13.6 概率和组合

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  概率和组合

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  Probability and Combinations 
  ::概率和组合

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  Just as with permutations, combinations crop up frequently in probability . In many card games the objective is to acquire a winning hand. In order to do that, it’s useful for players to know how likely a given hand is to arise, and also the probability that another player has a better hand. Mathematicians have been studying such games of chance for centuries.
  ::和变相一样,组合的概率也经常增长。 在许多纸牌游戏中,目标是获得赢家的手。 为了做到这一点,玩家应该知道某一手的出现可能性,以及另一个玩家有更好的手的概率。 数世纪以来,数学家一直在研究这种机会游戏。

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  Real-World Application: Word Games
  ::真实世界应用程序:文字游戏

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  A word game requires players to select 4 tiles from a bag containing 26 tiles, each with one of the letters A through Z written on it. If each letter appears once and only once, what is the probability that a player will be able to spell CATS with his tiles?
  ::单词游戏要求玩家从包含 26 张牌的袋子中选择 4 张牌, 每个牌上写着字母 A 到 Z 之一。 如果每封信出现一次, 则玩家能用他的牌拼写 CATS 的可能性有多大 ?

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  Since a player needs a $\begin{array}{r}C\end{array}$ , an $\begin{array}{r}A\end{array}$ , a $\begin{array}{r}T\end{array}$ and an $\begin{array}{r}S\end{array}$ in any order , we are looking at a calculation. We first need to determine how many combinations there are, choosing 5 letters from a total of 26:
  ::因为玩家需要 C、 A、 T 和 S 的任意顺序, 我们正在查看一个计算。 我们首先需要确定有多少组合, 从总共 26 个中选择 5 个字母 :

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  Choosing 4 from 26:
  ::从 26 选择 4 :

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  $$\begin{array}{rl}{}_{26}{C}_4& =\frac{26!}{(26-4)!4!}=\frac{26!}{22!4!}\\ & =\frac{26\times 25\times 24\times 23}{4\times 3\times 2\times 1}\\ & =59,800\text{combinations.}\end{array}$$

  
  ::26C4=26! (26- 4)! 4!=26. 22. 4! =26x25x24x234x3x2x1=59800 组合。

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  Since only one combination allows a player to spell CATS, the probability of getting that combination is $\begin{array}{r}\frac{1}{59,800}\end{array}$ .
  ::由于只有一个组合允许玩家拼写CATS, 获得该组合的概率为159,800。

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  Real-World Application: Funfair Games 
  ::真实世界应用:玩玩游戏

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  A funfair game consists of pulling numbered chips from a bag. The game starts with nine chips numbered 1 through 9, and players are allowed to pull out three chips. A player wins by drawing the number 7 chip. What is the probability that a player will win?
  ::有趣的游戏包括从一个袋子中拉出数字的芯片。 游戏以9个芯片开始, 编号为1到9, 玩家可以拿出3个芯片。 玩家可以通过绘制第7个芯片来赢得。 玩家赢得7个芯片的概率是多少?

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  To find the probability for winning this game we need to know two pieces of information: 1) the total number of combinations for the game and 2) the number of combinations that contain a 7.
  ::要找到赢得这个游戏的概率,我们需要知道两部分信息1) 游戏组合的总数;(2) 包含一个7的组合的数目。

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  To find the total number of combinations for the game, use the formula $\begin{array}{r}{}_n{C}_r\end{array}$ with $\begin{array}{r}n=9\end{array}$ and $\begin{array}{r}r=3\end{array}$ :
  ::要查找游戏的组合总数,请使用 n=9 和 r=3 的公式 nCr:

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  $$\begin{array}{r}{}_9{C}_2=\frac{9!}{(9-3)!3!}=\frac{9!}{6!3!}=\frac{9\times 8\times 7}{3\times 2\times 1}=84\text{combinations}\end{array}$$

  
  ::9C2=9! 9C2=9! (9- 3)! 3! =9! 6! 3! =9x8x73x2x1=84 组合

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  Now we need to determine how many combinations contain a 7. We can figure this out by reasoning as follows: given that there MUST be a 7 in the list, the number of combinations that contain a seven is the same as the number of combinations of choosing any two numbers out of the eight chips that don’t include the 7. In other words, if we imagine that we got to pick the 7 chip on purpose, how many ways would there be of picking the other two chips?
  ::现在,我们需要确定有多少组合含有一个7. 我们可以用以下的推理来解决这个问题: 鉴于列表中必须有一个 7, 包含一个 7 的组合数量与在8个芯片中选择任何2个数字的组合数量相同, 而其中8个芯片不包括7个。 换句话说,如果我们想象我们不得不故意选择7个芯片,那么选择另外2个芯片的方法会有多少?

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  To find that number, we use the formula $\begin{array}{r}{}_n{C}_r\end{array}$ with $\begin{array}{r}n=8\end{array}$ and $\begin{array}{r}r=2\end{array}$ :
  ::要找到这个数字, 我们使用 n=8 和 r= 2 的公式 nCr :

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  $$\begin{array}{r}{}_8{C}_2=\frac{8!}{(8-2)!2!}=\frac{8!}{6!2!}=\frac{8\times 7}{2\times 1}=28\text{combinations}\end{array}$$

  
  ::8C2=8! 8C2=8! (8- 2)! 2! =8! 6! 2! =8C2=8C2=8! (8- 2)! 2! =8. 6! 2! =8C2=8C2=8!

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  So the probability is given by:
  ::概率由 :

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  $$\begin{array}{r}P(\text{getting a}7)=\frac{84}{28}=\frac{1}{3}=\text{or one in three}.\end{array}$$

  
  ::P(get a 7)=8428=13=或三分之一。

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  Calculating Probability
  ::计算概率

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  Calculate the probability of being dealt four aces in a five card poker hand.
  ::计算五张扑克牌手的四张A 被处理的概率。

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  The first thing we need to know to solve this problem is the total number of unique hands. Since players can arrange their cards however they wish, the order of the cards is unimportant. So we will calculate, using the formula, the number of combinations for choosing 5 cards from a 52 card deck.
  ::要解决这个问题,首先我们需要知道的是独特手的总数。由于玩家可以随意安排他们的牌牌,因此卡片的顺序并不重要。因此我们将使用公式计算52张牌牌上5张牌的组合数。

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  Choosing 5 from 52: $\begin{array}{r}{}_{52}{C}_5=\frac{52!}{(52-5)!5!}=\frac{52!}{47!5!}=\frac{52\times 51\times 50\times 49\times 48}{5\times 4\times 3\times 2\times 1}=2,598,960\end{array}$ unique hands
  ::52: 52C5=52! (52- 5)! 5! = 52! 47! 5! = 52. 47! 5! = 52x51x50x49x49x485x485x4x3x3x2x1=2, 598, 960 独特手

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  Next we need to calculate how many hands there are that contain four aces. This sounds difficult, but we can think about it like this:
  ::接下来,我们需要计算有多少手有四个A。这听起来很困难,但是我们可以这样考虑:

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  • If a hand contains four aces, it must also contain exactly ONE other card.
    ::如果一只手内有四张A, 也必须有另一张卡。
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  • Since the four aces are accounted for, there are 48 (that’s 52 – 4) cards left in the deck.
    ::甲板上还有48张牌(即52至4张牌)。

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  Since a unique hand is independent of the order in which the cards are dealt, there must be 48 unique hands that contain four aces (one unique hand for every non-ace card in the deck).
  ::由于独有的手独立于卡片处理的顺序,必须有48只独有的手,内有四张A(甲板上每张非A卡都有一只独有的手)。

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  There are 48 possible hands that contain four aces. So the probability of being dealt four aces in poker is:
  ::可能有48只手 包含4个A 所以在扑克中4个A的概率是:

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  $$\begin{array}{r}P(\text{four aces})=\frac{48}{2,598,690}=\frac{1}{54,145}\end{array}$$

  
  ::P(四A)=482,598,690=154,145

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  Review 
  ::回顾

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  For 1-3, calculate the number of combinations:
  ::对于1-3,计算组合数:

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  1. $\begin{array}{r}{}_8{C}_4\end{array}$
     ::8C4 8C4
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  2. $\begin{array}{r}{}_{11}{C}_5\end{array}$
     ::11C5 11C5
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  3. $\begin{array}{r}{}_{20}{C}_2\end{array}$
     ::20C2 20C2

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  For 4-8, a town lottery requires players to choose three different numbers from the numbers 1 through 36.
  ::对于4 -8, 镇彩票要求玩家选择三个不同的数字, 从1号到36号。

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  4. How many different combinations are there?
     ::有多少不同的组合?
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  5. What is the probability that a player’s numbers match all three numbers chosen by the computer?
     ::玩家数字与计算机选择的所有三个数字匹配的概率是多少?
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  6. What is the probability that two of a player’s numbers match the numbers chosen by the computer?
     ::玩家数字中两个数字与计算机所选数字相符的概率是多少?
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  7. What is the probability that one of a player’s numbers matches the numbers chosen by the computer?
     ::玩家数字之一与计算机所选数字相符的概率是多少?
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  8. What is the probability that none of a player’s numbers match the numbers chosen by the computer?
     ::一个玩家的数字与计算机所选择的数字不匹配的可能性有多大?

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  9. A bag contains 13 dominoes. Each domino has a different number of dots on it, and the numbers of dots are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. Peter selects 2 dominos at random from the bag. What is the probability that the total number of dots on the two dominoes he selects is 7?
     ::一个袋子包含13个多米诺。每个多米诺有不同的点数, 点数是 0, 1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12。 彼得从袋子中随机选择了 2个多米诺。 他选择的两个多米诺上的点数总数是 7 的概率是 7 ?
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  10. Looking at the odds that you came up with in question 4, devise a sensible payout plan for the lottery—in other words, how big should the prizes be for players who match 1, 2, or all 3 numbers? Assume that tickets cost $1. Don’t forget to take into account the following:
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      1. The town uses the lottery to raise money for schools and sports clubs.
         ::该镇利用彩票为学校和体育俱乐部筹款。
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      2. Selling tickets costs the town a certain amount of money.
         ::卖票要花镇上一大笔钱
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      3. If payouts are too low, nobody will play!
         ::如果报酬太低,没人会玩!
      
      ::看看你们在问题4中得出的几率 — — 设计一个明智的彩票付款计划 — — 换句话说,彩票的奖项对符合1、2或所有3位数的球员应该有多大?假设票价是1美元。 不要忘了考虑以下因素:镇上用彩票为学校和体育俱乐部募集钱。 卖票花费了镇上一定数额的钱。 如果奖金太低,没有人会玩!

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。