13.4 概率和变异

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  概率和变相

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  Probability and Permutations 
  ::概率和变相

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  When we talk about probability , we are talking about chance. We may use words like ‘possible’ to indicate something has a low to medium chance of happening, or ‘probable’ to indicate something has a high chance of happening.
  ::当我们谈论概率时,我们谈论的是机会。我们可以用“可能性”这样的词来表示某事发生的可能性低到中等,或者“可能性”表示某事发生的可能性大。

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  When we perform calculations in probability we are generally looking at a situation to find the number of favorable outcomes in relation to the total number of all possible outcomes. In mathematics, a favorable outcome simply means the outcome we are looking to solve for. Over the course of 100 days a child may get ice-cream on 25 of those days. A mathematician would look at that data and conclude that the fraction of days when the child gets ice-cream is one-fourth. He may also conclude that on any day the child has a “one in four chance” of receiving ice-cream. This is one way to think about probability: the number of times something favorable happens divided by the total number of outcomes. We have no way of knowing for sure if the child is going to get ice-cream today, but we can estimate the chance that he will by looking at the number of times he did get ice-cream and the number of times he did not.
  ::当我们以概率进行计算时,我们一般都在研究一种情况,以找到相对于所有可能结果的总数而言的有利结果的数量。在数学中,有利的结果只是意味着我们期待的结果。在100天的时间里,一个儿童可能在25天里得到冰淇淋。一位数学家将研究这些数据,并得出结论,儿童得到冰淇淋的天数的一小部分是四分之一。他还可以得出结论,在任何一天里,儿童有接受冰淇淋的“四分之一的机会 ” 。这是思考概率的一种方式:一个有利的事情发生的次数与结果的总数分开。我们无法肯定儿童今天是否能得到冰淇淋,但我们可以估计他通过查看他得到冰淇淋的次数和没有得到的机会。

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  Real-World Application: Movies 
  ::真实世界应用程序:电影

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  Consider the example where Nadia and Peter are going to watch two movies on a rainy Saturday. Nadia will choose the first movie, and Peter gets to choose the second. The four movies they have to choose from are The Lion King, Aladdin, Toy Story and Pinocchio. Given that Peter will choose a different movie than Nadia, what is the probability that Aladdin is the second movie they will watch?
  ::想想Nadia和Peter将在雨季星期六看两部电影的例子。Nadia将选择第一部电影,Peter将选择第二部电影。他们必须选择的四部电影是狮子王、阿拉丁、玩具故事和皮诺奇奥。 鉴于Peter将选择与Nadia不同的电影,阿拉丁是他们将要观看的第二部电影的概率是多少?

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  There are a total of 12 possible outcomes. Out of those 12, Aladdin was the second movie 3 times. So the probability is as follows:
  ::总共12个可能的结果。在这12个中,阿拉丁是第二部电影,3次。概率如下:

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  $\begin{array}{r}P(\text{Aladdin being second})=\frac{3}{12}=\frac{1}{4}=\end{array}$ one in four or 25%.
  ::P( 添加为第二位)=312=14=四分之一或25%。

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  Real-World Application: "21 Hearts"
  ::真实世界应用: "21心"

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  The card game “21 hearts” consists of dealing two cards to a player and counting the points as follows: face cards (King, Queen, Jack) are worth 10 points, and number cards are worth their face value except for aces , which are worth 11. The maximum score is 21. If the game is played only with the 13 cards in the hearts suit, what is the probability that a player will score 21?
  ::牌游戏“21红心”包括向玩家发放两张牌,并计算以下点数:脸牌(King,Queen,Jack)值10点,数字牌值面值,A值11分的牌除外,最高分是21分,如果游戏只用13张牌玩,玩家会得21分的概率是多少?

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  To find the answer, we need to know two pieces of information: 1) the total number of permutations it’s possible to get in the game and 2) the number of permutations that will score 21.
  ::为了找到答案,我们需要知道两部分信息1) 游戏中可能的变异总数;(2) 得21分的变异数目。

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  To find the total number of permutations for the game, use the formula $\begin{array}{r}{}_n{P}_r\end{array}$ with $\begin{array}{r}n=13\end{array}$ and $\begin{array}{r}r=2\end{array}$ :
  ::要找到游戏的变换总数, 请使用 n=13 和 r= 2 的公式 nPr :

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  $$\begin{array}{r}{}_{13}{P}_2=\frac{13!}{(13-2)!}=\frac{13!}{11!}=13\times 12=156\text{permutations}\end{array}$$

  
  ::13P2=13!(13-2)!=13!11!=13x12=156变式

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  Now we need to determine how many of those permutations score 21. We can do that by simply listing all ways to score 21. Remember, of course, that as we are talking about permutations, (ace, king) and (king, ace) each count as separate hands! The winning permutations are:
  ::现在,我们需要确定其中多少人得21分,我们只要列出所有得21分的方法就可以做到这一点。 当然,记住,当我们谈论变异,(王,王)和(王,王)时, 每一个都算为单独的手! 获胜的变异是:

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  $$\begin{array}{rl}& (\text{ace, king})(\text{king, ace})(\text{ace, queen})(\text{queen, ace})\\ & (\text{ace, jack})(\text{jack, ace})(\text{ace},10)(10,\text{ace})\end{array}$$

  
  :ace, king), ace(ce, queen), ace(ce, jake), ace(ce, jake), Ace(ce, ce), a(10,ace))

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  There are 8 winning hands, so the probability of scoring 21 is given by:
  ::得分21的概率如下:

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  $\begin{array}{r}P(21)=\frac{8}{156}=\frac{2}{39}=\end{array}$ two in thirty- nine  or approximately 5%.
  ::P(21)=8156=239=三十九分之二或大约5%。

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  Sometimes when looking at probability (especially in games of chance), there are too many wining permutations to calculate directly. In such circumstances it can be useful to remember that a player must either win or lose – it’s impossible to do both! In these circumstances remember that:
  ::有时候,当看概率(特别是在机会游戏中)时,太多的赢取式变数可以直接计算。 在这种情况下,记住玩家要么赢要么输可能有用 — — 两者都不可能。 在这些情况下,记住:

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  $$\begin{array}{r}P(\text{winning})+P(\text{losing})=1\end{array}$$

  
  ::P(获)+P(关闭)=1

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  Since it’s certain that you’ll either win or lose, the probability that you’ll win and the probability that you’ll lose add up to 1.
  ::既然确定你要么赢要么输,那么你赢的概率和输的概率加起来等于1。

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  So to calculate the probability of losing we can use:
  ::因此,为了计算损失概率,我们可以使用:

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  $$\begin{array}{r}P(\text{winning})=1-P(\text{losing})\end{array}$$

  
  ::P(获)=1-P(关闭)

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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 throwing three darts at a board with 16 numbered squares in a $\begin{array}{r}4\times 4\end{array}$ grid. The squares are numbered 1 through 16 and no number is repeated or omitted. In order to win a player needs to score 9 or more. If a dart hits a square that has already been taken or if a dart misses the board the player must throw the dart again. What are the chances of winning the game?
  ::有趣的游戏包括向一个板块投掷三只飞镖, 4x4 网格中有16个编号的方块。 方块编号为 1 至 16, 没有重复或省略数字 。 要赢得玩家得分9 或以上 。 如果飞镖击中已经拍摄的方块, 或者飞镖错过了棋盘, 玩家必须再次扔出飞镖 。 赢得游戏的机会是什么 ?

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  There are too many winning permutations to list easily, but there are only a few losing permutations . If you need a score of 9 or more to win, then you’ll lose with a score of 8 or less. There are only four combinations of numbers that add up to 8 or less (remember that you can’t repeat a number), and they can occur in any of the following orders:
  ::有太多的胜者可以轻易地列出,但只有少许的败者。 如果你需要9分或9分以上才能获胜,那么8分或更少的得分就会输。 数字组合只有四组,加起来等于或少于8分(记住不能重复一个数字 ) , 它们可以在以下任何一项命令中出现:

  $$\begin{array}{rl}& \mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{(}\mathbf{=}\mathbf{6}\mathbf{)}1,3,22,1,32,3,13,1,23,2,1\\ & \mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{4}\mathbf{(}\mathbf{=}\mathbf{7}\mathbf{)}1,4,22,1,42,4,14,1,24,2,1\\ & \mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{5}\mathbf{(}\mathbf{=}\mathbf{8}\mathbf{)}1,5,22,1,52,5,15,1,25,2,1\\ & \mathbf{1}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{4}\mathbf{(}\mathbf{=}\mathbf{8}\mathbf{)}1,4,33,1,43,4,14,1,34,3,1\end{array}$$

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  So there are 24 permutations totaling less than 9.
  ::因此,有24个总变差总数不到9个。

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  The total number of ways to choose 3 numbers out of 16 is
  ::选择16个中3个数字的方法总数

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  $$\begin{array}{r}{}_{16}{P}_3=\frac{16!}{16-3!}=\frac{16!}{13!}=16\times 15\times 14=3360\text{permutations}\end{array}$$

  
  ::16P3=16! 16- 3!=16! 13!=16x15x14=3360 变换

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  So the probability of losing the game is $\begin{array}{r}\frac{24}{3360}\end{array}$ , or $\begin{array}{r}\frac{1}{140}\end{array}$ . That means the probability of winning is:
  ::所以输掉比赛的概率是243360或1140。这意味着赢的概率是:

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  $\begin{array}{r}P(\text{winning})=1-P(\text{losing})=1-\frac{24}{3360}=1-\frac{1}{140}=\frac{139}{140}\end{array}$ or approximately 99.3% .
  ::P(获奖)=1-P(获奖)=1-243360=1-1-1140=139140,或大约99.3%。

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  Example
  ::示例示例示例示例

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  Example 1
  ::例1

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  What is the probability that a randomly generated 3-letter arrangement of the letters $\begin{array}{r}M\end{array}$ , $\begin{array}{r}M\end{array}$ , and $\begin{array}{r}O\end{array}$ will result in spelling the word $\begin{array}{r}MOM\end{array}$ ?
  ::随机生成的字母M、M和O的3字母安排,导致MOM这个词拼写的可能性有多大?

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  To find the probability, we first must describe the sample space . It might be tempting to think that the only possibilities are $\begin{array}{r}MMO\end{array}$ , $\begin{array}{r}MOM\end{array}$ and $\begin{array}{r}OMM\end{array}$ . However, we must treat each $\begin{array}{r}M\end{array}$ as an distinct letter. Let's think of them as $\begin{array}{r}{M}_1\end{array}$ and $\begin{array}{r}{M}_2\end{array}$ . Since we have 3 distinct objects and we want to order all three of them, this is:
  ::要找到概率, 我们首先必须描述样本空间。 认为唯一的可能性是 MMO、 MOM 和 OMM。 但是, 我们必须将每个 MM 视为一个不同的字母。 让我们把它们想象成 M1 和 M2 。 因为有三个不同的对象, 我们想要订购所有三个对象,

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  $\begin{array}{r}{}_3{P}_3=3!=3\cdot 2\cdot 1=6\end{array}$
  ::3P3=3!=321=6

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  Thus, there are 6 ways to order the 3 letters. Let's see what that looks like:
  ::因此,有6种方法可以订购这3个字母。让我们看看这看起来像什么:

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  $$\begin{array}{r}{M}_1{M}_{2}O\\ MOM=M_{1}O{M}_2\\ M_2{M}_{1}O\\ MOM=M_{2}O{M}_1\\ O{M}_1{M}_2\\ O{M}_2{M}_1\end{array}$$

  
  ::M1M2OM = M1OM2M2M2M1OM = M2OM1OM1M2M2OM2M2MOM1

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  2 of these 6 orderings spell $\begin{array}{r}MOM\end{array}$ , which means the probability of spelling $\begin{array}{r}MOM\end{array}$ is:
  ::在这6个命令中,有2个命令拼写MOM,这意味着拼写MOM的概率是:

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  $\begin{array}{r}P(MOM)=\frac{2}{6}\approx 0.67\end{array}$ .
  ::P(MOM)=260.67。

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  Review
  ::回顾

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  For 1-5, find the number of permutations.
  ::1 -5,找到变换次数。

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  1. $\begin{array}{r}{}_5{P}_2\end{array}$
     ::5P2 5P2
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  2. $\begin{array}{r}{}_9{P}_4\end{array}$
     ::9P4
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  3. $\begin{array}{r}{}_{11}{P}_5\end{array}$
     ::11P5 11P5
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  4. How many ways can you plant a rose bush, a lavender bush and a hydrangea bush in a row?
     ::有多少种方法可以种玫瑰灌木, 熏衣草和花朵灌木在一排?
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  5. How many ways can you pick a president, a vice president, a secretary and a treasurer out of 28 people for student council?
     ::在28个学生会的会员中, 有多少方法可以挑选一名总统、一名副总统、一名秘书和一名司库?

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  For 6-10, find the probabilities.
  ::6 -10,找到概率。

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  6. What is the probability that a randomly generated 3-letter arrangement of the letters A,E,L, Q and U will result in spelling the word EQUAL?
     ::随机生成的字母A、E、L、Q和U的3字母安排 导致EQUAL这个词拼写的可能性有多大?
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  7. What is the probability that a randomly generated 3-letter arrangement of the letters in the word SPIN ends with the letter N?
     ::随机生成的3字母的字母安排在SPIN一词中以字母N结尾的概率是多少?
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  8. A bag contains eight chips numbered 1 through 8. Two chips are drawn randomly from the bag and laid down in the order they were drawn. What is the probability that the 2-digit number formed is divisible by 3?
     ::一个袋子里有8个芯片,编号为1至8。两个芯片是从袋子中随机抽取的,并按抽取顺序排列。2位数的组成数字除以3的可能性有多大?
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  9. A prepaid telephone calling card comes with a randomly selected 4-digit PIN, using the digits 1 through 9 without repeating any digits. What is the probability that the PIN for a card chosen at random does not contain the number 7?
     ::预付电话呼叫卡带有随机选择的4位数 PIN,使用数字 1至9,不重复任何数字。随机选择的卡的 PIN 不包含数字 7 的概率是多少?
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  10. Janine makes a playlist of 8 songs and has her computer randomly shuffle them. If one song is by Little Bow Wow, what is the probability that this song will play first?
      ::Janine制作了一个8首歌的播放列表, 并让她的电脑随机地打乱了这些歌曲。 如果其中一首歌是小鲍Wow, 那么这首歌首播的可能性是多少?

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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 。