### 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Total probability and Bayes formulas

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Total probability formula

Theorem 8. Let $(\Omega, \mathcal{F}, P)$ be a probability space, $A \in \mathcal{F}$, and an events $H_{1}, H_{2}, \ldots \in \mathcal{F}$ are pairwise disjoint $\left(H_{i} H_{j}=\varnothing, i \neq j\right)$, have positive probabilities $\left(P\left(H_{i}\right)>0\right)$ and satisfy the condition $\sum_{i=1}^{\infty} H_{i}=\Omega$.

Then the probability of the event $A$ can be found by the total probability formula
$$P(A)=\sum_{i=1}^{\infty} P\left(H_{i}\right) P\left(A / H_{i}\right) .$$
Proof. It follows from the conditions of theorem that
$$A=\sum_{i=1}^{\infty}\left(A H_{i}\right),$$
and the events of a sequence $A H_{1}, A H_{2}, \ldots$ are pairwise disjoint.
Then, by the countable additivity axiom, the probability
$$P(A)=\sum_{i=1}^{\infty} P\left(A H_{i}\right) .$$
Further, by the multiplication of probabilities formula,
$$P\left(A H_{i}\right)=P\left(H_{i}\right) P\left(A / H_{i}\right) .$$
Substituting these probabilities into the previous formula, we obtain the desired formula (16).

The total probability formula is usually applied in those cases when (for one reason or another) it is easier to find conditional probabilities $P\left(A / H_{i}\right)$ and probabilities $P\left(H_{i}\right)$ than to directly calculate the probability $P(A)$.

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|The Bayes formulas

Theorem 9. Let an events $A$ and $H_{1}, H_{2}, \ldots$ satisfy the conditions of the theorem 8 . Then, if $P(A)>0$, then the following Bayes formulas take place:
$$P\left(H_{i} / A\right)=\frac{P\left(H_{i}\right) P\left(A / H_{i}\right)}{\sum_{j=1}^{\infty} P\left(H_{j}\right) P\left(A / H_{j}\right)}, \quad i=1,2, \ldots$$
Proof. By the conditional probability formula
$$P\left(H_{i} / A\right)=\frac{P\left(H_{i} A\right)}{P(A)}$$

Furthermore, by the probabilities multiplication formula, a numerator of the formula $(25)$ is equal to $P\left(H_{i} A\right)=P\left(H_{i}\right) P\left(A / H_{i}\right)$, and (by the total probability formula) the probability $P(A)$ of the denominator of $(25)$ is equal to the expression of denominator of $(24)$.

The general scheme of application of the Bayes formula to the solution of practical problems is as follows.

Let the event $A$ can occur under different conditions, with respect to the nature of which the assumptions (hypotheses) $H_{1}, H_{2}, \ldots$ can be made, and for whatever reasons we know the probabilities $P\left(H_{i}\right)$ of these hypotheses. Let it also be known that the hypothesis $H_{i}$ gives a probability $P\left(A / H_{i}\right)$ to the event $A$. If an experiment, in which an event $A$ has occurred, is made, this should cause a reassessment of the hypotheses $H_{i}$ probabilities – Bayes’ formulas qualitatively solve this problem.

Usually, the probabilities of hypotheses $P\left(H_{i}\right)$ are called a priori (pre-experimentally determined ) probabilities, but $P\left(A / H_{i}\right)$ – a posterior (determined after the experiment) probabilities.

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Tasks for independent work

1. Three numbers are selected from the set of numbers ${1,2, \ldots, n}$ in the random selection scheme without returning.

Find a conditional probability that third number will be in the interval between first and second numbers, given that first number less than second one.

1. Three dice are tossed.
If you know that a different number of points fell on the dice, then what is the probability that the “six” fell on one of them?
2. It is known that when throwing five dice, a «ll) fell out on at least one of them.
What is the probability that in this case a « $\$ » fell out at least twice?
3. Three dice are tossed.
Find the probability of falling out six points on all dice, if you know that:
a) $\operatorname{six}$ points fell out on one die;
b) $s i x$ points fell out on the $1^{\text {st }}$ die;
c) $s i x$ points fell out on two dice;
d) the same number of points fell out at least on two dice;
e) the same number of points fell out on all dice;
f) $\operatorname{six}$ points fell out at least on one die.
1. Four balls are randomly placed in four boxes.
If it is known that the first two balls were placed into different boxes, then what is the probability that there are exactly three balls into some box?
2. It is known that with the random placement of seven balls into seven boxes, exactly two boxes remained empty.
Prove that then the probability of placing three balls in one of the boxes is $1 / 4$.
3. From an urn containing $m$ white and $n$ – $m$ black balls, $r$ balls are extracted according to the random selection scheme without return. An event $A_{0}^{(i)}\left(A_{1}^{(i)}\right)-i$-th extracted ball is black (white).
Find the conditional probabilities
$$P\left(A_{1}^{(s+1)} / A_{q_{1}}^{(1)} A_{q_{2}}^{(2)} \ldots A_{q_{s}}^{(s)}\right), \quad q_{i}=0 \text { or } q_{i}=1 .$$
How will these probabilities change if the balls were extracted by random selection with a return?
4. Let a sample space be a union of a set, consisting of all $r$ ! permutations, obtaining from the elements $a_{1}, a_{2}, \ldots, a_{r}$ and sets $\omega_{j}=\left(a_{j}, a_{j}, \ldots, a_{j}\right), j=1,2, \ldots, r$. We assume that each permutation has a probability $1 /\left(r^{2}(r-2)\right.$ !), and each sequence $\omega_{j}$ – a probability $1 / r^{2}$. Let the event $A_{k}$ means that the element $a_{k}$ is in its own ( $k$-th) place $(k=1, \ldots, r)$.

Show that in this case an events $A_{1}, A_{2}, \ldots, A_{r}$ are pairwise independent, but for different indexes $i, j, k$ an events $A_{i}, A_{j}, A_{k}$ aren’t independent.

1. There are 3 white, 5 black and 2 red balls in some urn. Two players alternately retrieve one ball without returning from this urn. The winner is the one who takes out the white ball first. If a red ball appears, then a draw is declared.

Consider following events: $A_{1}=$ {the player, who starts the game, wins}, $A_{2}={$ the second participant wins $}, \mathrm{B}={$ the game ended in a draw $}$.
Find the probabilities of an events $A_{1}, A_{2}, \mathrm{~B}$.

1. Die is tossed twice. Let $\xi_{1}$ and $\xi_{2}$ be the number of points, occurred on $1^{\text {st }}$ and $2^{\text {nd }}$ die (respectively).
Find the probabilities of following events:
$A_{1}=\left{\xi_{1}\right.$ is divided into $2, \xi_{2}$ is divided into 3$}$;
$A_{2}=\left{\xi_{1}\right.$ is divided into $3, \xi_{2}$ is divided into 2$}$;
$A_{3}=\left{\xi_{1}\right.$ is divided into $\left.\xi_{2}\right} ;$
$A_{4}=\left{\xi_{2}\right.$ is divided into $\left.\xi_{1}\right}$;
$A_{5}=\left{\xi_{1}+\xi_{2}\right.$ is divided into 2$}$;
$A_{6}=\left{\xi_{1}+\xi_{2}\right.$ is divided into 3$}$.
Find all pairwise independent events $A_{i}, A_{j}(i, j$ are different).

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Tasks for independent work

1. 从一组数字中选择三个数字1,2,…,n在随机选择方案中没有返回。

1. 掷了三个骰子。
如果您知道骰子上的点数不同，那么“六”点落在其中一个上的概率是多少？
2. 众所周知，当投掷五个骰子时，至少有一个骰子会掉出一个«ll)。在这种情况下，«  » 掉出至少两次
的概率是多少？
3. 掷了三个骰子。
如果你知道，求所有骰子掉出 6 点的概率：
a)六一个骰子点掉了；
b)s一世X点掉了1英石 死;
C）s一世X两个骰子点数掉了；
d) 至少两个骰子点数相同；
e) 所有骰子的点数相同；
F）六点数至少在一个骰子上掉了下来。
1. 四个球被随机放置在四个盒子里。
如果已知前两个球被放入不同的盒子中，那么某个盒子中恰好有三个球的概率是多少？
2. 众所周知，将七个球随机放入七个盒子中，正好有两个盒子是空的。
证明那么在其中一个盒子里放三个球的概率是1/4.
3. 从一个包含米白色和n – 米黑球，r根据随机选择方案提取球，不返回。一个事件一种0(一世)(一种1(一世))−一世-th 提取的球是黑色（白色）。
找到条件概率
磷(一种1(s+1)/一种q1(1)一种q2(2)…一种qs(s)),q一世=0 或者 q一世=1.
如果球是通过随机选择和回报提取的，这些概率将如何变化？
4. 设一个样本空间是一个集合的并集，由所有r！排列，从元素中获取一种1,一种2,…,一种r并设置ωj=(一种j,一种j,…,一种j),j=1,2,…,r. 我们假设每个排列都有一个概率1/(r2(r−2)!)，以及每个序列ωj– 一个概率1/r2. 让事件一种ķ意味着元素一种ķ是在它自己的（ķ-th) 地方(ķ=1,…,r).

1. 一些瓮中有3个白球、5个黑球和2个红球。两名球员交替取一个球而不从这个瓮中返回。获胜者是最先取出白球的人。如果出现红球，则宣布平局。

1. 骰子被扔了两次。让X1和X2是点数，发生在1英石 和2nd 死（分别）。
求下列事件的概率：
A_{1}=\left{\xi_{1}\right.$分为$2，\xi_{2}$分为3$}A_{1}=\left{\xi_{1}\right.$分为$2，\xi_{2}$分为3$};
A_{2}=\left{\xi_{1}\right.$分为$3，\xi_{2}$分为2$}A_{2}=\left{\xi_{1}\right.$分为$3，\xi_{2}$分为2$};
A_{3}=\left{\xi_{1}\right.$分为$\left.\xi_{2}\right} ；A_{3}=\left{\xi_{1}\right.$分为$\left.\xi_{2}\right} ；
A_{4}=\left{\xi_{2}\right.$分为$\left.\xi_{1}\right}A_{4}=\left{\xi_{2}\right.$分为$\left.\xi_{1}\right};
A_{5}=\left{\xi_{1}+\xi_{2}\right.$分为2$}A_{5}=\left{\xi_{1}+\xi_{2}\right.$分为2$};
A_{6}=\left{\xi_{1}+\xi_{2}\right.$分为3$}A_{6}=\left{\xi_{1}+\xi_{2}\right.$分为3$}.
查找所有成对独立事件一种一世,一种j(一世,j是不同的）。

## 广义线性模型代考

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