Activity for DNB
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #282943 |
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— | over 3 years ago |
| Edit | Post #282943 | Initial revision | — | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282942 |
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| Edit | Post #282942 |
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| Edit | Post #282942 | Initial revision | — | over 3 years 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) |
— | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282890 | Initial revision | — | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282605 |
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— | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282889 | Initial revision | — | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282888 | Initial revision | — | over 3 years 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) |
— | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282873 |
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| Edit | Post #282873 |
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| Edit | Post #282873 | Initial revision | — | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282771 | Initial revision | — | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282645 |
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| Edit | Post #282645 | Post undeleted | — | over 3 years ago |
| Edit | Post #282645 |
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| Edit | Post #282642 |
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| Edit | Post #282645 | Post deleted | — | over 3 years ago |
| Edit | Post #282666 | Initial revision | — | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282643 |
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| Edit | Post #282643 |
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| Edit | Post #282643 | Post undeleted | — | over 3 years ago |
| Edit | Post #282643 |
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| Edit | Post #282643 | Post deleted | — | over 3 years ago |
| Edit | Post #282645 |
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| Edit | Post #282645 |
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| Edit | Post #282645 |
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| Edit | Post #282645 |
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| Edit | Post #282645 |
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| Edit | Post #282644 |
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| Edit | Post #282645 |
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| Edit | Post #282645 | Initial revision | — | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282644 |
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— | over 3 years ago |
| Edit | Post #282644 | Initial revision | — | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282643 | Initial revision | — | over 3 years 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) |
— | over 3 years ago |
| Edit | Post #282642 |
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— | over 3 years ago |