Tag Info

New answers tagged

3

The probability of NOT getting one of those $8$ numbers for four rolls would be $$x=\frac{92}{100}\cdot\frac{92}{100}\cdot\frac{92}{100}\cdot\frac{92}{100}$$ So the probability of hitting it at least once would be $(1-x)$


0

It is clearly possible if $m+1 = 2^k$ for some $k \in \mathbb{N}$, by tossing $k$ fair coins. It is impossible for all other $m$ if you require a deterministic procedure that always halts, simply because the number of possibilities for $n$ fair coin tosses is $2^n$ which is only divisible by powers of $2$. If you want determinism but do not mind ...


1

we know the mean $E[\epsilon_j]$=0 so the variance $E[\epsilon_j^2] - (E[\epsilon_j])^2$ is $E[\epsilon_j^2]$ which is 1 due to independence $var(Z_i)=\sum_{j=1}^i var(\epsilon_j)=\sum_{j=1}^i 1 = i$


0

Assuming $X$ and $Y$ are independent, the distribution of $\dfrac{Y^2}{X^2}$ (for $z>0$) is given as: $$ \begin{eqnarray*} P\left(\dfrac{Y^2}{X^2}<z\right){}={}P\left(Y^2<z\,X^2\right)&{}={}&P\left(Y<\sqrt{z}\,X\right)\,,\,\,\,\,\,\,\mbox{ since }Y\,,\,X\,,z>0\newline ...


0

Consider the function $g(x,y)=\left(\dfrac{y}{x},x\right)$. Define $u=\dfrac{y}{x}$ and $v=x$. Then $uv=y$. Then the inverse of $g$ will be $h(u,v)=(v,uv)$. Let's call the pdf of a random variable by $f$ (you call it $p$). By the transformation theorem we know that $f_{ZX}(z,x)=f_{XY}(h(z,x))|\mathbf J(h)|$, where $|\mathbf J(h)|$ is the absolute value of ...


1

I agree with Joel's answer, but it doesn't address the question of random directions. See the first sentence of http://arxiv.org/pdf/math/0701658.pdf which was subsequently published. For rational billiards (ie, all angles a rational multiple of $\pi$), the dynamics is ergodic in almost all directions, and in particular it is uniformly distributed in ...


2

We make an interpretation of your problem. We are choosing $n+1$ times, with replacement, a number from the set $\{0,1,2,\dots,100\}$, where all numbers are equally likely to be chosen. Let random variables $X_1, X_2,\dots, X_n,X_{n+1}$ be the sequence of numbers chosen. We want the probability that $X_{n+1}$ is different from all the $X_i$, where $1\le ...


0

By definition, an unbiased coin is fair. So, I suspect the answer is "don't know". Let $F=${"the coin is fair"} and $B=${"the coin is biased"} and $P\left ( B\cup \bar{B} \right )=P\left ( B \right ) + P\left ( \bar{B} \right ) = 1$ (obvious). Then $$P\left ( F | B \cup \bar{B} \right ) = \frac{P\left ( F\cap \left ( B \cup \bar{B} \right ) \right ...


0

Let $E$ be the expected number of wrong guesses. With probability ${1\over9}$ the first guess is correct. With probability ${8\over9}$ the first guess is wrong, and $E$ more wrong ones are to be expected. It follows that $$E={1\over9}\cdot 0+{8\over9}\cdot(1+E)\ ,$$ from which we deduce that $E=8$.


4

You could theoretically guess wrong an arbitrarily large number of times. The probability that you guess wrong $n-1$ times and get it right on the $n^\text{th}$ time is $$P(\text{takes } n \text{ guesses}) = \frac{1}{9}\left(\frac{8}{9}\right)^{n-1}$$ We can find the expected value of the number of guesses it will take to get it right as an infinite sum: ...


0

Let the event Y be guessing the successful number. Then we can say that Y follows a geometric distribution: http://en.wikipedia.org/wiki/Geometric_distribution And I think it is safe to say that a person could make any countable number of incorrect guesses. Because the number changes after each guess, each guess is an independent event. That wikilink has ...


2

The numer of wrong guesses may be arbitrary large. In the case from your example $$ P(W=n)=\left(\frac89\right)^n\frac19, $$ where $W$ means the numer of wrong guesses.


4

Take your paper and cut it into $100$ smaller (equal sized) pieces. Write each of the numbers $1$ through $100$ on the pieces. Then mix up the pieces and pick one at random. If you don't want to do $100$ slips, use $10$ slips, labelled $0,1,...,9$. Pick twice with replacement for the two digits. If you get $00$, that's $100$.


0

Because $-1$ is not a quadratic residue modulo a prime $p\equiv 3 \mod 4$ and the product of two non-residues is a residue. A (non-zero) quadratic residue $r=d^2\mod p$ has the two square roots $\pm d$ which differ by a factor of $-1$. If $d$ is a non-residue, $-d$ is a residue.



Top 50 recent answers are included