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19

Peter's answer gets to the heart of the matter relatively quickly, but I feel it would also be best to demonstrate where the $6/\pi^2$ comes from, seemingly out of nowhere to the uninitiated. So, it should be obvious that $a/b$ is in simplest form if and only if $a,b$ are coprime, i.e. $\gcd(a,b) = 1$. Well, what does that mean? It means that $a,b$ share no ...


16

The sentence "The probability of picking a number at random and having it be prime is zero." is, mathematically speaking, either very sloppy or plain wrong (depending on who you ask) and, if you ask me, is a clear demonstration of why there's always a bit of tension between mathematicians and physicists. We call them sloppy, they call us hair splitters. The ...


9

The only way to get $k$ as an answer is if all rolls are $1$, so the probability is $\frac{1}{6^k}$. To get $k+1$ all rolls must be $1$ except for one roll which is a $2$; there are therefore $k$ different ways to do this and the probability is $\frac{k}{6^k}$.


9

Expectation is easy enough (Variance seems like more work). We have four states, according to how much of $ABC$ is complete. Thus the states are $\emptyset, A, AB, ABC$. Of course, Start is $\emptyset$ and End is $ABC$. For a state $\mathscr S$ we denote by $E[\mathscr S]$ the expected number of steps, given that you are starting from $\mathscr S$. The ...


7

Formally $X_i$ are real valued (measurable) functions i.e $X_i\colon \Omega\to \mathbb{R}$ for $i=1,2$ where $\Omega$ is the sample space. So $$ Y_i=e^{X_i} $$ for $i=1,2$ as an equality of functions means that $$ Y_i(\omega)=e^{X_i(\omega)} $$ for all $\omega\in \Omega$. Hence since $X_i(\omega)$ is a real number it follows using the first part of your ...


7

Your are correct, and the proof is rather simple (not requiring the wall of text you wrote :) $$\begin{align}P(\neg B|C)&=\frac{P(\neg B \land C)}{P(C)} &\text{by definition} \\&= \frac{P(C) - P(B\land C)}{P(C)} & \text{Because $B\land C$ and $\neg B\land C$ form a partition of $C$} \\&=\frac{P(C)}{P(C)}-\frac{P(B\land C)}{P(C)}&\...


7

Because it gets rounded up to the nearest half hour, the expected fee is simply $$$150\left(\frac{0.5^2}{16}-\frac{0^2}{16}\right)+$300\left(\frac{1^2}{16}-\frac{0.5^2}{16}\right)+\cdots+$1200\left(\frac{4^2}{16}-\frac{3.5^2}{16}\right)$$ (the probability it lands in each half hour multiplied by the amount received in that half hour) which is just $$$1200-...


6

(This is more like a comment with images) Here are some simulations of the values $c = c(p)$ using the grid of size $1000\times1000$ and $500$ steps together with some fitting curves. $\hspace{8em}$ The data clearly deviates from the polynomial $2p^2$, and although the above plot may seem to suggest that $c(p)$ assumes a nice closed form, I believe that ...


6

Complementary counting! Your answer is $1-P(\text{only lentils})-P(\text{only chickpeas})$. This would then be $1-\displaystyle\frac{3}{10}\cdot\frac{2}{9}\cdot\frac{1}{8}-\frac{7}{10}\cdot\frac{6}{9}\cdot\frac{5}{8}=\boxed{0.7}$


6

This question is equivalent to ask for the probability that $a$ and $b$ are coprime. If $a$ and $b$ are random integers below $x$, then the probability $P(x)$ that $a$ and $b$ are coprime , satisfies $$\lim_{x\rightarrow\infty} P(x)=\frac{6}{\pi^2}\approx0.6079$$


6

If the first flip is head, then the die won't produce coins showing head, hence we will have exactly one head. (You could just leave the die alone). If the first flip is tails, then there are exactly three out of eight outcomes from the three extra coins that result in exactly one head. Hence $$p=\frac12\cdot 1+\frac12\cdot\frac38 $$


6

This is similar to the question, "does $\displaystyle \int \frac{f(x)}{g(x)} dx= \frac{\int f(x) dx}{\int g(x) dx}?$" The answer is no, and this will resolve your issue. We know that the probability of surviving given that he faced blank rounds is the probability of both divided by the probability of surviving. So the expression we're really looking for here ...


5

You calculated the probability that all three solve it. The probability that none solve it is $\frac23×\frac34×\frac45=\frac25$, and the probability that at least one does (the question asks for this) is the complement, or $\frac35$.


5

Since $(Z-1)^2\le XY$ is equivalent to $1-Z\le \sqrt{XY}$ or $1-\sqrt{XY}\le Z$, it follows that $$\eqalign{\mathbb{P}((Z-1)^2\le XY)&=\mathbb{P}(1-\sqrt{XY}\le Z)\cr &=\int_{x=0}^1\int_{y=0}^1\int_{z=1-\sqrt{xy}}^1dzdydx\cr &=\int_{x=0}^1\int_{y=0}^1\sqrt{xy}dydx\cr &=\int_{x=0}^1\sqrt{x}dx\int_{y=0}^1\sqrt{y}dy=\frac{4}{9}. }$$


5

I'm not sure if this is an answer, but I can't fit what I want to say in a comment. It's not a matter of what we are told, but of what is known. How we come by the information is important. If someone, known to be truthful, looks at the hand, and we ask her, "Are there at least two Kings in the hand," and she answer "Yes," then we are certainly in the ...


5

Sometimes a figure is worth 1000 words:


5

Your observation that every second jump you will be at an even position simplifies the problem a lot, because the only even position where you haven't already won is $0$. So when you are at position $0$, the probability that you will win in two steps is $(1-p)^2+p^2$ because you need to make two jumps in the same direction. If you don't win, then after two ...


5

It looks like you are trying to calculate the number of states and divide the number of states fitting some criteria by the total number of states. One thing to note with your approach is that not all of the states you sum up are equally likely. In particular if I asked what is the probability of the first coin flip being heads you would get 6/14, but we ...


5

There are multiple ways to flip $10$ coins and obtain $5$ heads. One possible way would be HHHHHTTTTT and another, equally likely one is HTHTHTHTHT. So how many sequences are there? It turns out that if we have $10$ flips, we can choose $5$ of them to be heads, and the rest would be tails, hence $10$ choose $5$.


5

You need to choose the $2r$ shoes from $n$ different pairs. So there are $\binom{n}{2r}$ ways to pick the different pairs, and once that is done, there are $2^{2r}$ ways to pick one shoe from all those pairs. So the number of "good" cases is $\binom{n}{2r}\cdot 2^{2r}$.


5

The intuition is that in an optimal strategy, the picker should be indifferent to what the guesser chooses. Suppose we just take $n=3$ for simplicity. Suppose the picker chooses $1$ with probability $p_1$, chooses $2$ with probability $p_2$, and $3$ with probability $p_3$. The selection of $p_1, p_2, p_3$ constitutes the picker's strategy. The ...


4

Yes, they are different. Suppose $p$ is irrational. Then for every fixed $n$, the even $A_n=p$ would never happen (because $A_n(\omega)$ must be one of $0,\frac1n,\frac2n,\dots,\frac{n-1}n,1$ for each $\omega$). Similarly if $p\in\mathbb{Q}\cap(0,1)$ with least denominator $m$, then $A_n=p$ would never happen for $n$ which are not a multiple of $m$. So $\...


4

Let $Z$ denote a standard normal random variable, and recall the well-known approximation $$ \mathbb P(Z\geq n)\approx \frac{e^{-n^2/2}}{n\sqrt{2\pi}},\qquad (1) $$ where I am using the symbol $a_n\approx b_n$ to mean that $\lim_{n\to\infty}\frac{a_n}{b_n}=1$ throughout this posting. Warmup: standard log-normal case. Let $X=e^Z$ be a standard log-normal ...


4

By symmetry, a particular order is just as likely as its reversal. And in the reversal, we pick at least one red and one green ball before we pick a blue one. But the probability of this is easy to calculate: The probability that red is the first colour to be picked, followed by green, is $$\frac{10}{60}\times \frac{20}{50}=\frac{1}{15}$$ And the ...


4

One of the possible applications of this fact is that it can be used to prove, that if $G$ is a non-abelian finite group, then $|\{g \in G | g^2 = e\}| \leq \sqrt{\frac{5}{8}}|G|$. The proof of this fact by Geoff Robinson can be found here. However, to avoid being accused of posting a link-only answer, I will quote the corresponding part of their post: ...


4

Hint You have $$\int_2^x \frac{dt}{t (\log t)^{1+2\epsilon}} = \frac{1}{(\log x)^{2 \epsilon}}-\frac{1}{(\log 2)^{2 \epsilon}}$$ Proving that $$\int_2^\infty \frac{dt}{t (\log t)^{1+2\epsilon}}$$ converges. You can then use a series / integral comparison.


4

Both are incorrect. You need to learn when to add and when to multiply and what the significance of each action is in counting. Very loosely speaking, you multiply when you are describing different steps in a larger process of describing an arrangement and you add when you are describing entirely different categories of arrangements. It is a bit more ...


4

Preliminaries Compute scaled even moments about the mean: $$ \begin{align} a_m(n)\,2^n &=\sum_{k=0}^n(n-2k)^{2m}\binom{n}{k}\\ &=\sum_{k=0}^n(n-2k)^{2m-2}(n-2k)^2\binom{n}{k}\\ &=\sum_{k=0}^n(n-2k)^{2m-2}\left(n^2-4k(n-k)\right)\binom{n}{k}\\ &=n^2a_{m-1}(n)\,2^n-\sum_{k=0}^n(n-2k)^{2m-2}4k(n-k)\binom{n}{k}\\ &=n^2a_{m-1}(n)\,2^n-\sum_{k=...


4

The key idea is to use the Laplace transform $\tfrac1{\sqrt{\pi}}\int_{\mathbb{R}_+}\exp(-xt)t^{-1/2}=\tfrac1{\sqrt{x}}$ therefore, the sum in question (denoted by $S_n$) becomes, by the binomial theorem $$ 2^n S_n=(1/\sqrt{\pi}) \int_0^{\infty} t^{-1/2}\left((\exp(-t)+1)^n-1\right)\underbrace{=}_{ibp}C_n\underbrace{\int_0^{\infty}\sqrt{t}e^{-t}(1+e^{-t})^...


4

You have $X\sim \mathcal U(-1;3)$ and $Y=X^2$ Now $Y\in(0;1)$ when $X\in(-1;0)$ and also when $X\in(0;1)$. So this interval for $Y$ is mapped to by two intervals for $X$. Ie, for all $0\leq y\lt 1$ we have $\{Y\leq y\} = \{-\surd y\leq X\leq\surd y\}$ However $Y\in[1;9)$ when $X \in[1;3)$. So this interval for $Y$ is mapped to by only one interval for $...


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