独立与依赖

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Independence Versus Dependence 
::独立与依赖

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If the result of one event has no bearing on the probability of the second event, we call them . For example, if you flip a coin 3 times and get heads 3 times, what is the probability that the next flip will result in tails? Many people think that the previous run of heads somehow influences the flip to make tails more likely, but in reality the previous flips have no bearing on the outcome of the new flip – how could they? The coin doesn’t have a brain or a memory.
::如果一个事件的结果对第二个事件的概率没有影响,我们称之为“我们” 。 比如,如果你翻了一个硬币3次,然后换了3次头,那么下一个硬币会导致尾巴的概率是多少? 许多人认为,前一次头部的转动会以某种方式影响尾巴的概率,但在现实中,前一次转动对新翻动的结果没有影响 — — 怎么会有这种影响? 硬币没有大脑或记忆力。

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(The idea that tails is more likely after a run of heads is called the Gambler’s Fallacy, and it probably arises from people getting confused about something called prior probability . Now that you’ve learned a little about probability, you know that getting three heads and one tail is a little more likely than getting four heads on four coin flips, so before you flip the coin, you’d expect that it’s more likely you’ll get three heads than four heads. But after you’ve already flipped the coin three times, the chances of getting heads on the first three flips don’t matter any more, because you already got those three heads; the probability of getting those first three heads has gone from 12.5% to 100%! So the only probability that still matters is the probability of getting heads on the one flip remaining, which is just the same as it always is on a single coin flip: 50%.)
:所谓尾巴更有可能是在赌博者的赌局的赌局的赌局的赌局的赌局的赌局的赌局的赌局之后,它可能产生于人们对所谓的先期概率的迷惑。现在,你对概率略有了解,你知道获得三个头和一个尾的概率比在四个硬币的翻盘上获得四个头的可能性要小一些,所以在你翻硬币之前,你可能会期望它更有可能得到三个头而不是四个头。但是,在你已经翻了三个硬币之后,在头三个翻盘上获得头的可能性就不再重要了,因为你已经有了这三个头;把头三个头从12.5%上升到100 % 。 因此,唯一重要的可能性是把头移到剩下的一个硬币的概率,这与在一个硬币翻盘上总是一样:50 % ) 。 )

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Because one flip of the coin has no effect on the outcome of any other flips, each flip of the coin counts as an independent event .
::因为一枚硬币的翻转对任何其他硬币的翻转结果都没有影响,每一枚硬币的翻转都算作一个独立事件。

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To find the probability of multiple independent events happening together, we multiply the individual probabilities:
::为了发现一起发生多个独立事件的可能性,我们乘以个体概率:

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::独立事件:P(A和B)=P(A)P(B)

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Calculating Probability 
::计算概率

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1. Find the probability of rolling a 5 on a 6-sided die and getting heads if you flip a coin at the same time.
::1. 发现同时抛硬币时,六面形死亡和头部被击落的概率。

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Clearly the outcome of rolling a die has no effect on flipping a coin, so the two events are independent . So $\begin{array}{r}P(5\text{and heads})=P(5)\cdot P(\text{heads})=\frac{1}{6}\times \frac{1}{2}=\frac{1}{12}.\end{array}$
::显然滚动死亡的结果对翻硬币没有影响, 所以这两个事件是独立的。 所以 P( 5 和 头) = P(5)P( 头) = 16x12= 112 。

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2. Out of the 480 students in a school, 40 have art first period; also, 96 students have math first period. Find the probability that a student picked at random will either have math or art in first period.
::2. 在校学生480人中,40人首学艺术,96人首学数学,发现随机选取的学生在首学数学或艺术的概率。

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A student cannot take both math and art during the same period, so the events are not overlapping. If event A is having art first period and event B is having math first period, $\begin{array}{r}P(AandB)=0\end{array}$ . We want to find $\begin{array}{r}P(AorB)\end{array}$ .
::学生在同一期间不能同时学习数学和艺术, 因此事件不会重叠 。 如果事件 A 有艺术首期, 事件 B 有数学首期, P( A 和 B) = 0 。 我们希望找到 P( A 或 B ) 。

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$\begin{array}{r}P(AorB)=P(A)+P(B)-P(AandB)\end{array}$
::P(A或B)=P(A)+P(B)-P(A和B)

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$\begin{array}{r}P(AorB)=\frac{40}{480}+\frac{96}{480}-0\end{array}$
::P(A或B)=40480+96480-0

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$\begin{array}{r}P(AorB)=\frac{96}{480}\end{array}$
::P(A或B)=96480

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$\begin{array}{r}P(AorB)=\frac{1}{5}=20\mathrm{\%}\end{array}$
::P(A或B)=15=20%

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Finding the Probability of Dependent Events
::查找依赖事件的可能性

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If the result of one event influences the probability of the second, we call them dependent events . For example, if you pick two cards from a deck, the chances of getting an ace on the first pick is $\begin{array}{r}\frac{4}{52}=\frac{1}{13}\end{array}$ . If you keep that ace and draw again, the chance of getting another ace on your second pick is less: there are now only 3 aces left in the deck (of 51 cards), so the chance of getting an ace is $\begin{array}{r}\frac{3}{51}=\frac{1}{17}\end{array}$ . To find the probability of getting two aces we multiply the two individual probabilities: $\begin{array}{r}\frac{1}{13}\times \frac{1}{17}=\frac{1}{221}\end{array}$ .
::如果一个事件的结果影响第二个事件的概率,我们称之为附带事件。例如,如果你从甲板上选取两张牌,第一个取出A的可能性是452=113。如果你保留A,再抽一次,第二个取出A的可能性就更少了:现在甲板上只剩下三张A(51张),因此获得A的概率是351=117。要找到获得两张A的可能性,我们要将两种个人概率(113×117=1221)乘以113×117。

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Three cards are picked from a standard 52 card deck. The cards are not replaced. Find the probability of picking 3 queens.
::从标准的52张牌牌牌中选取三张牌。这些牌不替换。 找出选择 3 Q 的可能性 。

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There are 52 cards and 4 of them are queens, so the chance of getting a queen on the first pick is $\begin{array}{r}\frac{4}{52}=\frac{1}{13}\end{array}$ .
::共有52张卡片 其中4张是皇后 所以第一次选取皇后的机会是452=113

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Assuming you get a queen on the first pick, there are 51 cards remaining of which 3 are queens, so the chance of getting a queen on the second pick is $\begin{array}{r}\frac{3}{51}=\frac{1}{17}\end{array}$ .
::假设你第一次选一个皇后, 还有51张卡片, 其中3张是皇后, 所以第二次选一个皇后的机会是351=117。

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If you were successful on the second pick, there will be 50 cards remaining of which 2 are queens, so the chance of getting a queen on the third pick is $\begin{array}{r}\frac{2}{50}=\frac{1}{25}\end{array}$ .
::如果你在第二个选择中成功, 剩下的50张卡片, 其中2张是皇后, 所以在第三个选择中获得皇后的机会是250=125。

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The probability of picking 3 queens in a row is $\begin{array}{r}\frac{1}{13}\times \frac{1}{17}\times \frac{1}{25}=\frac{1}{5525}\end{array}$ or 1 in 5,525 .
::在行中选择 3 Q 的概率是 113x117x125=15525 或 5 525 1 。

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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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100 raffle tickets were sold and Peter bought 4 of them. There are 3 prizes, and winners are selected randomly from a hat containing all the numbers. Find the probability that Peter wins all 3 prizes.
::100张彩票被卖掉,彼得买了其中4张。有3个奖项,赢家是从包含所有数字的帽子中随机挑选出来的。看看Peter赢得所有3个奖项的可能性。

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For the first draw, Peter’s numbers account for 4 tickets out of 100: $\begin{array}{r}\frac{4}{100}=\frac{1}{25}\end{array}$
::首先,Peter的数字占100张票中的4张票:4100=125。

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For the second draw, Peter’s remaining numbers (assuming he won the first draw) account for 3 tickets out of 99: $\begin{array}{r}\frac{3}{99}=\frac{1}{33}\end{array}$
::在第二批中,Peter的其余号码(假设他赢了第一批)在99张399=133张票中占3张。

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For the first draw, Peter’s remaining numbers (assuming he won the first two draws) account for 2 tickets out of 98: $\begin{array}{r}\frac{2}{98}=\frac{1}{49}\end{array}$
::首先,Peter的其余号码(假设他赢了头两张)占98张票中的2张:298=149。

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The probability that Peter wins all 3 prizes is $\begin{array}{r}\frac{1}{25}\times \frac{1}{33}\times \frac{1}{49}=\frac{1}{40425}\end{array}$ or 1 in 40, 425 .
::彼得赢得所有三个奖项的概率是125x133x149=140425 或40,425分之一。

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Review 
::回顾

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For 1-6, determine whether the events are dependent or independent.
::1-6,确定事件是依附的还是独立的。

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1. Driving at night and falling asleep at the wheel.
   ::晚上开车,在车轮上睡着了
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2. Visiting the zoo and seeing a giraffe.
   ::去动物园看长颈鹿
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3. The next 2 cars you see are both red.
   ::接下来的两辆车都是红色的
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4. A coin tossed twice comes up heads both times.
   ::抛硬币两次 头两次出现。
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5. Being dealt 4 aces in a hand of poker.
   ::4个A在扑克牌手里被打4个A
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6. It is your birthday and it is a windy day.
   ::今天是你的生日 今天是狂风的一天

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For 7-10, a bag contains 10 colored marbles – 4 red, 4 blue and 2 green. Calculate the probability of:
::7-10时,一个包装有10个彩色弹珠——4个红色、4个蓝色和2个绿色。计算概率:

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7. Removing 2 green marbles in a row if you replace the marble each time.
   ::如果你每次换掉大理石 就会一排清除两颗绿色大理石
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8. Removing 2 green marbles in a row if you do not replace the marble each time.
   ::如果每次不更换大理石,则将两颗绿色大理石排成一行删除。
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9. Removing 3 marbles without replacing and getting all blue.
   ::清除三颗大理石,不更换, 得到所有蓝色。
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10. Removing 4 marbles without replacing and getting exactly 3 blue.
    ::清除四颗大理石,不替换 并获得精确的3蓝色。

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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 。