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2

There are ${^{10}C_{5}}$ ways to choose 5 of 10 people. That equals $252$. $${^{10}C_{5}} = \dfrac{10!}{5!\,5!} = 252$$ You want to count the ways to choose 2 of 4 women and 3 of 6 men. $${^{\Box}C_{\Box}}\cdot{^{\Box}C_{\Box}} = \dfrac{\Box!}{\Box!\,\Box!}\dfrac{\Box!}{\Box!\,\Box!} = \Box$$ Can you not fill in the boxes? Thank you, I can fill in ...


0

Bayes' Theorem states that: $$\begin{align} \mathsf P(A\mid B, C) & = \dfrac{P(B\mid A, C)\;P(A\mid C)}{P(B\mid C)} \\[2ex] \therefore p(\theta \mid \mathbf X,\alpha) & = \dfrac{p(\mathbf X\mid \theta, \alpha)\; p(\theta\mid \alpha)}{p(\mathbf X\mid \alpha)} \end{align}$$ Now $\mathbf X$ is a vector of data points $x_i$ each from a distribution ...


2

There are $7$ students, and the number of project assignments is $4+5+6=15$. Since every student did at least $2$ (presumably different!) projects, one overburdened student did $3$, and the others did $2$ each. The student who did $3$ can be chosen in $\binom{7}{1}$ ways. Now we must distribute the remaining projects, $2$ of each, between the remaining $6$ ...


1

$$D = b^2-4a = (4k)^2 - 4 \cdot 4 \cdot (k+2) = 16(k^2 - k - 2) = 16(k-2)(k+1)$$ We need $D \ge 0$ for real roots, which happens when $k \ge 2$ or $k \le -1$. Can you take it from here?


-1

It is much easier to prove if you consider the intersection of events ($\cap$). This is associative and commutative just like the set intersection operator. For any events A, B: $p(A|B) = \dfrac{p(A \cap B)}{p(B)}$ So we have $p(\theta | X,\alpha) = \dfrac{p(\theta \cap (X \cap \alpha))}{p(X \cap \alpha)} = \dfrac{p((\theta \cap X) \cap \alpha)}{p(X \cap ...


0

In the worst case, the machine does not work $365{\tiny .25}$ days in the year. At least one component fails every day, which has probability $(1 - 0.99^3)^{356.25} = 1.5\times 10^{-558}$ In the best case, the machine does not work $0$ days in the year. No component fails on any day, which has probability $0.99^{3\cdot365.25} = 1.6\times 10^{-5}$ On the ...


2

What you are doing is correct, except there was a typo where the '2' should be a '1'. $$ \begin{aligned} M(t)&=E(e^{tx})=\int_{-1}^{\infty} e^{tx} e^{-x-1} dx \\ &=e^{-1}\int_{-1}^{\infty} e^{(t-1)x} dx \\ &=e^{-1}\frac{1}{t-1}\left. e^{(t-1)x} \right|_{-1}^\infty \\ &=e^{-1}\frac{e^{1-t}}{1-t} \\ &=\frac{e^{-t}}{1-t} \\ \end{aligned} ...


0

You seem a little confused about terminology in general and the binomial and Poisson distributions in particular. The cure for that is to go back in your text or lecture notes to where these ideas are explained. (You will rarely find an exact formula that is 'tailor made' to answer a particular probability problem so that you can plug into the formula ...


2

As the comments mentioned the worst case scenario is it never works, best case is it works every day. Some information that may be of interest: Let $A$ be the event that $A$ works (thus $A^c$ means it does not work), and $M$ be that the machine works (define similarly for $B,C$). Assuming $A,B,$ and $C$ are independent, on any given day the chance the ...


0

It's a way of gathering things that sum to a constant. For example, \begin{align} \sum_{n=0}^{\infty}a_{n}z^{n}\sum_{n=0}^{\infty}b_{n}z^{n} & = \sum_{n=0}^{\infty}\left(\sum_{j+k=n}a_{j}b_{k}\right) z^{n} \\ & = \sum_{n=0}^{\infty}\left(\sum_{j=0}^{n}a_{j}b_{n-j}\right)z^{n}. \end{align} A Fourier transform can be viewed as an integral ...


0

In the case of $n$ persons the graph of possible friendships is made of ${n \choose 2}$ pairs. For each edge $e =\{a,b\}$ of this graph we consider a random variable $X_e$ if $X_e = 1$ this means that they are friends. If $X_e=0$ they are not friends. ($a,b$ are two individuals of our population). We assume that $X_e$ are independent random variables and ...


3

Hint: $E[X|X+Y=1] = E[Y|X+Y=1]$. Now use additivity of conditional expectations. Of course this assumes these conditional expectations exist.


1

It seems like you may be referring to the Law of Large Numbers. The Wikipedia page linked to gives a good explanation, from what I can see. See also: the relevant part of this SE answer for more information on the theorem, and why it is important, this SE question for another good, layman oriented explanation of the theorem, the top-rated answer to this SE ...


0

$P(A_1\cup A_2 \cup A_3)= P(A_1\cup A_2) + P(A_3) - P((A_1\cup A_2)\cap A_3 )$ $=P(A_1\cup A_2) + P(A_3) - P(A_1\cup A_2).P(A_3 )$ ..since $(A_1\cup A_2)$ and $(A_3)$ are independent Similarly, $P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2)=P(A_1)+P(A_2) - P(A_1).P(A_2)$ Now you can just do the calculation. I guess the final answer is 3/4


2

Possibly you're thinking of the Poisson distribution. Suppose you have a one-in-$1{,}000{,}000$ chance of success on each trial, and there are $3{,}600{,}000$ trials. The expected number of successes is then $3.6$. If we ask for the probability that there are exactly $5$ successes, we get $$ \frac{3.6^5 e^{-3.6}}{5!} = \frac{3.6^5 e^{-3.6}}{120} \approx ...


0

If you want a set of outcomes with equal probability, you need to account somehow for the fact that when the contestant chooses the prize door on the first guess, Monty has a choice of which door to open. This is best modeled as a random variable with two equally likely outcomes. We might as well assume Monty flips a coin and opens the leftmost possible ...


0

The number of subsets of a set with $n$ elements is $2^n$ since each element is either in the subset or it is not. Each of the $2^4 = 16$ subsets of $A$ except the empty set could be the range of $f: A \to A$. Hence, there are $15$ possible ranges for $f$. There are also $15$ possible ranges for $g: A \to A$. Hence, there are $15^2 = 225$ possible pairs ...


0

I think you might be able to follow this route ($q=1-p$): $\sum_k {N\choose k} (xp)^k (1-p)^{N-k}=(xp+q)^n$ $f(z)=(zp+q)^n(\frac{p}{z}+q)^n$ has your sum $\sum_k {N\choose k}^2 p^{2k} (1-p)^{2N-2k}$ as the free term (no $z$) Your sum is the integral $\frac{1}{2\pi i}\int\frac{f(z)}{z}dz$ over the unit circle. $z=e^{2\pi i t}$ reduces the integral to ...


1

"Randomly chosen" isn't clearly defined, but let's take it to mean that we randomly choose the values of $f(1),f(2),f(3), f(4)$ and similarly for $g$. We'll work case by case, indexed by the size of the range of $f$. I will not complete the calculation, it's a bit messy and I'll leave some of the arithmetic off. Case I: Range of $f$ has exactly 1 element. ...


0

$\newcommand{\var}{\operatorname{var}}$The conditional distribution of $X_D-X_{\overline D}$ given the event $C=c$ is normal with expected value $\Big(\mu+g(c)\Big) - g(c) = \mu$ and variance $1+1=2$, since \begin{align} \var(X_D - X_{\overline D} \mid C=c) & = \var(X_D \mid C=c) + \var( -X_{\overline D} \mid C=c) \\[10pt] & = \var(X_D \mid C=c) + ...


3

I believe that you are either misinterpreting the video, the video is wrong, or something else is going on because, if $\sum_{i = 0}^{n} \binom{n + i}{i} = \binom{2n + 1}{n + 1}$ and $\sum_{i = 0}^{n} \binom{n + i}{i} = \binom{n + k + 1}{k}$ for $0 \leq k \leq n$, that would imply $\sum_{i = 0}^{n} \binom{n + i}{i} = \binom{2n + 1}{n + 1} = \binom{n + k ...


0

Let $X$ be a random variable with density function $f_X(x)$, $g(x)$ be a monotone increasing function of $x$, and $Y = g(X)$. We seek the density function $f_Y(y)$ of $Y.$ In this instance it can be shown that $$f_Y(y) = f_X(g^{-1}(y)) \frac{dg^{-1}(y)}{dy},$$ where $g^{-1}$ is the inverse function of $g$. In your Question, $X \sim Exp(1),\;$$g(x) = ...


1

One more explanation, given a CS or CS-like background: Consider the set of all $n$-bit numbers, from $0$ through $2^n-1$, inclusive; there are thus $2^n$ of them, by inspection. (For instance, there are $2^3 = 8$ numbers from $0$ through $2^3-1 = 7: 0, 1, 2, 3, 4, 5, 6, 7$.) We could, also, count them as follows: How many of the numbers have no $1$ bits? ...


4

Another explanation is to observe that, since $\,\dbinom nk$ is the number of subsets of $k$ elements in a set of $n$ elements, the sum of all these coefficients ($n$ fixed) is but the number of all subsets of a set of $n$ elements – which is known to be $2^n$.


7

It's from the Binomial Theorem, expanding $(1+1)^n$


2

There is no coincidence here. We learn nothing from $A$ about $X$ only if the coin is fair (otherwise the fact that both tosses produced the same outcome enhances our belief that the outcome was the more likely one) or "one-sided" (i.e., $P(X=0)=0$ or $P(X=1)=0$). When the coin is not fair, the mutual information of $A$ and $X$ is non-zero: if $P(X=0)=p$, ...


0

You can use the approximate confidence interval for a proportion. In general it is used for normally distributed variables. But you have 3333 identical and independent distributed (iid) random variables. So you can apply the central limit theorem. The two sided limits are $\Large{\left[\hat p-z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat ...


0

If $a$ and $b$ are independent there is a very simple solution. First convince yourself that $$a + b \leq C \Leftrightarrow \mathrm{max}(a,b) \leq \frac{C}{2}$$ Then due to independence, $P\{\mathrm{max}(a,b) \leq \frac{C}{2}\} = P\{a \leq \frac{C}{2}\}P\{b \leq \frac{C}{2}\}$ I am assuming both $a$ and $b$ have uniform distributions. Then, $P\{a \leq ...


0

Can be done in $\min(O(A\log B),O(B\log A))$ using binary serach.


0

If anyone was curious, I was able to solve this using optimization by minimizing the objective function $$ (F(b) - F(a) - 0.95)^2 + (F(a) - (1 - F(b))^2 $$ where $F$ is the cdf. The first term requires that $P(a \leq X \leq b) = 0.95$ and the second requires that $P(X < a) = P(X > b)$, thereby "centering" this $95\%$ interval on the median. MATLAB's ...


2

The probability is in $\Omega\left(n^{-2}\right)$, and apparently also in $O\left(n^{-2}\right)$ (and thus in $\Theta\left(n^{-2}\right)$). I'll focus on odd $n$ because it makes things easier, but I believe with some slight complications similar considerations apply to even $n$. If we add another entry for $j=-1$ to $S$, we have $S_{-1}=-S_{n-1}=:\sigma$. ...


0

No. The following example is due to R. O. Davies, Separate approximate continuity implies measurability, Mathematical Proceedings of the Cambridge Philosophical Society, Vol. 73, Issue 03, May 1973, pp 461. Let $\kappa$ be a real valued measurable cardinal and $m: \mathcal{P}(\kappa) \to [0, 1]$ be a $\kappa$-additive diffused probability measure. Put ...


-1

It suffices to use the following property of integrals $$f \leq g \Rightarrow \int f(y)\, dy \leq \int g(y)\, dy$$ and note that $$1_{\{0 \leq y <t\}} y^2 Q(y)\leq t^2 1_{\{0 \leq y <t\}} Q(y)$$ the same observation will allow you to solve the other inequalities.


0

Let there be n women. To meet the criteria, men must either be in two blocks of 4 & 2 and positioned in two of the (n+1) gaps between women (including the ends), or in a block of 6 positioned in the (n+1) gaps, i.e. in $[(n+1)\cdot n + (n+1)]$ patterns, having 4!2! and 6! permutations respectively, thus $$\frac{[(n+1)\cdot n!]\cdot(n\cdot 4!\cdot2! ...


-1

Let us assume that the men and the women are distinguishable. From the data, you must have EITHER a group of 6 men together, arranged amongst $n$ women, OR a group of 4 men and a group of two men arranged so that there is at least one woman between these groups. In the first case, the number of arrangements is $$6!\times(n+1)!$$ In the second case the men ...


1

Suppose there are $x$ women. I assume them all to be the same, like you seem to do in the question. Case 1. If there is a block of exactly 4 men, we have two possibilities. Case 1.1. If the block of 4 men is at the end or the beginning of the row, we only need one women enclosing it. So we have the block $B$ consisting of 4 men and 1 woman, which can ...


1

Consider the men as $M, M,\ldots,M$ to get 4 aligned and no men alone you need to form 2 groups, one with $4$ men and one with $2$ men. let $G(4),G(2)$ be the group with $4$ and $2$ men respectively. A favorable occurrence is of the form $$W(n_1) G(4) W(n_2) G(2) W(n_3) $$ $$W(n_1) G(2) W(n_2) G(4) W(n_3) $$ or $$ W(N_1) G(4) G(2) W(N_2) $$ where ...


0

I think that they are almost ok but it depends on what do you exactly need. I see two potential problems among the kernel functions you mentioned (they arise because kernel functions are made for density estimation but you want something little bit different). 1) All of these functions are symetrical around $0$ but you have distances that are always ...


2

I'll take the statement "All players are equally skilled." to mean that every player has a $50\%$ chance to win against every other player. $7$ players have been eliminated by $p_5$; the probability that $p_1$ was not among them is $24/31$. If $p_1$ has not been eliminated, the identity of the winner of $p_5$'s branch of the tournament is irrelevant to her; ...


0

The prime numbers is the set of numbers derived by removing the set of composite nunbers from the set of natural numbers. The pattern that the natural numbers follow is obvious and easily understood. The set of composite numbers also follow a discernable pattern:it is the union of the factors of 2, the factors of 3 etc. Therefore the prime numbers as ...


0

As Did pointed out, we have to assume that $X$ and $Y$ are independent. Otherwise, consider a symmetric non-degenerated random variable $X$, $(X_n)$ and $(Y_n)$ independent identically distributed sequences where $X_1$ and $Y_1$ are distributed as $X$. Then $X_n\to X$ in distribution, $Y_n\to -X$ in distribution, but the sequence $(X_n+Y_n)_{n\geqslant 1}$ ...


2

For both parts you need to calculate conditional probabilities. For a) let $B$ denote the event that the biased coin is chosen at the final stage, and let $3H$ denote the event of obtaining three heads. We require $$p(B|3H)=\frac{p(B\cap3H)}{p(3H)}=\frac{\frac 16\times\frac{27}{64}}{\frac 16\times\frac{27}{64}+\frac 56\times\frac 18}=\frac{27}{67}$$ For ...


0

Your calculation of the probability distribution of the absolute difference is correct: $\Pr(D=1) = \frac34$ $\Pr(D=3) = \frac14$ You could have used these to find the expected absolute difference: $E[D]=1\times \frac34 +3 \times \frac14 = \frac32 .$


1

$$\begin{align} \mathsf P(X < Y) &=\int_1^3 \left\{\int_0^y f_X(x) \operatorname dx\right\}g_Y(y) \operatorname dy\\ &\neq \tfrac{1}{4} \int_1^3 \int_0^y \operatorname dx \operatorname dy \end{align}$$ Here's the problem.   The inner integral's upper bound should be $\min(2, y)$ because the support for $X$ is $(0;2)$.   Watch out for ...


1

Split the inteval and use Baye's Theorem to get $$P(X<Y) = P(X<Y | X<1) P(X<1) + P(X<Y|X\geq1)P(X\geq 1))$$ sind $X$ is uniform on $(0,2)$, we know that $P(X<1)=P(X>1) = \frac12$. Furthermore, $P(X<Y|X<1) = 1$ since $Y$ is uniform on $(1,3)$. This yields $$P(X<Y) = 1 \cdot \frac12 + P(X<Y|X\geq1)\frac12$$ Now $P(X<Y| ...


1

If you want to solve the problem using integrals then you should notice that you have wrong upper limit in the inner integral. It should be min(y,2).


9

This sketch might help. You want the red area as a proportion of the red and blue areas.


1

We .can draw the rectangle and it's interior $ 0 \leq x \leq 2$ and $1 \leq y \leq 3$. Then we can draw the line $y=x$. Let's look at our event. So we should draw the line $y=x$. Now, the region delimited is given by the triangle whose vertices are $(1,1)$, $(2,2)$ and $(2,1)$.The probanility is $\int\limits_{1}^{2}\int\limits_{1}^{x} \frac{1}{4}dydx$.


0

Answer: Divide the regions of X with respect to Y for the condition $X<Y$. For $0<X<1$, $P(X<Y) = \frac{1}{2}$ For $1<X<2$ and $1<Y<2$ $P(X<Y) = \int_{1}^{2}\int_{x}^{2} \frac{1}{2}\frac{1}{2}dydx = \frac{1}{8}$ For $1<X<2$, and $2<Y<3$ $P(X<Y) = \frac{1}{2}.\frac{1}{2}=\frac{1}{4}$ Thus $P(X<Y) = ...


0

Not a complete answer, but too long for a comment: Technically, the definition of a median is any number $m$ such that $\Pr(X\le m)\ge0.5$ and $\Pr(X\ge m)\ge0.5$. Note that the first one may also be written as $1-\Pr(X>m)\ge0.5$, with the inequality being strict. If your random variable is continuous, then the strictness of the inequality doesn't ...



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