Hot answers tagged probability
3
As $a,b$ are $n$-digit numbers, we have $10^{n-1}\le a,b<10^n$. If $b|a$, this implies $\frac ab\in\{1,2,\ldots,9\}$. So at most $9$ of the $9^n$ possible $b$ are divisors of $a$. $p(n)\le \frac 9{9^n}$.
While this estimate is far from sharp, it shows $p(n)\to0$ as $n\to\infty$.
Can we calculate $p(n)$ more precisely? Your forbidding zeroes is partly ...
3
You can prove that the $n$-dimensional Gaussian is invariant under transformation by $T$ for any orthogonal matrix $T$. (This is well known)
For orthogonal $T$ we have $\|TX\|=\|X\|$ hence $T\underline W = \sqrt n\frac{TX}{\|TX\|}$. Therefore the distribution of $W$ is also invariant under transformation by $T$
So the distribution of $W$ is invariant under ...
2
The statistical analysis given in the other answer seems fine.
But the discussion of integrals is unnecessary, I think. The standard Monte-Carlo technique for estimating $\pi$ is to generate $N$ random points in a square centered at the origin -- say the square $[-1,1] \times [-1,1]$. After generating each point $(x,y)$, you can check to see if $x^2 + y^2 ...
2
If you have $N$ independent random variables with densities $f_1,\ldots,f_N$, then the joint density is simply $$
f(x_1,\ldots,x_N) = f_1(x_1)\cdot\ldots\cdot f_N(x_N)
$$
The join density of $N$ independent random variables with $X_i \sim \textrm{Bin}(m,p)$ is thus $$
f(x_1,\ldots,x_N) = \prod_{i=1}^N \underbrace{\binom{m}{x_i} ...
1
Note: I wrote this answer before question 3 became a separate part of the question; this answer only answers questions 1 and 2. Question 3 seems harder.
Every directed graph arises in this way. Consider probability distributions over all $n!$ possible total orders of the values of the $X_i$, and associate a potential directed edge $i\to j$ with the ...
1
We calculate the probability of rolling at least one double-six in $24$ tosses. The probability we roll a double-six is, as you point out, $\frac{1}{36}$.
so, on any toss, the probability of not getting a double six is $\frac{35}{36}$.
The probability of "failure" $24$ times in a row is therefore $\left(\frac{35}{36}\right)^{24}$.
So the probability of ...
1
This question uses the inclusion exclusion principle:
$\Pr(F \cup S) = \Pr(F) + \Pr(S) - \Pr(F \cap S)\tag{1}$
$\cup$ is the math notation for "union" which is equivalent to "OR"
$\cap$ is the math notation for "intersection" which is equivalent to "AND"
Pr(F) = Probability of Females
Pr(S) = Probability of Sophomores
$\Pr(F \cap S)$ = Probability of ...
1
No, there are a few necessary conditions, I don't know if there's a sufficient one.
If $g(t) = \mathbb E(e^{tX})$ we must have $g(0) = 1$ and $\lim_{t\to -\infty}g(t) = \mathbb P(X=0) \in [0,1]$.
we also have $\mathbb E(X) = f'(0)$ and Var$(X) = f''(0) - f'(0)^2 >0$.
We also have by Jensen's inequality
$$\begin{array}{rl}g(t) &=\mathbb E(e^{tX}) ...
1
Let $X_1$ and $X_2$ be independent Bernoulli$(0.5)$ random variables ($1$ with probability $\frac 12$, $0$ with probability $\frac 12$) and set $X_3 = |X_1-X_2|$.
Then $X_1$, $X_2$ and $X_3$ are pairwise independent Bernoulli$(0.5)$ random variables, but $X_1 + X_2 + X_3$ is always even. This is not the same as the iid case.
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