Activity for DNB
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Question | — |
2 construals "of 100 patients presenting with a lump like the claimant’s in Gregg v Scott, 42 will be ‘cured’ if they are treated immediately." Are there official terms for these 2 different interpretations of the same statistic? >Lord Hoffmann and Baroness Hale advanced the following arguments against awarding damages for the pure loss of a chance of being cured: > >(1) The Hotson82 problem. In cases such as Gregg v Scott, it may well... (more) |
— | 12 months ago |
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How to intuit that if you destroy someone else's lottery ticket, you didn't deprive her of a chance of winning the lottery? Despite perusing the following at least 20 times, I'm still tempted (in dereliction of the authors' advice to resist this temptation) to answer $\color{red}{\text{"Yes – though it was a very small chance"}}$ to the question whether Friend deprived Ungrateful of a chance of winning the National Lott... (more) |
— | 12 months ago |
| Question | — |
25% probability that there was a chance of avoiding injury $\quad$ vs. $\quad$ 25% chance of avoiding injury I ask about merely the math behind the last sentence of footnote 71 quoted below. I quote the legalistic sentences thereinbefore for context, but they may be immaterial. How does "a 25% probability that there was a chance of avoiding injury" differ from "25% chance of avoiding injury"? Alas, my mi... (more) |
— | 12 months ago |
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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) |
— | almost 2 years ago |
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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) |
— | about 2 years ago |
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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) |
— | about 2 years ago |
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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) |
— | about 2 years ago |
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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) |
— | about 2 years ago |
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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) |
— | about 2 years ago |
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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) |
— | about 2 years ago |
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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) |
— | over 2 years ago |
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Why doesn't \$ work? Why isn't $1 rendering? Both `\$1` and `$\$$1` failed! (more) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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"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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |
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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) |
— | over 2 years ago |