diff --git a/8.md b/8.md index 7a6f09ca43cce6c067be59edd4f1e18b4dbeeb7f..129f332c6f46ce8e4ec6bcbb91d07377bc67ae79 100644 --- a/8.md +++ b/8.md @@ -702,27 +702,27 @@ combined.barh(0) 数学是准确发现概率的主要工具，尽管计算机也可用于此目的。模拟可以提供出色的近似，具有很高的概率。在本节中，我们将以非正式方式制定一些简单的规则来管理概率的计算。在随后的章节中，我们将回到模拟来近似复杂事件的概率。 -我们将使用标准符号 ![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Bevent%7D%29) 来表示“事件”发生的概率，我们将交替使用“几率”和“概率”两个字。 +我们将使用标准符号 ![](img/tex-8-1.gif) 来表示“事件”发生的概率，我们将交替使用“几率”和“概率”两个字。 ## 事件不会发生的时候 如果事件发生的概率是 40%，不发生的几率就是 60%。这个自然的计算可以这样秒速： -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Ban%20event%20doesn%27t%20happen%7D%29%20%7E%3D%7E%201%20-%20P%28%5Cmbox%7Bthe%20event%20happens%7D%29) +![](img/tex-8-2.gif) ## 所有结果等可能的时候 如果你投掷一个普通的骰子，一个自然的假设是，所有六个面都是等可能的。 那么一个面出现的概率可以很容易地计算出来。 例如，骰子显示偶数的几率是： -![](http://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Cmbox%7Bnumber%20of%20even%20faces%7D%7D%7B%5Cmbox%7Bnumber%20of%20all%20faces%7D%7D%20%7E%3D%7E%20%5Cfrac%7B%5C%23%5C%7B2%2C%204%2C%206%5C%7D%7D%7B%5C%23%5C%7B1%2C%202%2C%203%2C%204%2C%205%2C%206%5C%7D%7D%20%7E%3D%7E%20%5Cfrac%7B3%7D%7B6%7D) +![](img/tex-8-3.gif) 与之相似： -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Bdie%20shows%20a%20multiple%20of%203%7D%29%20%7E%3D%7E%20%5Cfrac%7B%5C%23%5C%7B3%2C%206%5C%7D%7D%7B%5C%23%5C%7B1%2C%202%2C%203%2C%204%2C%205%2C%206%5C%7D%7D%20%7E%3D%7E%20%5Cfrac%7B2%7D%7B6%7D) +![](img/tex-8-4.gif) 通常： -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Ban%20event%20happens%7D%29%20%7E%3D%7E%20%5Cfrac%7B%5C%23%5C%7B%5Cmbox%7Boutcomes%20that%20make%20the%20event%20happen%7D%5C%7D%7D%20%7B%5C%23%5C%7B%5Cmbox%7Ball%20outcomes%7D%5C%7D%7D) +![](img/tex-8-5.gif) 前提是所有的结果都是等可能的。 @@ -734,21 +734,21 @@ combined.barh(0) 有六种可能的颜色对：RB，BR，RG，GR，BG，GB（我们已经缩写了每种颜色的名字，就是它的第一个字母）。 所有这些都是抽样方案是等可能的，只有其中一个（GR）使事件发生。所以： -![](http://latex.codecogs.com/gif.latex?%24%24%20P%28%5Cmbox%7Bgreen%20first%2C%20then%20red%7D%29%20%7E%3D%7E%20%5Cfrac%7B%5C%23%5C%7B%5Cmbox%7BGR%7D%5C%7D%7D%7B%5C%23%5C%7B%5Cmbox%7BRB%2C%20BR%2C%20RG%2C%20GR%2C%20BG%2C%20GB%7D%5C%7D%7D%20%7E%3D%7E%20%5Cfrac%7B1%7D%7B6%7D) +![](img/tex-8-6.gif) 但是还有另外一种方法来得到答案，可以用两个阶段来思考这个事件。 必须首先抽取绿色纸条。几率是 1/3，也就是说在所有实验的大约 1/3 的重复中，先抽取了绿色纸条，但事件还没完成。在这 1/3 的重复中，必须再次抽取红色纸条。这个发生在大约 1/2 的重复中，所以： -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Bgreen%20first%2C%20then%20red%7D%29%20%7E%3D%7E%20%5Cfrac%7B1%7D%7B2%7D%20%7E%5Cmbox%7Bof%7D%7E%20%5Cfrac%7B1%7D%7B3%7D%20%7E%3D%7E%20%5Cfrac%7B1%7D%7B6%7D) +![](img/tex-8-7.gif) 这个计算通常按照事件顺序，像这样： -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Bgreen%20first%2C%20then%20red%7D%29%20%7E%3D%7E%20%5Cfrac%7B1%7D%7B3%7D%20%7E%5Ctimes%7E%20%5Cfrac%7B1%7D%7B2%7D%20%7E%3D%7E%20%5Cfrac%7B1%7D%7B6%7D) +![](img/tex-8-8.gif) 因数 1/2 叫做“假设第一次出现了绿色纸条，第二次出现红色纸条的条件几率”。 通常，我们拥有乘法规则： -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Btwo%20events%20both%20happen%7D%29%20%7E%3D%7E%20P%28%5Cmbox%7Bone%20event%20happens%7D%29%20%5Ctimes%20P%28%5Cmbox%7Bthe%20other%20event%20happens%2C%20given%20that%20the%20first%20one%20happened%7D%29) +![](img/tex-8-9.gif) 两个事件同时发生的概率，等于第一个事件发生的概率，乘上第一个事件发生的情况下第二个事件发生的概率。 @@ -762,11 +762,11 @@ combined.barh(0) 根据上面的计算，GR 和 RG 每个的几率都是 1/6。所以你可以通过把它们相加来计算一绿一红的概率。 -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Bone%20green%20and%20one%20red%7D%29%20%7E%3D%7E%20P%28%5Cmbox%7BGR%7D%29%20+%20P%28%5Cmbox%7BRG%7D%29%20%7E%3D%7E%20%5Cfrac%7B1%7D%7B6%7D%20+%20%5Cfrac%7B1%7D%7B6%7D%20%7E%3D%7E%20%5Cfrac%7B2%7D%7B6%7D) +![](img/tex-8-10.gif) 通常，我们拥有加法规则： -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Ban%20event%20happens%7D%29%20%7E%3D%7E%20P%28%5Cmbox%7Bfirst%20way%20it%20can%20happen%7D%29%20+%20P%28%5Cmbox%7Bsecond%20way%20it%20can%20happen%7D%29%20%7E%7E%7E%20%5Cmbox%7B%7D) +![](img/tex-8-11.gif) 事件发生的概率，等于以第一种方式发生的概率，加上以第二种方式发生的概率。 @@ -788,21 +788,21 @@ combined.barh(0) 得出这个答案的另一种方法是，弄清楚如果你不能得到至少一个正面，会发生什么事情：这两次投掷都必须是反面。所以： -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Bat%20least%20one%20head%20in%20two%20tosses%7D%29%20%7E%3D%7E%201%20-%20P%28%5Cmbox%7Bboth%20tails%7D%29%20%7E%3D%7E%201%20-%20%5Cfrac%7B1%7D%7B4%7D%20%7E%3D%7E%20%5Cfrac%7B3%7D%7B4%7D) +![](img/tex-8-12.gif) 要注意根据乘法规则： -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Bboth%20tails%7D%29%20%7E%3D%7E%20%5Cfrac%7B1%7D%7B4%7D%20%7E%3D%7E%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%20%7E%3D%7E%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5E2) +![](img/tex-8-13.gif) 这两个观察使我们能够在任何给定数量的投掷中找到至少一个正面的几率。 例如： -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Bat%20least%20one%20head%20in%2017%20tosses%7D%29%20%7E%3D%7E%201%20-%20P%28%5Cmbox%7Ball%2017%20are%20tails%7D%29%20%7E%3D%7E%201%20-%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B17%7D) +![](img/tex-8-14.gif) 而现在我们有能力找到在骰子的投掷中，六点至少出现一次的几率： -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Ba%20single%20roll%20is%20not%206%7D%29%20%7E%3D%7E%20P%281%29%20+%20P%282%29%20+%20P%283%29%20+%20P%284%29%20+%20P%285%29%20%7E%3D%7E%20%5Cfrac%7B5%7D%7B6%7D) +![](img/tex-8-15.gif) -![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7Bat%20least%20one%206%20in%20two%20rolls%7D%29%20%7E%3D%7E%201%20-%20P%28%5Cmbox%7Bboth%20rolls%20are%20not%206%7D%29%20%7E%3D%7E%201%20-%20%5Cleft%28%5Cfrac%7B5%7D%7B6%7D%5Cright%29%5E2%20%24%24%20and%20%24%24%20P%28%5Cmbox%7Bat%20least%20one%206%20in%2017%20rolls%7D%29%20%7E%3D%7E%201%20-%20%5Cleft%28%5Cfrac%7B5%7D%7B6%7D%5Cright%29%5E%7B17%7D) +![](img/tex-8-16.gif) 下表展示了，这些概率随着投掷数量从 1 增加到 50 而增加。 diff --git a/9.md b/9.md index 2a1c0844a020a62fa2b189e9be1b2664581de315..8c6934c0482b74a64691772797d06b867151620d 100644 --- a/9.md +++ b/9.md @@ -709,7 +709,7 @@ Table().with_column('Max Serial Number', maxes).hist(bins = every_ten) 这个估计的基本思想是观察到的序列号的平均值可能在1到`N`之间。 因此，如果`A`是平均值，那么： -![](http://latex.codecogs.com/gif.latex?A%20%7E%20%5Capprox%20%7E%20%5Cfrac%7BN%7D%7B2%7D%20%7E%7E%7E%20%5Cmbox%7Band%20so%7D%20%7E%7E%7E%20N%20%5Capprox%202A) +![](img/tex-9-1.gif) 因此，可以使用一个新的统计量化来估计飞机总数：取观测到的平均序列号并加倍。 diff --git a/img/tex-8-1.gif b/img/tex-8-1.gif new file mode 100644 index 0000000000000000000000000000000000000000..caa8a4000c037002a26f4e44af78f982761e09bf Binary files /dev/null and b/img/tex-8-1.gif differ diff --git a/img/tex-8-10.gif b/img/tex-8-10.gif new file mode 100644 index 0000000000000000000000000000000000000000..d454104d6e0c9f76bc855aad778b32ccaab8c0df Binary files /dev/null and b/img/tex-8-10.gif differ diff --git a/img/tex-8-11.gif b/img/tex-8-11.gif new file mode 100644 index 0000000000000000000000000000000000000000..fb1bcb8d40947a2a2fc37ad50dfad0d700fcd814 Binary files /dev/null and b/img/tex-8-11.gif differ diff --git a/img/tex-8-12.gif b/img/tex-8-12.gif new file mode 100644 index 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