Course: 11.2 独立活动-interactive

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独立活动

Disjoint and Overlapping Events
::分裂和重叠事件

When you toss a coin to calculate a probability , it is impossible to get a head and a tail at the same time. If you get a heads then the probability of tails is 0. Similarly, when you roll a single die, it is impossible to get an odd number and an even number at the same time. The above events are disjointed, or not connected, that is they have no outcomes in common. Events that don't have any outcomes in common are disjoint events .
::当您抛硬币来计算概率时, 无法同时得到一个头和尾。如果您得到一个头, 尾的概率是0。同样, 当滚动一个死时, 无法同时获得一个奇数和偶数。上述事件是互不关联的, 或没有连接, 也就是说它们没有共同的结果。没有共同结果的事件是脱节事件。

The for disjoint events shows no overlap between the two events.
::由于事与愿违,这两个事件之间没有重叠。

Disjoint events cannot happen at the same time. In other words, they are mutually exclusive . Put in formal terms: events $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ are disjoint if their intersection is an empty set :
::无法同时发生脱节事件。换句话说, 它们是相互排斥的。换句话说, 如果事件 A 和 B 的交叉点是空的, 事件 A 和 B 是脱节 :

\begin{aligned} P (A \cap B) = \emptyset \end{aligned}

::P(AB)________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

The events that are connected, that is they have some outcomes in common, are overlapping events .
::相互关联的事件,即具有一些共同结果的事件,是重叠的事件。

The Venn diagram for overlapping events shows some overlap between the two events.
::重叠事件的文恩图显示这两个事件之间有一些重叠。

INTERACTIVE

Independent Events (Warm Up)

Warm Up:
::温暖起来 :

You are playing a game of basketball, and you have just learned about disjoint events! The ball is passed to you and you have many choices of what you can do with it. Two of these choices are listed on the ball. Think about whether these events must happen independently or can happen simultaneously, and decide whether or not they are disjoint events.
::您正在玩篮球游戏, 您刚刚学会了脱节事件。球传给您, 您有很多选择。其中两个选择被列在球上。想想这些事件是否必须独立发生还是同时发生, 并决定它们是否是脱节事件。

Shoot the basketball into one of the "Disjoint Events" or "Non Disjoint Events" boxes by dragging it. Once the ball is in a valid field, a "Submit" button will appear. Click this button to reveal the correct answer and then click the subsequent "Next" button to reset the basketball.
::拖动篮球, 将篮球射入“ 脱节事件” 或“ 脱节事件” 的框中。一旦球进入一个有效的字段, 就会出现一个“ 上交” 按钮。单击此按钮以显示正确的答案, 然后单击随后的“ 下一步” 按钮重置篮球。

Determining Probability
::确定概率

Recall that the probability of an event is the chance of it happening. Probabilities can be written as fractions or decimals between 0 and 1, or as percents between 0% and 100%. T o find the probability of an event if all the outcomes in the event have an equal chance of occurring, find the number of outcomes in the event and divide by the number of outcomes in the sample space .
::提醒注意事件概率是发生事件的概率。概率可以以零到一之间的分数或小数或小数表示, 或者以0%到100%的百分率表示。要找到事件概率, 如果事件的所有结果都有同样的机会发生, 请找到事件结果的数量, 并用样本空间的结果数量来区分。

$\begin{aligned} P (E) = \frac{# of outcomes in E}{# of outcomes in sample space} \end{aligned}$
::样本空间结果E#结果结果的E(E)+_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Consider a dartboard. To determine the probability of landing a dart in a green zone, find the number of green sectors and divide by the total number of sectors on the dartboard. Similarly, if each of the letters of the word 'PROBABILITY' is written on a card and all the cards are well-shuffled and placed face down. To determine the probability of drawing a vowel card, find the number of vowel cards and divide by the total number of cards .
::考虑一个飞镖板。要确定在绿区降落飞镖的概率, 请在飞镖板上找到绿区数和区数。同样, 如果“ PROBABIBILIBY” 字词的字母都写在一张卡片上, 并且所有的卡片都被打乱了, 并面部被放倒。要确定绘制公誓卡的概率, 找到公誓卡数, 并用卡片总数来分隔。

A spinner is divided into 5 equal sectors of different colors. What is the probability of the pointer landing on the green, when the spinner is spun?
::旋转器被分为五等分层, 颜色不同。当旋转器被旋转时, 指针在绿色上着陆的概率是多少 ?

The probability of the pointer landing on the green is .
::指针在绿地上着陆的概率是

Consider the experiment of flipping a coin two times and recording the sequence of heads and tails.
::考虑两次翻硬币的实验记录头部和尾部的顺序

The sample space is $\begin{aligned} S = {H H, H T, T H, T T}, \end{aligned}$ which contains four outcomes.
::样本空间是SHH,HT,TH,TT}, 包含四个结果。

Let $\begin{aligned} A \end{aligned}$ be the event that heads comes up exactly once .
::让一个事件头一出现正好一次。

\begin{aligned} A = {H T, T H} \end{aligned}

::啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊,啊

Therefore: $\begin{aligned} P (A) \end{aligned}$ $\begin{aligned} = \frac{2}{4} = \frac{1}{2} \end{aligned}$
::因此:P(A)=24=12

To find the probability of a single event, all you need to do is count the number of outcomes in the event and the number of outcomes in the sample space. Probability calculations become more complex when you consider the combined probability of two or more events.
::要找到单一事件的概率, 您只需要计算事件结果的数量和样本空间的结果数量。如果考虑到两个或两个以上事件的综合概率, 概率计算会变得更加复杂。

Two events are independent if one event occurring does not change the probability of the second event occurring. Two events are dependent if one event occurring causes the probability of the second event to go up or down. Two events are independent if the probability of $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ occurring together is the product of their individual probabilities:
::如果一个事件的发生并不改变第二个事件的发生概率,两个事件是独立的。如果一个事件的发生导致第二个事件的发生概率上升或下降,两个事件是独立的。如果A和B的发生概率是其个别概率的产物,两个事件是独立的。

$\begin{aligned} P (A \cap B) = P (A) P (B) \end{aligned}$ if and only if $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ are independent events.
::P(AB) = P(A)P(B) ,条件是且仅在A和B是独立活动的情况下。

In some cases it is pretty clear whether or not two events are independent. In other cases, it is not at all obvious. You can always test if two events are independent by checking to see if the probabilities satisfy the relationship above.
::在某些情况下,两个事件是否独立是显而易见的。在另一些情况下,它根本不是显而易见的。你总是可以通过检查两种事件是否独立来测试它们是否独立,检查其概率是否满足上述关系。

CK-12 PLIX Interactive
::CK-12 PLIX 交互式互动

D etermining Independence
::确定独立性

1. Consider the experiment of flipping a coin two times and recording the sequence of heads and tails.
::1. 考虑两次翻硬币的试验,并记录头部和尾部的顺序。

The sample space is $\begin{aligned} S = {H H, H T, T H, T T}, \end{aligned}$ which contains four outcomes. Let $\begin{aligned} C \end{aligned}$ be the event that the first coin is a heads . Let $\begin{aligned} D \end{aligned}$ be the event that the second coin is a tails .
::样本空间是 SHH、HT、TH、TT}, 它包含四个结果。让 C 成为第一个硬币是头的事件。让 D 成为第二个硬币是尾巴的事件。

List the outcomes in events $\begin{aligned} C \end{aligned}$ and $\begin{aligned} D \end{aligned}$ .
::列出活动C和D的结果。

$\begin{aligned} C = {H H, H T} . \end{aligned}$ $\begin{aligned} D = {H T, T T} . \end{aligned}$ Note that the outcomes in the two events overlap. This does NOT mean that the events are not independent!
::注意这两个事件的结果重叠。这并不意味着事件不是独立的!

Take a guess at whether or not you think the two events are independent.
::猜猜你是否认为这两个事件是独立的。

If you get heads on the first coin, that shouldn't have any effect on whether you get tails for the second coin. It makes sense that the events should be independent.
::如果你在第一个硬币上找到头,那不应该对第二个硬币的尾巴是否得到尾巴有任何影响。事件应该是独立的,这是有道理的。

Find $\begin{aligned} P (C) \end{aligned}$ and $\begin{aligned} P (D) \end{aligned}$ .
::查找 P(C) 和 P(D) 。

$\begin{aligned} P (C) = \frac{2}{4} = \frac{1}{2} . P (D) = \frac{2}{4} = \frac{1}{2} \end{aligned}$ .
::P(C)=24=12 P(D)=24=12。

1b) Find $\begin{aligned} P (C \cap D) \end{aligned}$ - Are the two events independent?
::1(b) 找到P(CD) - 这两项活动是否独立?

$\begin{aligned} C \cap D \end{aligned}$ is the event of getting heads first and tails second . $\begin{aligned} C \cap D = {H T} \end{aligned}$ .
::CD是头部第一,尾部第二的事件 CD是头部第一,尾部第二

$\begin{aligned} P (C \cap D) & = \frac{1}{4} . \\ P (C) P (D) & = (\frac{1}{2}) (\frac{1}{2}) = \frac{1}{4} . \end{aligned}$
::P(CD)=14.P(C)P(D)=(12)(12)=14。

Because $\begin{aligned} P (C \cap D) = P (C) P (D), \end{aligned}$ the events are independent.
::因为P(CD)=P(C)P(D),事件是独立的。

2. Consider the experiment of flipping a coin two times and recording the sequence of heads and tails.
::2. 考虑两次翻硬币的试验,并记录头部和尾部的顺序。

The sample space is $\begin{aligned} S = {H H, H T, T H, T T} \end{aligned}$ , which contains four outcomes. Let $\begin{aligned} C \end{aligned}$ be the event that the first coin is a heads . Let $\begin{aligned} E \end{aligned}$ be the event that both coins are heads .
::样本空间是 SHH、HT、TH、TT}, 它包含四个结果。让 C 成为第一个硬币是头的事件。让 E 成为两个硬币都是头的事件。

List the outcomes in events $\begin{aligned} C \end{aligned}$ and $\begin{aligned} E \end{aligned}$ .
::列出活动C和E的结果。

$\begin{aligned} C = {H H, H T} \end{aligned}$ . $\begin{aligned} E = {H H} \end{aligned}$ .
::嘘,嗨,嗨

Take a guess at whether or not you think the two events are independent.
::猜猜你是否认为这两个事件是独立的。

If you get heads on the first coin, then you are more likely to end up with two heads than if you didn't know anything about the first coin. It seems like the events should NOT be independent.
::如果您在第一个硬币上找到头, 那么与您对第一个硬币一无所知的情况相比, 您更有可能以两个头结束。似乎事件不应该是独立的。

Find $\begin{aligned} P (C) \end{aligned}$ and $\begin{aligned} P (E) \end{aligned}$ .
::查找 P(C) 和 P(E) 。

$\begin{aligned} P (C) = \frac{2}{4} = \frac{1}{2} \end{aligned}$ . $\begin{aligned} P (E) = \frac{1}{4} \end{aligned}$ .
::P(C)=24=12;P(E)=14。

2b) Find $\begin{aligned} P (C \cap E) \end{aligned}$ . Are the two events independent?
::2(b) 寻找P(C)-E. 这两项活动是否独立?

$\begin{aligned} C \cap E \end{aligned}$ is the event of getting heads first and both heads. This is the same as the event of getting both heads, since if you got both heads then you definitely got heads first. $\begin{aligned} C \cap E = {H H} \end{aligned}$ .
::CE是头先头、两头先头的活动。这和把头兼头一样,因为如果你两头兼头,你肯定先头兼头。 CEHH}。

$\begin{aligned} P (C \cap E) & = \frac{1}{4} . \\ P (C) P (E) & = (\frac{1}{2}) (\frac{1}{4}) = \frac{1}{8} . \end{aligned}$
::P(CE)=14.P(C)P(E)=(12)(14)=18。

Because $\begin{aligned} P (C \cap E) \neq P (C) P (E) \end{aligned}$ , the events are NOT independent.
::因为 P(CE)P(C)P(E),事件并不独立。

3. Consider the experiment of flipping a coin two times and recording the sequence of heads and tails.
::3. 考虑两次翻硬币的试验,并记录头部和尾部的顺序。

The sample space is $\begin{aligned} S = {H H, H T, T H, T T} \end{aligned}$ , which contains four outcomes. Let $\begin{aligned} E \end{aligned}$ be the event that both coins are heads . Let $\begin{aligned} F \end{aligned}$ be the event that both coins are tails .
::样本空间是 SHH、HT、TH、TT}, 它包含四个结果。让E 成为两个硬币都是头的事件。让F 成为两个硬币都是尾巴的事件。

List the outcomes in events $\begin{aligned} E \end{aligned}$ and $\begin{aligned} F \end{aligned}$ .
::列出活动E和F的结果。

$\begin{aligned} E = {H H} \end{aligned}$ . $\begin{aligned} F = {T T} \end{aligned}$ .
::来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来

Take a guess at whether or not you think the two events are independent.
::猜猜你是否认为这两个事件是独立的。

If you get both heads, then you definitely didn't get both tails. It seems like the events should NOT be independent.
::如果你得到两个头,那么你肯定没有得到两个尾巴。似乎事件不应该是独立的。

Find $\begin{aligned} P (E) \end{aligned}$ and $\begin{aligned} P (F) \end{aligned}$ .
::查找P(E)和P(F)。

$\begin{aligned} P (E) = \frac{1}{4} \end{aligned}$ . $\begin{aligned} P (F) = \frac{1}{4} \end{aligned}$ .
::P(E)=14 P(F)=14。

3b) Find $\begin{aligned} P (E \cap F) \end{aligned}$ . Are the two events independent?
::3(b) 寻找P(E)-F. 这两项活动是否独立?

$\begin{aligned} E \cap F \end{aligned}$ is the event of getting two heads and two tails . This is impossible to do because these two events are disjoint. $\begin{aligned} E \cap F = {} \end{aligned}$ .
::EF是两个头和两个尾巴之间的事件。这是不可能的,因为这两个事件是脱节的。 EF。 EF。

$\begin{aligned} P (E \cap F) & = 0. \\ P (E) P (F) & = (\frac{1}{4}) (\frac{1}{4}) = \frac{1}{16} . \end{aligned}$
::PEF=0.PP(E)P(F)=(14)(14)=116。

Because $\begin{aligned} P (E \cap F) \neq P (E) P (F) \end{aligned}$ , the events are NOT independent.
::因为P(E)F)P(E)P(E)P(F),事件并不独立。

$\begin{aligned} P (G) = \frac{1}{3} \end{aligned}$ and $\begin{aligned} P (H) = \frac{1}{2} \end{aligned}$ . If $\begin{aligned} P (G \cap H) = \frac{1}{4} \end{aligned}$ , are events $\begin{aligned} G \end{aligned}$ and $\begin{aligned} H \end{aligned}$ independent?
::P( G) = 13 和 P( H) = 12。如果 P( G) = 14, 事件 G 和 H 是独立的吗 ?

The events $\begin{aligned} G \end{aligned}$ and $\begin{aligned} H \end{aligned}$ are .
::事件G和H是。

Examples
::实例实例实例实例

Example 1
::例1

Two events are disjoint if they have no outcomes in common. Consider two events $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ that are disjoint. Can you say whether or not events $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ are independent?
::两个事件如果没有共同的结果,就会脱节。认为两个A和B事件是脱节的。你能说事件A和B是否独立吗?

If $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ are disjoint, then $\begin{aligned} A \cap B = {} \end{aligned}$ and $\begin{aligned} P (A \cap B) = 0 \end{aligned}$ .
::如果A和B脱节,则A和B和P=0。

For the two events to be independent, $\begin{aligned} P (A) P (B) = P (A \cap B) \end{aligned}$ . This means that $\begin{aligned} P (A) P (B) = 0 \end{aligned}$ . By the zero product property, the only way for $\begin{aligned} P (A) P (B) = 0 \end{aligned}$ is if $\begin{aligned} P (A) = 0 \end{aligned}$ or $\begin{aligned} P (B) = 0 \end{aligned}$ .
::要使两个事件独立,P(A)P(B)=P(A)B。这意味着P(A)P(B)=0。根据零产品属性,P(A)P(B)=0的唯一途径是P(A)=0或P(B)=0。

In other words, two disjoint events are independent if and only if the probability of at least one of the events is 0.
::换句话说,如果而且只有在至少一个事件的概率为0的情况下,两个脱节事件是独立的。

Example 2
::例2

Consider the experiment of tossing a coin and then rolling a die. Event $\begin{aligned} A \end{aligned}$ is getting a tails on the coin. Event $\begin{aligned} B \end{aligned}$ is getting an even number on the die. Are the two events independent? Justify your answer using probabilities.
::考虑一下抛硬币然后滚死实验。事件A会得到硬币的尾巴。事件B会得到死亡的偶数。这两个事件是独立的吗? 用概率来说明答案的理由。

The sample space for the experiment is $\begin{aligned} S = {H 1, H 2, H 3, H 4, H 5, H 6, T 1, T 2, T 3, T 4, T 5, T 6} \end{aligned}$ . Next consider the outcomes in the events. $\begin{aligned} A = {T 1, T 2, T 3, T 4, T 5, T 6} \end{aligned}$ . $\begin{aligned} B = {H 2, H 4, H 6, T 2, T 4, T 6} \end{aligned}$ . $\begin{aligned} A \cap B \end{aligned}$ $\begin{aligned} = {T 2, T 4, T 6} \end{aligned}$ .
::实验的样本空间是SH1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T4,T5,T6}。接下来考虑活动的结果。AT1,T2,T3,T4,T5,T6},BH2,H4,H6,T2,T4,T6}。AT2,T4,T6}。

$\begin{aligned} P (A) = \frac{6}{12} = \frac{1}{2} \end{aligned}$
::P(A)=612=12

$\begin{aligned} P (B) = \frac{6}{12} = \frac{1}{2} \end{aligned}$
::P(B)=612=12

$\begin{aligned} P (A) P (B) = (\frac{1}{2}) (\frac{1}{2}) = \frac{1}{4} \end{aligned}$
::P(A)P(B)=(12)(12)=14

$\begin{aligned} P (A \cap B) = \frac{3}{12} = \frac{1}{4} \end{aligned}$
::P(AB)=312=14

The events are independent because $\begin{aligned} P (A) P (B) = P (A \cap B) \end{aligned}$ .
::这些事件是独立的,因为P(A)P(B)=P(A)B。

Example 3
::例3

Consider the experiment of rolling a pair of dice. Event $\begin{aligned} C \end{aligned}$ is a sum that is even and event $\begin{aligned} D \end{aligned}$ is both numbers are greater than 4 . Are the two events independent? Justify your answer using probabilities.
::想象一下滚动一对骰子的实验。事件C是一个偶数, 事件D是一个偶数, 事件D两个数字都大于 4 。这两个事件独立吗? 使用概率来解释答案是否合理。

First find the sample space and the outcomes in each event:
::首先找到每个活动的样本空间和结果 :

$\begin{aligned} S = { & (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \\ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \\ (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), \\ (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \\ (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), \\ (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} \\ C = { & (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), \\ (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), \\ (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)} \\ D = { & (5, 5), (5, 6), (6, 5), (6, 6)} \\ C \cap D = { & (5, 5), (6, 6)} \end{aligned}$
::S[1,1,1,(1,2,2,2,2,2,2,2,2,2,5,5,6,6,6,2,2,2,3,3,3,3,3(3,3,3,3,3,3,3,3,3,3,3,3,4,3,4(4,4,5,5,5,4,5,4,4,4,5,6,6,5,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,7,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,5,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4

Next find the probabilities:
::下一个发现概率 :

$\begin{aligned} P (C) = \frac{18}{36} = \frac{1}{2} \end{aligned}$
::P(C)=1836=12

$\begin{aligned} P (D) = \frac{4}{36} = \frac{1}{9} \end{aligned}$
::P(D)=436=19

$\begin{aligned} P (C) P (D) = (\frac{1}{2}) (\frac{1}{9}) = \frac{1}{18} \end{aligned}$
::P(C)P(D)=(12)(19)=118

$\begin{aligned} P (C \cap D) = \frac{2}{36} = \frac{1}{18} \end{aligned}$
::P(CD)=236=118

The events are independent because $\begin{aligned} P (C) P (D) = P (C \cap D) \end{aligned}$ .
::这些事件是独立的,因为P(C)P(D)=P(C)D。

Summary

Overlapping events are events that have outcomes in common.
::重叠事件是具有共同结果的事件。

Disjoint or mutually exclusive events cannot both occur in a single trial of a given experiment.
::在一个特定试验的单一试验中,不能同时发生脱节或相互排斥的事件。

The probability of $\begin{aligned} E \end{aligned}$ is $\begin{aligned} P (E) = \frac{# outcomes in E}{# outcomes in sample space} \end{aligned}$
::E的概率是样本空间E#结果中的P(E)结果

Two events are independent if one event occurring does not change the probability of the second event occurring.

$\begin{aligned} P (A \cap B) = P (A) P (B) \end{aligned}$

::如果发生一个事件并不改变发生第二个事件的概率,则两个事件是独立的。 P(AB)=P(A)P(B)

Dependent events are events where one outcome impacts the probability of the other.
::依附事件是指一种结果影响另一种结果概率的事件。

Review
::审查审查审查审查

Consider the experiment of flipping a coin three times and recording the sequence of heads and tails. The sample space is $\begin{aligned} S = {H H H, H H T, H T H, T H H, H T T, T H T, T T H, T T T} \end{aligned}$ , which contains eight outcomes. Let $\begin{aligned} A \end{aligned}$ be the event that exactly two coins are heads. Let $\begin{aligned} B \end{aligned}$ be the event that all coins are the same. Let $\begin{aligned} C \end{aligned}$ be the event that at least one coin is heads. Let $\begin{aligned} D \end{aligned}$ be the event that all coins are tails.
::将硬币翻转三次并记录头和尾的顺序的实验。样本空间是 SHHH、 HHT、 HHT、 HHH、 HT、 THT、 TTT}, 包含八个结果。如果事件恰好是两枚硬币的头, 请将B 当作所有硬币都一样的事件。让C 至少一个硬币是头的事件。让 D 成为所有硬币都是尾巴的事件。

1. List the outcomes in each of the four events. Which of the two events are complements?
::1. 列出这四个活动的成果,其中哪一个是补充性活动?

2. Find $\begin{aligned} P (A), P (B), P (C), P (D) \end{aligned}$ .
::2. 查找P(A)、P(B)、P(C)、P(D)。

3. Find $\begin{aligned} P (A \cap C) \end{aligned}$ . Are events $\begin{aligned} A \end{aligned}$ and $\begin{aligned} C \end{aligned}$ independent? Explain.
::3. 找到P(AC):事件A和C是否独立?解释。

4. Find $\begin{aligned} P (B \cap D) \end{aligned}$ . Are events $\begin{aligned} B \end{aligned}$ and $\begin{aligned} D \end{aligned}$ independent? Explain.
::4. 找到P(BD):事件B和D是否独立?解释。

5. Find $\begin{aligned} P (B \cap C) \end{aligned}$ . Are events $\begin{aligned} B \end{aligned}$ and $\begin{aligned} C \end{aligned}$ independent? Explain.
::5. 找到P(BC):事件B和C是否独立?解释。

6. Create two new events related to this experiment that are independent. Justify why they are independent using probabilities.
::6. 创建与这一实验有关的两个独立的新事件,说明为什么它们使用概率独立。

Consider the experiment of drawing a card from a deck. The sample space is the 52 cards in a standard deck. Let $\begin{aligned} A \end{aligned}$ be the event that the card is red. Let $\begin{aligned} B \end{aligned}$ be the event that the card is a spade. Let $\begin{aligned} C \end{aligned}$ be the event that the card is a 4. Let $\begin{aligned} D \end{aligned}$ be the event that the card is a diamond.
::从甲板上绘制一张牌的实验。试样空间是标准甲板上的52张牌。请将牌标为红色的A。请将牌标为红色的B。请将牌标为黑板的B。请将牌标为4的C。请将牌标为钻石的D。

7. Describe the outcomes in each of the four events.
::7. 说明这四个活动的成果。

8. Find $\begin{aligned} P (A), P (B), P (C), P (D) \end{aligned}$ .
::8. 查找P(A)、P(B)、P(C)、P(D)。

9. Find $\begin{aligned} P (A \cap B) \end{aligned}$ . Are events $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ independent? Explain.
::9. 找到P(AB):事件A和B是否独立?

10. Find $\begin{aligned} P (B \cap C) \end{aligned}$ . Are events $\begin{aligned} B \end{aligned}$ and $\begin{aligned} C \end{aligned}$ independent? Explain.
::10. 找出P(BQ)C. B事件和C事件是否独立?解释。

11. Find $\begin{aligned} P (A \cap D) \end{aligned}$ . Are events $\begin{aligned} A \end{aligned}$ and $\begin{aligned} D \end{aligned}$ independent? Explain.
::11. 找到P(AD):事件A和D是否独立?

12. $\begin{aligned} P (A) = \frac{1}{4} \end{aligned}$ and $\begin{aligned} P (B) = \frac{1}{8} \end{aligned}$ . If $\begin{aligned} P (A \cap B) = \frac{1}{16} \end{aligned}$ are events $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ independent?
::12. P(A)=14和P(B)=18。如果P(A)=116事件A和B是独立的?

13. $\begin{aligned} P (A) = \frac{1}{4} \end{aligned}$ and $\begin{aligned} P (B) = \frac{1}{8} \end{aligned}$ . If $\begin{aligned} P (A \cap B) = \frac{1}{32} \end{aligned}$ are events $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ independent?
::13. P(A)=14和P(B)=18.如果P(A)=132事件A和B是独立的?

14. What is the difference between disjoint and independent events?
::14. 脱节事件和独立事件有什么区别?

15. Two events are disjoint, and both have nonzero probabilities. Can you say whether the events are independent or not?
::15. 两个事件是脱节的,两个事件都是非零概率的,你能说这些事件是否独立吗?

16. Bag A contains 5 yellow, 6 blue, and 4 white marbles. Bag B contains 8 green, 5 black, and 4 red marbles. Create a probability problem that would show that is you pick two marbles from the bags, that these events would be independent?
::16. Bag A包含5个黄色、6个蓝色和4个白色大理石; Bag B包含8个绿色、5个黑色和4个红色大理石; 产生一个概率问题,表明你从袋子中摘取了2个大理石,这些事件是独立的?

17. Consider the outcomes for the tossing of 3 coins. Let A represent the outcomes where the first coin tossed is a head. Let B represent the outcomes where there is at least 2 heads tossed. Create a problem, using notation found in the concept, that would illustrate two independent events.
::17. 考虑3个硬币的抛掷结果。让 A 代表第一个抛掷的硬币是头的结果。让 B 代表至少有2个头抛掷的结果。使用在概念中发现的符号, 产生一个可以说明两个独立事件的问题。

18. You and your friend are in separate lines for tickets to the concert. Describe the outcome(s) that would make these event disjoint. What about independent? Is it possible to be both disjoint and independent? Why or why not?
::18. 你和你的朋友单独排队参加音乐会,描述导致这些音乐会脱节的结果。关于独立呢?是否可能既脱节又独立?为什么或为什么不独立呢?

Review (Answers)
::审查(答复)

To see the answer key for this book, go to the and click on the Answer Key under the ' ' option.
::要查看本书的答案键, 请在“ ” 选项下点击答案键。

Section outline

General

独立活动

Disjoint and Overlapping Events ::分裂和重叠事件

Determining Probability ::确定概率

CK-12 PLIX Interactive ::CK-12 PLIX 交互式互动

D etermining Independence ::确定独立性

Examples ::实例实例实例实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::审查审查审查审查

Review (Answers) ::审查(答复)

Disjoint and Overlapping Events
::分裂和重叠事件

Determining Probability
::确定概率

CK-12 PLIX Interactive
::CK-12 PLIX 交互式互动

D etermining Independence
::确定独立性

Examples
::实例实例实例实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::审查审查审查审查

Review (Answers)
::审查(答复)