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# System prepended metadata

title: 機率期中考考古 2019

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# 機率期中考考古 2019
==TODO: 11==
![](https://i.imgur.com/bsz4T7p.png)

a. Shou-de Lin
b. Simpon's Paradox is to said that it is possible for a batter A having higher batting average than batter B at every season in his career and yet B could have a better overall batting average at the end of their careers.
只要在打擊率較高的那一年打較多場，之後的平均就會較高
->學分修比較多那個學期gpa要比較高(可是怎麼辦我這學期修超級多@@)
c. ==N(0, 1) = ChiSquare(1)?==: 可看筆記：）
(不確定) With sets that have a relatively large mean, they are both approximates of the Binomal Distributuon. The normal distribution can actually be used to approximate a poisson distribution if you set their means and variances the same.
[網路上的資料](https://www.quora.com/What-is-the-relation-between-normal-distribution-and-poisson-distribution)
我自己的理解是因為Poisson Distribution是固定單位時間內偶發事件的機率分佈，所以如果expected value極大，則可以用來估計二項分布，而根據中央極限定理可以得知，二項分布執行很多次之後會接近常態分佈（就把兩個連起來了？！）
d. P(A...Z) = P(A) $\times$ P(B|A) $\times$ P(C|A..B) $\times$ ... $\times$ P(Z|A...Y)

![](https://i.imgur.com/mgd5BRW.png)

方法一：
* $F_{Y}$(y) = P(Y $\leq$ y) = P($\frac{1}{X}$ $\leq$ y) = P(X $\geq$ $\frac{1}{y}$) = 1 - P(X < $\frac{1}{y}$) = 1 - $F_{X}$($\frac{1}{y}$) = 1 - $\frac{1}{y^{3}}$ = 1 - $y^{-3}$
* $f_{Y}$(y) = 3$Y^{-4}$
(應該是 $X \leq \frac{1}{y}$，因為 $X < 0$)
方法二：
* Let Y = u(x) = $\frac{1}{X}$, X = v(y) = $\frac{1}{Y}$
* V'(y) = - $Y^{-2}$
* $f_{Y}$(y) = f(V(y)) $\times$ |V'(y)| = 3($\frac{1}{y})^{2}$ $\times$ |-$\frac{1}{y^{2}}$| = 3$Y^{-4}$

![](https://i.imgur.com/qofCP6e.png)

題目翻譯：在0~1中random取一個數字，做6次，取6次中最大的
(不確定)F(x)代表的是x是最大值的機率，則代表其他5次都是小於x，一次小於X的機率為P(X $\leq$ x) = x，6次則為$x^{6}$，f(x) = F'(x) = 6$x^{5}$
==$F(x) = x^6, f(x) = 6x^5$==

![](https://i.imgur.com/DZExMlq.png)

exponential distribution: expected value = $\theta$, P(X $\leq$ $\theta$) = F($\theta$) = 1 - $e^{-\frac{\theta}{\theta}}$ = 1 - $\frac{1}{e}$

![](https://i.imgur.com/1EYyqYY.png)

$\frac{280000+90000}{280000+90000+135000}$ = $\frac{74}{101}$

![](https://i.imgur.com/TYloUQy.png)

額 這個要寫什麼 我不會啦QQ
radom experiment that can be performed multiple times: 在相處的不同時間點和階段告白？？？
event based in it: 接受或拒絕？？？
可是事實上是表白失敗一次之後的機率就是0了吧？

![](https://i.imgur.com/E3SEKrL.png)

$\frac{P(E)}{P(E)+P(F)+P(G)}$
==$\lim_{n \to \infty}\Sigma_{i=0}^n(1-P(E)-P(F)-P(G))^iP(E) = \dfrac{P(E)}{P(E)+P(F)+P(G)}$==
![](https://i.imgur.com/T1XX0di.png)
$\int_{0}^{\pi/2} \frac{l}{2} sin\theta d\theta$ = $\frac{l}{2}$

p = $\frac{l/2}{\pi/4}$

$\pi$ = $\frac{2l}{p}$
![](https://i.imgur.com/4q5mi8W.png)

輸光:(
[相關資料](http://datagenetics.com/blog/may42015/index.html)

![](https://i.imgur.com/BKGqc54.png)
如果要算的是第k次紅燈和第k+1次紅燈：
(綠紅)or(紅綠)：$\frac{1}{3} \times \frac{2}{3} \times (\frac{1}{2})^{k}$
(紅紅)：$\frac{2}{3} \times (\frac{1}{2})^{k} \times \frac{2}{3} \times (\frac{1}{2})^{k+1}$

$F(x) = P(X<=x) = \dfrac{x}{1-w}
\\f(x) = \dfrac{1}{1-w}
\\E[X] = \dfrac{1-w}{2}
\\答案 = \dfrac{2}{9}E[紅綠]+\dfrac{2}{9}E[綠紅]+\dfrac{1}{9}E[紅紅]
\\E[紅綠]=E[綠紅]=\dfrac{1}{2}
\\E[紅紅]=否知:(，可能是\frac{3}{4}$

![](https://i.imgur.com/AJLLerY.png)

$\frac{1}{99}$

通式如下：若為丟x顆中y顆

$C^{x-2}_{y-1}$ $\frac{(x-y-1)!(y-1)!}{(x-1)!}$ = $\frac{1}{x-1}$

Let $P(X, Y)$ be the possibility that shoot $X$ balls and 投進 $Y$ 球

claim: $P(X, Y) = \frac{1}{X - 1}$

can be proved by 數學歸納法

$P(X, 1) = P(X - 1, 1) \times \frac{X - 2}{X - 1}$

$P(X, X - 1) = P(X - 1, X - 2) \times \frac{X - 2}{X - 1}$

$P(X, Y) = P(X - 1, Y) \times (1 - \frac{Y}{X - 1}) + P(X - 1, Y - 1) \frac{Y - 1}{X - 1}$

![](https://i.imgur.com/uSKiQZI.png)

![](https://i.imgur.com/aDWmaH9.png)
再次輸光:(
假設成功n次，對於每個正整數n, 滿足$x_1<x_2<...<x_{n-1}>x_n$的機率為$\frac{1}{n!}$
因為$1<=n<\infty，E[n] = \Sigma_1^{\infty}n\times\frac{1}{n!}=e$ :) 
下面才是對ㄉ!

> 真的是這樣ㄇ？ OAO
> $P(x_1 < x_2 \dots < x_{n - 1} > x_n)$ 應該是 $\frac{1}{(n - 1)!} - \frac{1}{n!}$ 吧 (?)
> 所以 $E(X) = \sum_{n = 2}^{\infty} n \times (\frac{1}{(n - 1)!} - \frac{1}{n!}) = \sum_{n = 2}^{\infty} \frac{1}{(n - 2)!} = e$
> 好像合理!