统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Operations on events

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统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Certain event

Certain event. I.e. $\Omega \subseteq \Omega$, so $\Omega={\omega}$ is an event and this event will necessarily happen as a result of experiment. Such an event is called a certain event.

Thus, a certain event (designation: $\Omega$ ) is an event that will necessarily occur as a result of experiment.

Impossible event is an event that will never happen as a result of experiment. An impossible event is denoted by $\varnothing$ (an empty set).

Sum (union) of events. Sum (union) of events $A$ and $B$ (designation: $A \cup B$ ) is an event consisting of elementary events belonging to at least one of the events $A$ and $B$ :
$$A \bigcup B={\omega \in \Omega: \omega \in A \text { or } \omega \in B} .$$
Thus, the sum $A \cup B$ of events $A$ and $B$ is an event, which will occur if and only if at least one of them occurs.

Product of events. A product (intersection) of events $A$ and $B$ (designation: $A \cap B$ or $A B$ ) is an event which consists of elementary events belonging to $A$ and $B$ :
$$A \cap B={\omega \in \Omega: \omega \in A, \omega \in B} .$$
So, a product $A \cap B$ of events $A$ and $B$ is an event, which occurs if and only if events $A$ and $B$ occur simultaneously.

Difference of events. $A$ difference of events $A$ and $B$ (designation: $A \backslash B$ ) is an event, which consists of elementary events belonging to $A$ but not belonging to $B$ :
$$A \backslash B={\omega \in \Omega: \omega \in A, \omega \notin B}$$
So, a difference $A \backslash B$ of events $A$ and $B$ is an event, which occurs if and only if an event $A$ occurs and $B$ doesn’t occur.

Opposite event. An event opposite to event $A$ (designation: $\bar{A}$ ) is an event, which consists of all elementary events not belonging to $A$ :
$$\bar{A}={\omega \in \Omega: \omega \notin A}$$
So, an opposite event $\bar{A}$ occurs if and only if an event $A$ doesn’t occur. Implication of one event from another. If all elementary events belonging to an event $A$ also belong to an event $B$, then it is said that an event $A$ implies an event $B$ (designation: $A \subseteq B$ ):
$$A \subseteq B \Leftrightarrow \omega \in A \Rightarrow \omega \subseteq B$$
So, $A \subseteq B$ (an event $A$ implies an event $B$ ) means that each time an event $A$ occurs, an event $B$ will also occur.

统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Elements of combinatorics

Let’s consider some finite sets $A$ and $B$ which consist of $n$ and $m$ elements $(|A|=n<\infty,|B|=m<\infty)$ :
$$A=\left{a_{1}, a_{2}, \ldots, a_{n}\right}, \quad B=\left{b_{1}, b_{2}, \ldots, b_{m}\right} .$$
We define a new set (the Cartesian product) $A \times B$ as follows:
$$A \times B=\left{\left(a_{i}, b_{j}\right): a_{i} \in A, b_{j} \in B\right}$$
Then the number of elements of a set (Cartesian product) is $|A \times B|=|A| \cdot|B|=n \cdot m$, because all elements of this set can be arranged in $n$ rows of $m$ elements in each in the following way:
\begin{aligned} &\left(a_{1}, b_{1}\right),\left(a_{1}, b_{2}\right), \ldots,\left(a_{1}, b_{m}\right), \ &\left(a_{2}, b_{1}\right),\left(a_{2}, b_{2}\right), \ldots,\left(a_{2}, b_{m}\right), \ &\left(a_{n}, b_{1}\right),\left(a_{n}, b_{2}\right), \ldots,\left(a_{n}, b_{m}\right) \end{aligned}
This statement can be generalized in the following sense.
Theorem 1. Let some finite sets be given:
$$\begin{gathered} A_{1}=\left{a_{11}, a_{12}, \ldots, a_{1 n_{1}}\right}, \quad A_{2}=\left{a_{21}, a_{22}, \ldots, a_{2 n_{2}}\right}, \ldots, A_{m}=\left{a_{m 1}, a_{m 2}, \ldots, a_{m n_{n}}\right} \ \left(\left|A_{k}\right|=n_{k}<\infty, k=1,2, \ldots, m\right) . \end{gathered}$$

We define a new set (the Cartesian product $A_{1} \times A_{2} \times \ldots \times A_{m}$ of sets $A_{1}, A_{2}, \ldots, A_{m}$ ) as follows:
$$A_{1} \times A_{2} \times \ldots \times A_{m}=\left{\left(a_{1 i_{1}}, a_{2 i_{2}}, \ldots, a_{m i_{n}}\right): a_{k_{k}} \in A_{k}, k=1,2, \ldots, m ; i_{k}=1,2, \ldots, n_{k} ;\right}$$
Then
$$\left|A_{1} \times A_{2} \times \ldots \times A_{m}\right|=\left|A_{1}\right|\left|A_{2}\right| \ldots\left|A_{m}\right|=n_{1} n_{2} \ldots n_{m}$$
Proof. For $m=2$ it is the above statement. In the case $m=3$ the number of triples $\left(a_{1 i_{1}}, a_{2 i_{2}}, a_{3 i_{3}}\right)$, according to the proved statement, is equal to the product of the number of pairs $\left(a_{1 i_{i}}, a_{2 i_{2}}\right)$ by the number of elements $a_{3 i_{3}}$, i.e.
$$\left(n_{1} \cdot n_{2}\right) \cdot n_{3}=n_{1} \cdot n_{2} \cdot n_{3}$$
Now, to prove the theorem definitively, it suffices to use induction. Theorem 1 can be formulated differently as follows.

统计代写|概率论作业代写Probability and Statistics代考5CCM241A|The paradox of de Mere

1. The paradox of de Mere. Which event is more likely when throwing three dice: the sum of the points dropped is 11 (eleven) or 12 (twelve)?

De Mere considered these events to be equally probable and justified this with the following reasoning.

The event that «the sum of the dropped points is 11 (eleven)» can occur as a result of the following combinations:
$$(6,4,1),(6,3,2),(5,5,1),(5,4,2),(5,3,3),(4,4,3)$$

where, ex, $(6,4,1)$ means that «6» occurred on the $1^{\text {st }}$ dice, $« 4 »-$ on the $2^{\text {nd }}$ dice and «1»- on the $3^{\text {rd }}$ one, etc.

On the other hand, the event «the sum of dropped points is 12 (twelve)» can also occur as a result of the following six combinations:
$$(6,5,1),(6,4,2),(6,3,3),(5,5,2),(5,4,3),(4,4,4) \text {. }$$
Consequently, these events are equally probable.
Here, the mistake of de Mere is that the possible outcomes that he considered are not equally probable.

For example, the event $(6,4,1)$ can occur in $3 !=6$ cases: $(6,4,1),(6,1,4), \ldots$, $(1,4,6)$. At the same time, for example, a combination $(4,4,4)$ can occur only in one case. In modern language, de Mere incorrectly constructed the space of elementary events corresponding to the given problem.
The solution of the problem. We define $\Omega$ as
$$\Omega={(i, j, k): i, j, k=\overline{1,6}}=\Omega_{0} \times \Omega_{0} \times \Omega_{0},$$
where $\Omega_{0}={1,2,3,4,5,6}$.
Let’s introduce the events:
$A_{11}={$ the sum of points is equal to 11$}, A_{12}={$ the sum of points is equal to 12$}$.
Hence
\begin{aligned} &A_{11}={(i, j, k) \in \Omega: i+j+k=11}, \ &A_{12}={(i, j, k) \in \Omega: i+j+k=12}, \end{aligned}
We have
$$\left|A_{11}\right|=27, \quad\left|A_{12}\right|=25, \quad|\Omega|=6^{3}=216,$$
Therefore
$$P\left(A_{11}\right)=\frac{27}{216}>\frac{25}{216}=P\left(A_{12}\right) .$$

1. From the general population $\Omega_{0}=\left{a_{1}, a_{2}, \ldots, a_{n}\right}$ (for example, from an urn with numbered balls) of size $n$, a random sample with replacement of size $r$ is extracted.
a) Find the probability that the extracted sample is a sample without replacement (that is, all the extracted balls have different numbers).
b) Find the probability that the first sample element is the first element of the general population, the second sample element is the second element of the general population (that is, the first ball extracted from the urn is the ball No. 1 and the second ball is the ball No. 2).

统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Elements of combinatorics

A=\left{a_{1}, a_{2}, \ldots, a_{n}\right}, \quad B=\left{b_{1}, b_{2}, \ldots, b_{m} \对} 。A=\left{a_{1}, a_{2}, \ldots, a_{n}\right}, \quad B=\left{b_{1}, b_{2}, \ldots, b_{m} \对} 。

A \times B=\left{\left(a_{i}, b_{j}\right): a_{i} \in A, b_{j} \in B\right}A \times B=\left{\left(a_{i}, b_{j}\right): a_{i} \in A, b_{j} \in B\right}

(一种1,b1),(一种1,b2),…,(一种1,b米), (一种2,b1),(一种2,b2),…,(一种2,b米), (一种n,b1),(一种n,b2),…,(一种n,b米)

\begin{聚集} A_{1}=\left{a_{11}, a_{12}, \ldots, a_{1 n_{1}}\right}, \quad A_{2}=\left{a_{ 21}, a_{22}, \ldots, a_{2 n_{2}}\right}, \ldots, A_{m}=\left{a_{m 1}, a_{m 2}, \ldots, a_ {m n_{n}}\right} \left(\left|A_{k}\right|=n_{k}<\infty, k=1,2, \ldots, m\right) 。\结束{聚集}\begin{聚集} A_{1}=\left{a_{11}, a_{12}, \ldots, a_{1 n_{1}}\right}, \quad A_{2}=\left{a_{ 21}, a_{22}, \ldots, a_{2 n_{2}}\right}, \ldots, A_{m}=\left{a_{m 1}, a_{m 2}, \ldots, a_ {m n_{n}}\right} \left(\left|A_{k}\right|=n_{k}<\infty, k=1,2, \ldots, m\right) 。\结束{聚集}

A_{1} \times A_{2} \times \ldots \times A_{m}=\left{\left(a_{1 i_{1}}, a_{2 i_{2}}, \ldots, a_{ m i_{n}}\right): a_{k_{k}} \in A_{k}, k=1,2, \ldots, m ; i_{k}=1,2, \ldots, n_{k} ;\right}A_{1} \times A_{2} \times \ldots \times A_{m}=\left{\left(a_{1 i_{1}}, a_{2 i_{2}}, \ldots, a_{ m i_{n}}\right): a_{k_{k}} \in A_{k}, k=1,2, \ldots, m ; i_{k}=1,2, \ldots, n_{k} ;\right}

|一种1×一种2×…×一种米|=|一种1||一种2|…|一种米|=n1n2…n米

(n1⋅n2)⋅n3=n1⋅n2⋅n3

统计代写|概率论作业代写Probability and Statistics代考5CCM241A|The paradox of de Mere

1. 德梅尔悖论。掷三个骰子时，哪个事件更有可能发生：掷出的点数之和是 11（十一）还是 12（十二）？

De Mere 认为这些事件同样可能发生，并通过以下推理证明了这一点。

“丢分之和为 11（十一）”的事件可能由于以下组合而发生：
(6,4,1),(6,3,2),(5,5,1),(5,4,2),(5,3,3),(4,4,3)

(6,5,1),(6,4,2),(6,3,3),(5,5,2),(5,4,3),(4,4,4).

Ω=(一世,j,ķ):一世,j,ķ=1,6¯=Ω0×Ω0×Ω0,

|一种11|=27,|一种12|=25,|Ω|=63=216,

1. 来自普通人群\Omega_{0}=\left{a_{1}, a_{2}, \ldots, a_{n}\right}\Omega_{0}=\left{a_{1}, a_{2}, \ldots, a_{n}\right}（例如，从带有编号的球的瓮中）大小n, 替换大小的随机样本r被提取。
a) 求抽取的样本是无放回样本的概率（即抽取的所有球的个数不同）。
b) 求第一个样本元素是总人口的第一个元素，第二个样本元素是总人口的第二个元素的概率（即从瓮中取出的第一个球是1号球，第一个球是1号球）第二个球是 2 号球）。

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