Reputation
Next tag badge:
989/1000 score
359/200 answers
Badges
24 302 601
Newest
 Necromancer
Impact
~6.1m people reached

3h
comment Counting matchings, the modern way
Line up the people in order of student number. The first person in the list chooses a mate, and they flip a coin to decide who will be the bride. The first person not so far chosen does the same, and so on. So we get $(2k-1)(2)(2k-3)(2)\dots$.
4h
comment Find $E(X_1X_2 \mid X_2 X_3)$ for i.i.d. symmetric Bernoulli random variables $X_k$
Note that $X_2X_3$ is irrelevant.
4h
revised When randomly distributing n points amongst m people, what are the odds that one certain person will get a certain amount of points?
added 121 characters in body
4h
answered When randomly distributing n points amongst m people, what are the odds that one certain person will get a certain amount of points?
4h
comment Stating the induction hypothesis
I would strongly avoid 3). To me it says assume there exists an $n$ such that $\dots$. Then $n$ has been quantified away and is no longer free. Something like suppose that $k$ is a natural number such that $\dots$ seems fine.
6h
comment Probability, expected frequency and resultant distribution skewed or not?
What has mean $2$? I suspect the problem has been incompletely transcribed.
8h
comment defective component and probability
Part (a) is entirely different from your previous question. Here the superscript $C$ (which should probably be $c$) refers to the complement and not to the event $C$. In particular, the first equality is true. (The second is not.)
10h
comment Finding conditional probability using Baytes formula
$\Pr(B|A_2)=0.03$.
10h
comment Find elements $x,y$ where $x\ne \pm1$ and $y\ne \pm 1$ in the field $\mathbb{Q}(\sqrt{5})$ satisfying $xy=19$.
Trivial in the field. In the ring of integers of this field, use $8+3\sqrt{5}$ and its conjugate.
10h
comment What is the maximum value of $M$ when $T$ is set of $\{2,4,8,16,… 2^n\}$ and $S$ is subset of $T$ by given conditions
If you are using $2^n-2$, and $n\ge 3$, the maximum number of numbers in $T$ such that no sum of two is greater than $2^n-2$ is $n-1$. The biggest set $S$ uses everybody but $2^n$ Your "let $m$ be $M$ number" is not clear. And the entire problem, as written, is far too easy. I wonder whether it was originally not in English, and got mistranslated.
16h
comment What is the maximum value of $M$ when $T$ is set of $\{2,4,8,16,… 2^n\}$ and $S$ is subset of $T$ by given conditions
But later in the post you added three, and then four. And the two you added was for the case $n=2$, and my comment above said explicitly $n\ge 3$.
19h
comment Straight Lines and Cordinate Geometry
$(x-1)^2+(y-2)^2+(x-3)^2+(y-4)^2=25$. Expand, and compare with the given equation.
19h
comment What is the maximum value of $M$ when $T$ is set of $\{2,4,8,16,… 2^n\}$ and $S$ is subset of $T$ by given conditions
The problem says the sum of no two elements of $S$ is greater than (I think now) $2^n-2$. That just means (for $n\ge 3$) that we must leave out $2^n$, all the others can be allowed into $S$. Too simple!
19h
comment What is the maximum value of $M$ when $T$ is set of $\{2,4,8,16,… 2^n\}$ and $S$ is subset of $T$ by given conditions
I assumed you meant $2^{n-2}$. Then it is $5$. But maybe you meant $2^n-2$. Then the answer is different, but looking at a few examples will give it to you.
20h
comment What is the maximum value of $M$ when $T$ is set of $\{2,4,8,16,… 2^n\}$ and $S$ is subset of $T$ by given conditions
Let $n=8$, so we are looking at $2,4,8,16,32,64,128,256$. Note that $256$ can't be in $S$, neither can $128$, neither can $64$, but everybody else can be.
20h
comment Right triangle trigonometry help?
By similar triangles $a/d=200/150$.
20h
comment How can I prove $e^{i\theta}=\cos\theta+i\sin\theta$?
One way is to use power series.
20h
comment Closed-form expression for $\binom{n}{1}+3\binom{n}{3}+5\binom{n}{5}+\cdots$
A full solution that uses that has been posted.
20h
comment Closed-form expression for $\binom{n}{1}+3\binom{n}{3}+5\binom{n}{5}+\cdots$
Note that for example $5\binom{n}{5}=n\binom{n-1}{4}$.
20h
comment Proving inequalities with the Archmedian property
If $\epsilon$ is positive then $\frac{1}{n}\lt \epsilon$ if and only if $n\gt \frac{1}{\epsilon}$.