13.1 概率测量

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  概率测量

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  Measurement of Probability 
  ::概率测量

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  A sample space is the set of all possible outcomes for an event . In tossing a coin, the sample space consists of 2 outcomes – getting heads, and getting tails. Each of these outcomes ( heads and tails ) could be considered an event . Each event has 1 matching element in the sample space .
  ::样本空间是事件所有可能结果的一组。 在抛硬币时,样本空间包含两个结果 — — 获取头部和尾部。 每一个结果(头部和尾部)都可以被视为一个事件。 每个事件在样本空间中都有一个匹配元素。

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  For example, the roll of a single die has 6 possible outcomes: the die can show any number from 1 to 6. We can say that the sample space for rolling a single die contains 6 outcomes: {1, 2, 3, 4, 5, 6}. If we say we are interested in rolling a six, then rolling a six is our event and this event has 1 matching element in the sample space: {6}. If, on the other hand, we are interested in rolling an even number, then rolling an even number is our event, and this event has 3 matching elements in the sample space: {2, 4, 6}.
  ::例如,单次死亡的滚动有6种可能的结果:死亡可以显示从1到6的任何数字。我们可以说滚动单一死亡的样本空间包含6种结果 : {1, 2, 3, 4, 4, 5, 6}。如果我们说我们有兴趣滚动一个六,那么滚动一个六就是我们的活动,而这个活动在样本空间里有一个匹配元素: {6}。另一方面,如果我们有兴趣滚动一个偶数,那么滚动一个偶数就是我们的活动,而这个活动在样本空间里有三个匹配元素: {2, 4, 6}。

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  Real-World Application: Rolling Dice 
  ::真实世界应用程序:滚骰

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  A pair of standard, 6-sided dice are rolled, and the total of the numbers that come up determines a player’s score. Find the sample space of possible outcomes, and determine how many outcomes result in a score of 5.
  ::一对标准、六面骰子被滚动,产生的数字总数决定了玩家的得分。 找出可能结果的样本空间,并确定有多少结果得出5分。

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  The scores a player can get are those in the following set: {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. But the sample space isn’t just this set of 11 events. For example, there’s only one way to score 12: the player has to roll a six on each of the dice. But to score 5, the player can roll a 1 and a 4, or a 2 and a 3. Furthermore, there are 2 possibilities for each of these combinations (imagine one die is red, and 1 die is green: we could roll 1 on the red die and 4 on the green or we could roll 4 on the red die and 1 on the green). Even though the dice we actually use may appear identical, they’re still separate entities, and it does make a difference which one rolls which number. So there are 4 ways a player can score 5:
  ::玩家可以得到的分数是以下组合中的分数 : {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} 。 但样本空间不仅仅是这组11个事件。 例如,只有一种方法可以得分12:玩家必须在每个骰子上滚动六分。 但是,要得分5,玩家可以滚1和4, 或2和3。 此外,每种组合都有两种可能性(想象一个人死亡是红色的,1人死亡是绿色的:我们可以在红色死亡上滚1,4人死亡是绿色的,或者我们可以在红色死亡上滚4,1人死亡是绿色的)。即使我们实际使用的骰子看起来可能是一样的,它们仍然是不同的实体,并且确实可以产生一个数字的区别。 因此有4种方法可以使玩家得分5:

  $$\begin{array}{r}(1\mathrm{\&}4)(2\mathrm{\&}3)(3\mathrm{\&}2)(4\mathrm{\&}1)\end{array}$$

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  To find the full sample space, we must consider all possible outcomes. The best way to do this is with a table (the outcomes that give a score of 5 are highlighted):
  ::为了找到完整的样本空间,我们必须考虑所有可能的结果。 这样做的最佳方式是用一张表格(突出显示得分为5的结果):

  

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  The sample space (shown above) has 36 outcomes, 4 of which result in a score of five.
  ::抽样空间(如上所示)有36个结果,其中4个结果得5分。

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  Notice that the number of outcomes in the sample space when you roll 2 dice is the product of the number of outcomes when you roll one die and the number of outcomes when you roll the other die. That is, there are 6 possible outcomes for one die and 6 outcomes for the other die, so there are $\begin{array}{r}6\times 6=36\end{array}$ possible outcomes when you roll both dice together. This property will be important later.
  ::请注意, 当您滚动 2 骰子时, 样本空间中的结果数是您滚动一死时结果数和滚动另一死时结果数的结果数的产物。 也就是说, 一个死时可能有6个结果, 另一个死时可能有6个结果, 所以当您同时滚动双骰子时, 就会有 6x6=36 的可能结果。 此属性稍后将很重要 。

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  Find Theoretical Probability of an Event
  ::查找事件理论上的概率

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  The theoretical probability of an event is a measure of how likely a given outcome (the event ) is for a particular experiment , such as tossing a coin. If the experiment were carried out a nearly infinite number of times, the probability of a particular event would be the ratio of how many times a particular outcome occurred to how many times the experiment was performed.
  ::事件的理论概率是衡量某一结果(事件)对某一实验(例如抛硬币)的概率的尺度。如果实验进行的次数几乎是无限的,特定事件的概率将是某一结果发生多少次与实验进行多少次的比率。

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  We write the probability that a particular event, $\begin{array}{r}E\end{array}$ , occurs as:
  ::我们写下一个特定事件(E)发生的概率如下:

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  $$\begin{array}{r}P(E)\end{array}$$

  
  ::P(英)项

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  For example, when tossing a coin we may only be interested in getting heads. We could denote that probability as:
  ::例如,在抛硬币时,我们可能只有兴趣获得头部。我们可以表示这种概率如下:

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  $$\begin{array}{r}P(Heads)\text{or simply}P(H)\end{array}$$

  
  ::P(主管)或简便P(H)

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  We know that the sample space for tossing a single coin has two elements: Heads and Tails (or $\begin{array}{r}H\end{array}$ and $\begin{array}{r}T\end{array}$ ). Each is as likely as the other to occur, so we know that:
  ::我们知道,抛掷一枚硬币的样本空间有两个要素:正面和反面(或H和T)。 每个都有可能发生,另一个可能发生,所以我们知道:

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  $$\begin{array}{r}P(H)=\frac{1}{2}\end{array}$$

  
  ::P(H)=12

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  To find the probability of a particular event, we look at how many possible outcomes would contribute to that event, and divide that number by the total number of outcomes in the sample space.
  ::为了找到某一具体事件的概率,我们研究有多少可能的结果有助于这一活动,并将这一数字除以抽样空间结果的总数。

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  $$\begin{array}{r}P(E)=\frac{EventSpace}{SampleSpace}\end{array}$$

  
  ::P(E) = Event SpaceSample空间

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  When we roll a single 6-sided die, we know that the chances of rolling a 3 are 1 in 6. This is, in effect, the probability of rolling 3:
  ::当我们滚动一个单一的六方死亡, 我们知道滚动3的机会是六分之一, 这实际上是滚动3的概率是三分之一:

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  $$\begin{array}{r}P(3)=\frac{1}{6}\end{array}$$

  
  ::P(3)=16

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  Calculating Probability
  ::计算概率

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  Four coins are tossed simultaneously. What is the probability of getting three or more tails?
  ::四个硬币同时被丢弃。 获得三个或更多尾巴的概率是多少?

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  We’ll start by listing all the possible outcomes in a table. But how many elements will the table have?
  ::首先,我们将在表格中列出所有可能的结果。 但表格中有多少元素?

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  Remember that when we rolled 2 dice together, we found the number of outcomes by multiplying the number of outcomes for one die by the number of outcomes for the other die. So if we’re flipping four coins, it makes sense to multiply the number of outcomes for each of the four coins together. There are 2 outcomes for each coin, so when all four coins are flipped there should be $\begin{array}{r}2\times 2\times 2\times 2=16\end{array}$ outcomes. Let’s organize them as follows:
  ::记住当我们一起滚动 2 骰子时, 我们通过将一个死亡的结果数乘以另一个死亡的结果数而发现结果数。 因此, 如果我们翻转四个硬币, 将四个硬币的每个硬币的结果数乘以一个死亡的结果数, 那么将四个硬币的每个硬币的结果数乘以四个硬币。 每个硬币有两个结果, 所以当所有四个硬币被翻转时, 结果数应该为 2x2x2x2=16。 把它们组织如下:

  

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  Once we fill out the table, we see that there are indeed 16 possible outcomes, and 5 of those outcomes match our event . So the probability of getting three or more tails is $\begin{array}{r}P(3\text{or more tails})=\frac{5}{16}\end{array}$ .
  ::一旦我们填写了表格,我们就会发现确实有16个可能的结果,其中5个结果与我们的活动相吻合。因此获得3个或3个以上尾巴的概率是P(3个或3个以上尾巴)=516。

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  Find Odds For and Against an Event
  ::查找和反对事件的风险

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  When we talk of probability , we generally think (as we’ve seen in this lesson) of the ratio of the number of times our event occurs to the number of times the experiment was carried out.
  ::当我们谈论概率时, 我们一般认为(正如我们从这个教训中看到的那样)我们事件发生次数与实验进行次数之比。

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  Another way to talk about the chances of an event occurring is with odds . You may have heard of the phrases “fifty-fifty” or “even odds” to describe an unpredictable situation like the chances of getting heads when you toss a coin. The phrase means that the coin is as likely to come up tails as it is heads (each event occurring 50% of the time). The odds of an event are given by the ratio of the number of times the event occurs to the number of times the event does not occur. In sample space terms it means:
  ::谈论某一事件发生概率的另一种方式是矛盾的。 您可能听说过“ 50- 50” 或“ 偶数” 等词来描述一种无法预测的情况, 比如扔硬币时获得头的机会。 这个词意味着硬币与头一样有可能出现尾巴( 每个事件发生50%的时间)。 事件发生的概率取决于事件发生次数与没有发生事件次数的比例。 在抽样空间中,它意味着:

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  $$\begin{array}{r}\text{Odds}=\frac{\text{number of matching events in sample place}}{\text{number of non-matching events in sample place}}\end{array}$$

  
  ::Odds = 抽样地点的匹配事件数; 抽样地点的非匹配事件数; 抽样地点的非匹配事件数; 抽样地点的非匹配事件数;

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  whereas we would describe probability as
  ::而我们将概率描述为

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  $$\begin{array}{r}\text{Probability}=\frac{\text{number of matching events in sample place}}{\text{number of total events in sample place}}\end{array}$$

  
  ::概率=抽样地点的匹配事件数目

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  To avoid confusion with probability, odds are usually left as a ratio such as 1:5, which would be read as “one to five”. When probability is read as a ratio, it’s usually written as a fraction like $\begin{array}{r}\frac{1}{5}\end{array}$ , which would usually be read as “one in five.”
  ::为了避免与概率混为一谈,概率通常被留作1:5之类比例,即1比5。 当概率被解读为比例时,概率通常被写成像15的分数,通常被解读为“五分之一 ” 。 当概率被解读为比例时,概率通常被写成像15分之一的分数,通常被解读为“五分之一 ” 。

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  Finding the Odds 
  ::寻找奇数

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  Find the odds of the following events:
  ::发现以下事件的几率:

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  The key to finding odds is looking at how many outcomes result in the event and how many do not :
  ::寻找不和的关键在于观察这次事件的结果有多少,有多少没有:

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  a) Tossing a coin and getting heads.
  :a) 抛硬币和砍头。

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  The sample space consists of 2 outcomes: 1 heads, 1 not-heads. The odds of getting heads are 1 : 1 ( one to one , or even ).
  ::样本空间由2个结果组成:1个头,1个非头。获得头的可能性为1:1(一对1,甚至一个)。

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  b) Rolling a die and getting a 3.
  :b) 滚动死亡并获得3分。

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  The sample space consists of 6 outcomes: 1 three and 5 not-three. The odds of getting three are 1 : 5 ( one to five ).
  ::样本空间包括6个结果:1 3 和 5 而非 3。 获得 3 的概率是 1 : 5 ( 1 至 5 ) 。

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  c) Tossing 4 coins and getting exactly 3 tails. 
  :c) 抛掷4枚硬币,并获得确切的3个尾巴。

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  Look back at the example where we found the sample space for tossing 4 coins. The sample space consists of 4 outcomes where exactly 3 tails came up and 12 outcomes when they did not. So the odds of getting 3 tails are 3 : 12 = 1 : 4 ( one to four ).
  ::举个例子来回顾一下, 我们找到了用于抛掷 4 个硬币的样本空间。 样本空间由 4 个结果组成, 其中 3 个尾巴正好出现, 12 个结果没有出现。 因此获得 3 个尾巴的概率是 3 : 12 = 1 : 4 ( 1 至 4 ) 。

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  Look carefully at part b above. This illustrates the need to avoid confusion between odds and probability. We know that the probability of getting a 3 is $\begin{array}{r}P(3)=\frac{1}{6}\end{array}$ or “one in six,” but the odds describes the same event with the ratio 1 : 5 or “one to five” .
  ::仔细看上文b部分。 这说明需要避免混淆概率和概率。 我们知道获得3的概率是P(3)=16或“六分之一 ” , 但概率是1比5或“一比五”的同一事件。

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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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  Determine the probability of scoring 5 as the combined score of a roll of 2 dice.
  ::确定评分5的概率为滚动 2 骰子的合并得分。

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  We have just seen (in example 2) that there are 4 ways we can score 5 if we roll 2 dice: {(1 & 4), (2 & 3), (3 & 2), (4 & 1)}. The sample space consists of $\begin{array}{r}6\times 6=36\end{array}$ elements, so the probability of rolling a 5 with 2 dice is $\begin{array}{r}P(5)=\frac{4}{36}=\frac{1}{9}\end{array}$
  ::我们刚刚看到(例2),如果我们滚动2骰子,我们有4种方法可以得5分:{(1和4)、2和3、3和3、3和2、4和1)。样本空间由 6x6=36 元素组成, 因此滚动 5 和 2 dice 的概率是 P(5)= 436=19

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  Review
  ::回顾

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  For 1-4, find the number of outcomes in the sample space of:
  ::1-4,在下列抽样空间中找到结果数目:

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  1. Tossing 3 coins simultaneously.
     ::一起扔3个硬币
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  2. Rolling 3 dice and summing the score.
     ::3骰子滚动 并打中得分。
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  3. Rolling 3 dice and interpreting the result as a 3 digit number.
     ::将3骰子滚动,并将结果解译为3位数。
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  4. Pulling a card from a standard 52-card deck.
     ::从标准52卡牌牌牌上拉一张牌

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  For 5-8, find the theoretical probability:
  ::对于 5-8, 找到理论概率 :

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  5. Tossing 3 coins simultaneously and getting 2 or more heads.
     ::同时扔3个硬币, 得到2个或2个以上的头。
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  6. Rolling 3 dice, summing the score and getting 17.
     ::3骰子滚动, 曲折得分和得到17。
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  7. Rolling 3 dice and interpreting the result as a 3 digit number, and getting 333.
     ::3张骰子滚动3张骰子 将结果解译为3位数 收到333张
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  8. Pulling a club from a standard 52-card deck.
     ::从标准的52卡牌甲板上拉一个俱乐部

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  For 9-12, find the odds:
  ::对于9 -12, 找出可能性:

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  9. Tossing 3 coins simultaneously and getting 2 or more heads.
     ::同时扔3个硬币, 得到2个或2个以上的头。
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  10. Rolling 3 dice, summing the score and getting 12.
      ::3号骰子滚动, 曲折得分和得到12。
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  11. Rolling 3 dice and interpreting the result as a 3 digit number, and not getting 333.
      ::3张骰子滚动3张骰子, 将结果解译为3位数, 但没有收到333张。
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  12. Pulling a club from a standard 52-card deck.
      ::从标准的52卡牌甲板上拉一个俱乐部

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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 。