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5

I agree with you and Nate Eldredge that the limit doesn't exist. However, if we replace $0$ with $+\infty$, then the limit exists and equals $e^{-a}$. Let $F$ denote the cdf of a standard normal, and let $\;f=F'$ be the pdf. Then $$\lim_{x \to \infty} \frac{P(X>x+\frac{a}{x})}{P(X>x)} = \lim_{x \to \infty} \frac{1-F(x+\frac{a}{x})}{1-F(x)}$$Now as $x ...


3

I understand the question but I am not sure how to solve it. For example, if we flip HHHTTTTT then the next three must be heads because of the question. This however seems counterintuitive. I believe that there are $2^{10}$ possible strings, but I am unsure of how to count all possible strings that begin with HHH. You don't understand the question. It ...


2

We do a formal conditional probability calculation. Let $A$ be the event the first $3$ tosses are heads, and let $B$ be the event we have an equal number of heads and tails in the $10$ tosses. We want $\Pr(A|B)$. By the definition of conditional probability, we have $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$ We calculate the two probabilities on the right. ...


2

The number of sequences of $10^6$ digits equally distributed is given by the multinomial coefficient $$\binom{10^6}{\underbrace{10^5,10^5,\ldots,10^5}_{10\text{ times}}}$$ And the total number of sequences is $10^{10^6}$. Then, the probability is $$\frac{(10^6)!}{(10^5)!^{10}\cdot10^{10^6}}$$


2

Your approach is correct. The problem is with the question. The values are not consistent. It can be proved that given $P(B)$ and $P(B \mid A)$, the value of $P(A \cup B)'$ should be at least $\frac{7}{15}$.


2

Added: The probability is $\frac{5}{8}$. Graham pretty much nailed it, but didn’t want to take away your chance to work out the answer on your own. Since you’ve marked his answer correct, maybe it’s not so bad to give a full answer here. First note that after $n$ balls are selected, and the $n$-th replacement of $1+4$ balls is made, the number of balls in ...


2

Hint: If you remove the known element $k+1$ from such a subset it becomes an arbitrary $r$-element subset of $[k]$. Second Hint: An $(r+1)$-element subset of $[n+1]$ is an $(r+1)$-element subset of $[n+1]$ that has $k+1$ as maximal elelment for exactly one $k\in\{r,\ldots,n\}$.


2

I agree with your analysis: the limit does not exist, since the right- and left-sided limits differ. I suspect that either the original problem was erroneous, or perhaps you made a mistake when copying it.


2

Each admissible selection of $5$ players can be encoded as a string of length $12$ consisting of $5$ ones and $7$ zeros with at least one zero between any two ones. Therefore we begin with the $7$ zeros: $$\cdot0\cdot0\cdot0\cdot0\cdot0\cdot0\cdot0\cdot\quad.$$ The eight dots represent slots into each of which a single one may be inscribed. There are ...


2

We supposed to have only one identical pair in $k$ numbers.So we have $k-1$ different numbers in our groups,right? First, we have $P(n,k-1)$ ways to select $k-1$ different numbers from $n$ different numbers and after that we have $C(k,2)$ different states for selecting the paired number.


1

We are looking for a real number $t$ such that $$ P(X\leq t)=0.75 $$ or equivalently $$ \int_0^tc e^{-cx}dx = 0.75 $$ $$ \left[-e^{-cx}\right]_0^t = 0.75 $$ $$ -e^{-ct}+1 = \frac34 $$ $$ t=\frac{\ln 4}c. $$ We apply the precedent result, putting $c:=\dfrac45$, to get $$ t=\frac{\ln 4}{4/5}\approx1.73 $$ that is about 1 hour and 44 minutes to see 75% of ...


1

$B= \{1,3,4,5,6\}^2$. $A= \{ (i,j) | i+j = 7, i, j \in \{1,2,3,4,5,6\} \}$.


1

We may exploit the Lagrange identity: $$ (a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2 \tag{1}$$ to state: $$ I_z=\int_{-\infty}^{+\infty}\frac{dx}{(1+x^2)(1+(z-x)^2)}=\int_{-\infty}^{+\infty}\frac{dx}{(1+x(z-x))^2+(z-2x)^2}\tag{2}$$ and by replacing $x$ with $x+\frac{z}{2}$ in the last integral, we get: $$ I_z = ...


1

This random walk, like many others, can be modeled by using a Markov chain. See Chapter 11 of the on-line probability book by Grinstead and Snell (it can be freely downloaded). You need to know a bit of linear algebra to understand Markov chain theory.


1

This is correct, other than getting your computation wrong. Check wythatoras' comment for the correct value. As there is only a single instance of 4 and 5 in the permuted set, 9! does not double count on the possible positions of 4 or 5. The probability of never rolling 6 is similarly accounted for- if you roll three 1s, two 2s, two 3s, a 4 and a 5, no dice ...


1

Let us consider the simplest case where $X_{i}$ are discrete random variables. In this case we can explicitly write down everything in a few cases. Let $X, Y$ be two biased fat coins with probability $0.3/0.7$ of head and tail respectively. Then individually you have $$ E(X)=E(Y)=0.3*1+0.7*0=0.3 $$ But you can also calculate it in a different way. Rather ...


1

If $A$ means three heads at the beginning, and $B$ means $5$ heads and $5$ tails, we want to compute $$p(A/B)=\frac{p(A\cap B)}{p(B)}$$ And $$p(A\cap B)=\frac{\binom72}{2^{10}}$$ $$p(B)=\frac{\binom{10}5}{2^{10}}$$


1

One can frequently read that the product of two multivariate Gaussian pdfs, $f_1(x)$*$f_2(x)$, is itself a Gaussian, This is an incorrect statement no matter how many times or in how many different places you have read it; indeed it does not hold even for univariate Gaussian random variables. If the link you have provided does indeed say exactly what ...


1

\begin{align} \Pr(Y<140) & = \Pr\left( \frac{Y-150}{\sqrt{120}} < \frac{140-150}{\sqrt{120}} \right) \\[10pt] = {} & \Pr\left(Z<\frac{140-150}{\sqrt{120}}\right) = \Phi\left(\frac{140-150}{\sqrt{120}}\right). \end{align}



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