calvin and Hobbes play a match consisting of a series of games, where Calvin has probability p of winning each game (independently)....
(more)
9 months ago
Edit Post #285430 Post edited:
9 months ago
Edit Post #285430 Post edited:
9 months ago
Edit Post #285430 Post edited:
9 months ago
Edit Post #285430 Initial revision 9 months ago
Question Intuitively, why does $p$ vary inversely with $P(C_3 \mid D_2)$? But directly with $P(C_2 \mid D_3)$?
I'm seeking merely intuition here — NOT about the algebra that I know how to execute. Please see the fractions colored in red and orange at the bottom. 1. $P(C3 \mid D2) = \color{red}{\dfrac1{ 1 + p}}$ means that $p$ is inversely related with $P(C3 \mid D2)$. How can you intuit this inverse re...
(more)
9 months ago
Edit Post #285429 Initial revision 9 months ago
Question For (a variation) of the Monty Hall problem, what permits you to repaint doors 2 and 3? What permits you to replace p with $1 - p$?
Please see the text colored in red at the bottom. 1. Why can you haughtily just — and what legitimizes you to — "[i]magine repainting doors 2 and 3, reversing which is called which"? This is completely unworkable, half-baked because you can't reverse doors in the real game! 2. What legitimate...
(more)
9 months ago
Edit Post #285365 Post edited:
9 months ago
Edit Post #285365 Post edited:
9 months ago
Edit Post #285371 Post edited:
9 months ago
Edit Post #285371 Initial revision 9 months ago
Question How does $P(C > D \mid C = 2) \neq P(C > D \mid C \neq 2)$ prove that B > C depends on C > D?
I grok that $\color{limegreen}{P(C > D \mid C = 2) = P(D = 1 \mid C = 6) = 1/2}$, and $\color{red}{P(C > D \mid C \neq 2) = P(C > D \mid C = 6) = 1}$. But I don't grok the last sentence in the quotation below, colored in blue. How do these two probabilities prove that B > C DEPENDS ON C > D? >3...
(more)
9 months ago
Edit Post #285364 Post edited:
9 months ago
Edit Post #285365 Post edited:
9 months ago
Edit Post #285365 Post edited:
9 months ago
Edit Post #285365 Initial revision 9 months ago
Question Acceptable, usual to write $\ge 2$ pipes simultaneously?
I'm NOT asking for the solution to this exercise that's publicly accessible. Rather, pls see the green and red underlines. If I apply the author's green definition to the red underline, then $\tilde P({\color{red}{L \mid M2}}) \equiv P(\color{red}{L \mid M2} \quad \color{limegreen}{\mid M1})$. Is...
(more)
9 months ago
Edit Post #285364 Post edited:
9 months ago
Edit Post #285364 Initial revision 9 months ago
Question In general, does $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$?
In this question, $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$. But the author's solution didn't annunciate this, and doesn't unfurl how to compute $\color{red}{P(B|M)}$. Can I simply subtract as $\color{red}{P(B|M)} = 1 - \color{forestgreen}{P(A|M)}$? I hanker to avoid calculating $\color... (more) 9 months ago Edit Post #284738 Initial revision 11 months ago Question Can the bijection for the Lost Boarding Pass Probability Problem, be formulated or pictured? The two proofs below moot "bijection", but I don't see a formula. 1. Does the bijection have an explicit formula? 2. Can this bijection be pictorialized? My 16 y.o. kid doesn't understand the abstract, Daedalian phrasing below. hunter's answer dated Dec. 4 2013 >Claim 3: There is a biject... (more) 11 months ago Edit Post #282643 Post edited: about 1 year ago Edit Post #282966 Post edited: about 1 year ago Comment Post #284218 How does your answer address my question? If it doesn't, please delete your answer? You expounded the algebra that I grokked, but not the proof strategy that I was asking about. (more) about 1 year ago Edit Post #282645 Post edited: about 1 year ago Edit Post #283252 Post edited: about 1 year ago 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... (more) about 1 year ago 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)
Edit Post #283887 Post edited:
Edit Post #283887 Initial revision about 1 year 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)
Edit Post #283886 Post edited:
Edit Post #283886 Initial revision about 1 year 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)
Edit Post #283388 Post edited:
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)
Edit Post #283255 Post edited:
Edit Post #283255 Post edited:
Edit Post #283255 Post edited:
Edit Post #283255 Post edited:
Edit Post #283118 Post edited:
Edit Post #282942 Post edited:
Edit Post #282942 Post edited:
Edit Post #282942 Post edited:
Edit Post #283388 Post edited:
Edit Post #283260 Post edited:
Edit Post #283260 Post edited:
Edit Post #283260 Post edited:
Edit Post #283388 Post edited:
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)
Comment Post #283387 @Moshi I don't think "By definition" answers my question? Why not define these two probabilities the other way around?
(more)
Edit Post #283388 Post edited:
Edit Post #283388 Post edited:
Edit Post #283388 Initial revision about 1 year 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)
Edit Post #283387 Initial revision about 1 year 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)
Edit Post #283371 Initial revision about 1 year 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, $\dfrac{1 - (q/p)^i}{1 - (q/p)^N} ... (more) about 1 year ago Comment Post #283299 Sorry. I think I was editing the old question, but I copypasted my draft for my new question in the wrong box. What would you like me to do? Create a new question for the the Alice question? (more) about 1 year ago Comment Post #283317 "For this equation to hold (in general) the exponent must be k = 0 and not k + 1 = 1". What do you mean These are equivalent.$k = 0 \iff k + 1 = 1$. (more) about 1 year ago Comment Post #282945 I don't think you answered by second question above on the "simple geometric interpretation"? (more) about 1 year ago Edit Post #283297 Post edited: about 1 year ago Edit Post #283297 Post edited: about 1 year ago Edit Post #283297 Post edited: about 1 year ago Edit Post #283297 Initial revision about 1 year ago Question How to vaticinate calculating$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$? 2. Even after reading the solution, I wouldn't have prognosticated to calcul... (more) about 1 year ago Edit Post #283295 Post edited: about 1 year ago Edit Post #283295 Initial revision about 1 year ago Question In$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$? 2. Is this question related to the Fence Post Error? Have I committed it? Image alt text Tsitsikli... (more) about 1 year ago Edit Post #283260 Post edited: about 1 year ago Edit Post #283260 Post edited: about 1 year ago Edit Post #283291 Initial revision about 1 year ago Question Why was my question closed: If Alice must've have classes on at least 2 days, why do you need the intersection of 3 's? 1. I don't know why the MathJax isn't processing at https://math.codidact.com/posts/282645. 2. But why was it closed "as not constructive"? >This question cannot be answered in a way that is helpful to anyone. It's not possible to learn something from possible answers, except for the solution... (more) about 1 year ago Edit Post #283115 Post edited: about 1 year ago Comment Post #282945 1. Thanks for your answer. I agree that P(A) + P(B) can$>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? (more) about 1 year ago 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 (more) about 1 year ago 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? (more) about 1 year ago 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... (more) about 1 year ago Edit Post #283260 Post edited: about 1 year ago Edit Post #283260 Initial revision about 1 year ago 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)
Edit Post #283255 Post edited:
Edit Post #283255 Initial revision about 1 year 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)
Edit Post #283254 Post edited:
Edit Post #283253 Post edited:
Edit Post #283254 Post edited:
Edit Post #283254 Initial revision about 1 year 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)
Edit Post #283253 Post edited:
Edit Post #283253 Post edited:
Edit Post #283253 Post edited:
Edit Post #283253 Initial revision about 1 year 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)
Edit Post #283111 Post edited:
Edit Post #283252 Initial revision about 1 year 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)
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)
Edit Post #283118 Post edited:
Edit Post #283118 Initial revision about 1 year 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)
Edit Post #283116 Initial revision about 1 year 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)
Edit Post #283115 Initial revision about 1 year 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 $P(M2) \neq ... (more) about 1 year ago Edit Post #283113 Initial revision about 1 year ago Question In the Monty Hall problem, why can you just assume the contestant picked door 1? Why are you entitled to relabel the doors, or rewrite this solution with the door numbers permuted? My bafflement ought be obvious. 1. A contestant could've picked doors 2, 3. So you can't just assume he picked door 1. 2. Correct me if I'm wrong, but the game show didn't authorize contestants "to relabel the doors, or" permute the door numbers. So what permits you to do any of this in this solut... (more) about 1 year ago Comment Post #282600 That question got removed, and your link no longer works. (more) about 1 year ago Edit Post #283112 Post edited: about 1 year ago Edit Post #283112 Post edited: about 1 year ago Edit Post #283112 Post edited: about 1 year ago Edit Post #283112 Initial revision about 1 year ago Question If a 2nd test's independent from the 1st test, then why does$\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}$, that I underlined in red and purple, stem from? >### Example 2.6.1 (Testing for a rare disease, continued). >Fred, who tested posi... (more) about 1 year ago Edit Post #283111 Initial revision about 1 year ago Question Why's the true positive rate termed Sensitivity and true negative rate Specificity, not vice versa? 1. To wit, what's "Sensitive" about True Positive Rates, and "Specific" about True Negative Rates? 2. Why weren't these Metaphors) or Imports reversed? Why wasn't "Sensitive" termed to signify True Negative Rates, and "Specific" True Positive Rates instead? I learn best visually. Can these g... (more) about 1 year ago Edit Post #283110 Post edited: about 1 year ago Edit Post #283110 Initial revision about 1 year ago Question How do these 3 bell curves of Likelihood, Posterior, Prior pictorialize the Odds form of Bayes' rule? I learn best visually, and I found these graph. 1. Does it furnish intuition on Theorem 2.3.5 below? 2. E.g. Is the Likelihood Ratio always graphically left of Posterior and Prior? If so, why? >This can also be pictorially represented – the graph below shows the new posterior belief for a cer... (more) about 1 year ago Edit Post #282966 Initial revision about 1 year ago Question What's wrong with evaluating$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$into this exp... (more) about 1 year ago Edit Post #282943 Post edited: about 1 year ago Edit Post #282943 Initial revision about 1 year ago Question Intuitively, why does A, B independent$\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... (more) about 1 year ago Edit Post #282942 Post edited: about 1 year ago Edit Post #282942 Post edited: about 1 year ago Edit Post #282942 Initial revision about 1 year ago 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)$? >#... (more) about 1 year ago Comment Post #282892 Thanks. Can you pls elaborate why terms like$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. (more) about 1 year ago Edit Post #282890 Initial revision about 1 year 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)
Edit Post #282605 Post edited:
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)
Edit Post #282889 Initial revision about 1 year 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)
Edit Post #282888 Initial revision about 1 year 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)
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)
Edit Post #282873 Post edited:
Edit Post #282873 Post edited:
Edit Post #282873 Initial revision about 1 year 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)
Edit Post #282771 Initial revision about 1 year 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)
Edit Post #282645 Post edited:
Edit Post #282645 Post undeleted about 1 year ago
Edit Post #282645 Post edited:
Edit Post #282642 Post edited:
Edit Post #282645 Post deleted about 1 year ago
Edit Post #282666 Initial revision about 1 year 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)
Edit Post #282643 Post edited:
Edit Post #282643 Post edited:
Edit Post #282643 Post undeleted about 1 year ago
Edit Post #282643 Post edited:
Edit Post #282643 Post deleted about 1 year ago
Edit Post #282645 Post edited:
Edit Post #282645 Post edited:
Edit Post #282645 Post edited:
Edit Post #282645 Post edited:
Edit Post #282645 Post edited:
Edit Post #282644 Post edited:
Edit Post #282645 Post edited:
Edit Post #282645 Initial revision about 1 year 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)
Edit Post #282644 Post edited:
Edit Post #282644 Initial revision about 1 year 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)
Edit Post #282643 Initial revision about 1 year 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)
Edit Post #282642 Post edited:
Edit Post #282642 Post edited:
Edit Post #282642 Post edited:
Edit Post #282642 Initial revision about 1 year 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)
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)
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)
Edit Post #282605 Post edited:
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)
Edit Post #282605 Post edited:
Comment Post #282611 Absolutely not! Manners please? I couldn't copy and paste from the second website.
(more)
Edit Post #282606 Post edited:
Edit Post #282614 Initial revision about 1 year 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)
Edit Post #282613 Initial revision about 1 year 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)
Edit Post #282612 Initial revision about 1 year 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)
Edit Post #282611 Initial revision about 1 year 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)
Comment Post #282602 Thanks. I fixed my typo.
(more)
Edit Post #282602 Post edited:
Edit Post #282609 Initial revision about 1 year 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)
Edit Post #282608 Initial revision about 1 year 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)
Edit Post #282607 Initial revision about 1 year 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)
Edit Post #280168 Post edited:
Edit Post #280168 Post edited:
Edit Post #280168 Post edited:
Comment Post #281319 Thanks. Does my edit [to my post] change your answer?
(more)
Edit Post #280168 Post edited:
Edit Post #282606 Initial revision about 1 year 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)
Edit Post #282605 Post edited:
Edit Post #282605 Post edited:
Edit Post #282605 Initial revision about 1 year 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)
Edit Post #282604 Initial revision about 1 year 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)
Edit Post #282603 Post edited:
Edit Post #282603 Initial revision about 1 year 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)
Edit Post #282602 Post edited:
Edit Post #282602 Initial revision about 1 year 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)
Edit Post #282134 Initial revision over 1 year 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)
over 1 year ago
Edit Post #281987 Initial revision over 1 year 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)
over 1 year ago
Edit Post #280741 Post edited:
over 1 year ago
Comment Post #280741 I posted at https://math.codidact.com/posts/280742 about the miffed Mathjax.
(more)
over 1 year ago
Edit Post #280742 Initial revision over 1 year 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)
over 1 year ago
Edit Post #280741 Initial revision over 1 year 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...
(more)
over 1 year ago
Edit Post #280168 Initial revision over 1 year 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...
(more)
over 1 year ago