Course: 6.6 计算条件概率 | The Way To Learn

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计算条件概率
Suppose you wanted to calculate the probability of pulling the King of Hearts, then the Jack of Diamonds, and then any of the four Aces, from a standard deck of 52 cards, in that order, and without replacing any cards between pulls. Would the probability be significantly different than if you put the cards back after drawing each time?
::假设你想要计算从52张牌的标准甲板上拉动心脏之王的概率,然后是钻石之王,然后是钻石之王,然后是四个A中的任何一个,按照这个顺序,而不是在拉动之间更换任何牌。概率是否与每次绘制之后再把牌放回去的可能性大不相同?

In this lesson we will discuss conditional probabilities that are different for each trial . We’ll return to this question after the lesson.
::在这个教训中,我们将讨论每个审判的有条件概率不同。在教训之后,我们将回到这个问题上来。

Conditional Probabilities
::有条件概率

In a previous lesson, we discussed compound probabilities and reviewed some situations involving the probability of multiple occurrences of the same event in a row. A standard example would be the probability of throwing a fair coin three times and getting three heads. In this lesson, we will be introducing a slightly more complex situation, where the coin may or may not be fair.
::在以往的一个教训中,我们讨论了复合概率,并审查了一系列涉及同一事件多次发生概率的一些情况。一个标准的例子就是投出公平硬币三次并获得三个头的概率。在这个教训中,我们将引入一个略为复杂的情况,硬币可能公平,也可能不公平。

A new concept we will be introducing in this lesson is the “given that” concept. The idea is that we sometimes need to calculate a probability with a specific condition, for example:
::我们将在这个教训中引入一个新的概念是“给这个”概念。我们的想法是,我们有时需要用一个特定的条件来计算概率,例如:

The probability of rolling a “2” on a standard die is $\begin{align*}\frac{1}{6}\end{align*}$ . What is the probability of rolling a “2”, given that I know already that I have rolled an even number? As described in the video above, this is a conditional probability , and we notate it this way: $\begin{align*}P(2|even)\end{align*}$ , which is read as “The probability of rolling a “2” given that we roll an even number”.
::在标准死亡上滚动“2”的概率是16,鉴于我知道我已经滚动了一个偶数,滚动“2”的概率是多少?正如上面的视频所描述的,这是一个有条件的概率,我们用这种方式把它记下来:P(2)-(even),它被解读为“滚动“2”的概率,因为我们滚动了一个偶数”。

The difference in calculations is:
::计算上的差异是:

$\begin{aligned} P (2) & = \frac{1}{6} (on number on the 6-sided die is a 2) \\ P (2 | e v e n) & = \frac{1}{3} (one number out of the three even numbers is a 2) \end{aligned}$
$\begin{align*}P(2)&=\frac{1}{6}( \text{on number on the 6-sided die is a 2}) \\ P(2|even)&=\frac{1}{3} ( \text{one number out of the three even numbers is a 2}) \\ \end{align*}$
::P(2)=16(六边死亡的号码是a 2)P(2)P(even)=13(三个偶数中有一个数字是a2)

To calculate a “given that” type of problem, we use the conditional probability formula :
::为了计算“给定的”问题类型,我们使用有条件的概率公式:

$\begin{aligned} P (A | B) = \frac{P (A \cap B)}{P (B)} \end{aligned}$
$\begin{align*}P(A|B)=\frac{P(A \cap B)}{P (B)}\end{align*}$
::P(AB)=P(AB)P(B)

This is read as: “The probability that $\begin{align*}A\end{align*}$ will occur, given that $\begin{align*}B\end{align*}$ will occur (or has occurred), is equal to the intersection of probabilities $\begin{align*}A\end{align*}$ and $\begin{align*}B\end{align*}$ divided by the probability of $\begin{align*}B\end{align*}$ alone.
::改为:“鉴于B将发生(或已经发生),A的发生概率等于概率A和B的交叉点除以B的单独概率。

We have practiced the use of the and the for calculating probabilities, here we will also be using those again, but this time we will need to combine them for some of the problems.
::我们已经实践了使用概率和计算概率的做法,我们在这里也将再次使用这些概率,但这次我们将需要将其中的一些问题结合起来。

For review:
::供审查:

Multiplication Rule for : $\begin{align*}P(A \ then \ B)= P(A)\times P(B)\end{align*}$
::乘法规则: P(A) = P(A) x(B)

Addition Rule for : $\begin{align*}P(A \ or \ B)=P(A)+P(B) \end{align*}$
::增加规则:P(A或B)=P(A)+P(B)

Addition Rule for : $\begin{align*}P(A \ or \ B)=P(A)+P(B)-P(A \ and \ B)\end{align*}$
::增加规则:P(A或B)=P(A)+P(B)-P(A和B)

Calculating Probability
::计算概率

1. What is the probability that you have pulled the Jack of Hearts from a standard deck, given that you know you have pulled a face card?
::1. 鉴于你知道你拉了一张脸卡,你把红心杰克从标准甲板拉出来的可能性有多大?

Let’s solve this using the conditional probability formula first (A), then check by looking at the question another way (B):
::让我们首先使用有条件概率公式(A)来解决这个问题,然后从另一个角度看问题(B):

A. The problem asks us to calculate the probability of a card being the Jack of Hearts, given that the card is a face card: $\begin{align*}P(Jack \ of \ Hearts|face \ card)\end{align*}$ . Apply the conditional probability formula: $\begin{align*}P(A|B)=\frac{P(A \cap B)}{P (B)}\end{align*}$ . Putting in the information from the problem gives us:
::A. 问题要求我们计算一张卡的概率为“红心之杰克”,因为牌是一张脸卡:P(“红心的杰克”脸卡。应用有条件的概率公式:P(AB)=P(AB)P(AB)P(B)。

$\begin{aligned} P (J a c k o f H e a r t s | f a c e c a r d) & = \frac{P (J a c k o f H e a r t s \cap f a c e c a r d)}{P (f a c e c a r d)} \\ P (J a c k o f H e a r t s | f a c e c a r d) & = \frac{(\frac{1}{52})}{\frac{12}{52}} \\ P (J a c k o f H e a r t s | f a c e c a r d) & = \frac{1}{52} \times \frac{52}{12} = \frac{1}{12} \\ P (J a c k o f H e a r t s | f a c e c a r d) & = \frac{1}{12} o r 8.33 % \end{aligned}$
$\begin{align*}P(Jack \ of \ Hearts|face \ card)&=\frac{P(Jack \ of \ Hearts \cap face \ card)}{P(face \ card)} \\ P(Jack \ of \ Hearts|face \ card)&= \frac{\left(\frac{1}{52}\right)}{\frac{12}{52}} \\ P(Jack \ of \ Hearts|face \ card)&=\frac{1}{52} \times \frac{52}{12}=\frac{1}{12} \\ P(Jack \ of \ Hearts|face \ card)&=\frac{1}{12} \ or \ 8.33 \%\end{align*}$
::P(红心的杰克)=P(红心的杰克)P(脸卡)P(红心的杰克)P(脸卡)P(红心的杰克)=(152)1252P(红心的杰克)=152×5212=112P(红心的杰克)=112或8.33%

B. The other way to view this is that we are looking for the probability of pulling the Jack of Hearts from the sample space including only face cards, which means we are looking for one specific card from a set including only 12 cards:
::B. 另一种观点是,我们正在寻找从样本空间拉动红心杰克(Jack of Hearts)的概率,其中只包括面孔卡,这意味着我们正在从一组中寻找一张特定的卡片,其中仅包括12张卡片:

$\begin{aligned} P (J a c k o f H e a r t s) & = \frac{1 J a c k o f H e a r t s}{12 f a c e c a r d s} = \frac{1}{12} \\ P (J a c k o f H e a r t s) & = \frac{1}{12} o r 8.33 % \end{aligned}$
$\begin{align*}P(Jack \ of \ Hearts)&=\frac{1 \ Jack \ of \ Hearts}{12 \ face \ cards}=\frac{1}{12} \\ P(Jack \ of \ Hearts)&=\frac{1}{12} \ or \ 8.33 \%\end{align*}$
::P(红心的杰克)=1心12的杰克脸卡=112P(红心的杰克)=112或8.33%

2. We calculate 8.33% both ways, looks like we got it!
::2. 我们计算了8.33%的两种方法, 看起来我们得到了!

What is the probability that you could roll a standard die and get a 6, then grab a deck of cards and pull the King of Clubs, keep it, and then pull the Jack of Hearts?
::有多少可能你滚一个标准死亡得到一个6 然后抓住一副牌并拉起俱乐部之王,保留它, 然后拉起心之杰克?

This one looks rather complex, but it can be seen as just three individual probabilities:
::这个看上去相当复杂,但可以认为它只是三种个别概率:

$\begin{align*}P(roll \ 6)=\frac{1 \ side \ with \ a \ 6}{6 \ sides}=\frac{1}{6}\end{align*}$
::P(roll 6) = 1 侧面66 侧面= 16

$\begin{align*}P(King)=\frac{1 \ King \ of \ Clubs}{52 \ cards}=\frac{1}{52}\end{align*}$
::P(King)=1 Clubs52卡片国王=152

$\begin{align*}P(Jack)=\frac{1 \ Jack \ of \ Hearts}{51 \ cards \ left \ after \ first \ pull}=\frac{1}{51} \end{align*}$
::P( Jack) = 1 Hearts51 的 Jack 在第一次拉出后留下的卡片 = 151

The overall probability can, and should, be calculated with the multiplication rule, since the 2 ^nd and 3 ^rd are dependent:
::总概率可以而且应该与乘法规则一起计算,因为第2和第3级取决于:

$\begin{aligned} P (r o l l 6 | K i n g | J a c k) & = P (r o l l 6) \times P (K i n g) \times P (J a c k) \\ P (r o l l 6 t h e n p u l l K i n g t h e n p u l l J a c k) & = \frac{1}{6} \times \frac{1}{52} \times \frac{1}{51} = \frac{1}{15912} O R .167 \times .019 \times .020 = .000063 \end{aligned}$
$\begin{align*}P(roll \ 6 | King | Jack)&=P(roll \ 6)\times P(King)\times P(Jack) \\ P(roll \ 6 \ then \ pull \ King \ then \ pull \ Jack)&=\frac{1}{6}\times \frac{1}{52}\times \frac{1}{51}=\frac{1}{15912} \ OR \ .167\times .019\times .020=.000063\end{align*}$
::P(roll 6KingQ}) = P(roll 6) ×P(King) ×P(Jack) P(roll 6 然后拉King 6 然后拉Jack) = 16×152×151= 115912 OR 167x.019x.020=. 00063

3. You reach into a bag containing 6 coins, 4 are ‘fair’ coins (they have an equal chance of heads or tails), and 2 are ‘unfair’ coins (they have only a 35% chance of tails). If you randomly grab a coin from the bag and flip it 3 times, what is the probability of getting 3 heads?
::3. 你触摸到一个装有6个硬币的袋子,4个是 " 公平 " 硬币(他们有相同的机会获得头部或尾部),2个是 " 不公平 " 硬币(他们只有35%的尾部机会),如果你随机从袋子中抢到硬币并翻了3次,获得3个头的可能性是多少?

We actually have two different situations here:
::我们这里实际上有两种不同的情况:

We flip a ‘fair’ coin 3 times and get 3 heads
::我们翻翻“公平”硬币3次,

We flip an ‘unfair’ coin and get 3 heads
::我们翻翻“不公平”硬币,

Since there are 4 fair coins, and 2 unfair coins, we can say the probability of: $\begin{align*}P(choose \ fair)=\frac{4}{6} \ or \ \frac{2}{3}\end{align*}$ and $\begin{align*}P(choose \ unfair)=\frac{2}{6}=\frac{1}{3}\end{align*}$ .
::由于有4个公平的硬币和2个不公平的硬币,我们可以说概率是:P(选择公平)=46或23,P(选择不公平)=26=13。

Note that $\begin{align*}P(choose \ unfair)\end{align*}$ would be the same thing as $\begin{align*}P(choose fair)^\prime\end{align*}$ , (see the apostrophe?) which is the complement of $\begin{align*}P(choose \ fair)\end{align*}$ . In other words: the probability of choosing an unfair coin is 100% minus the probability of choosing a fair coin.
::请注意 P( 选择不公平 ) 与 P( 选择公平 ) ( 选择不公平 ) ( 见代号? ) 是 P( 选择公平 ) 的补充。换句话说: 选择不公平硬币的概率是 100% 减去选择公平硬币的概率。

Let’s calculate the probabilities of flipping each 3 times using the multiplication rule:
::让我们使用乘法规则计算每3次翻转的概率:

The fair coin has a .5 chance of heads each flip: $\begin{align*}P(fair \ 3 \ heads)=.5\times .5\times .5=.125\end{align*}$
::公平硬币每张硬币头的概率为 0. 5 : P( 公平的 3 个头) = 5x.5x.5=. 125

The unfair coin has a .65 chance: $\begin{align*}P(unfair \ 3 \ heads)=.65\times .65\times .65= .275\end{align*}$
::不公平的硬币有一个.65的机会:P(不公平的3头)=.65×65×65=275

So now we can put them together to find the overall probability (the union) by applying the addition rule:
::因此,我们现在可以把它们组合在一起,通过适用附加规则,找到总概率(工会):

$\begin{aligned} P (3 h e a d s e i t h e r c o i n) & = .6 \bar{6} \times P (f a i r 3 h e a d s) + .3 \bar{3} \times P (u n f a i r 3 h e a d s) \\ P (3 h e a d s e i t h e r c o i n) & = .6 \bar{6} \times .125 + .3 \bar{3} \times .275 \\ P (3 h e a d s e i t h e r c o i n) & = .083 + .092 = .175 \end{aligned}$
$\begin{align*} P(3 \ heads \ either \ coin)&=.6 \overline{6}\times P(fair \ 3 \ heads)+.3 \overline{3}\times P(unfair \ 3 \ heads) \\ P(3 \ heads \ either \ coin)&=.6 \overline{6}\times .125+.3 \overline{3} \times .275 \\ P(3 \ heads \ either \ coin)&=.083+.092=.175\end{align*}$
::P( 3头硬币中的3个) =.66 P( 3个) +.33 P( 不公平3头) P( 3个) P( 3个) =.66 125+.33 275 P( 3个) =. 083+. 092=. 175

The probability that we can randomly grab a coin from the bag and flip three heads in a row with it is 17.5%
::我们可以随机从袋子里拿硬币并连续翻三个头的概率是17.5%

Earlier Problem Revisited
::重审先前的问题

Suppose you wanted to calculate the probability of pulling the King of Hearts, then the Jack of Diamonds, and then any of the four Aces, from a standard deck of 52 cards, in that order and without putting any back. Would the probability be significantly different than if you put the cards back after drawing each time?
::假设你想要计算从52张牌的标准甲板上拉动心脏之王的概率,然后是钻石之王,然后是钻石之王,然后是四个A中的任何一个,按照顺序,不放回任何回。概率是否与每次绘制牌后再放回牌的概率大不相同?

The probability would be different, but perhaps less different than you might think, at least as a percentage. Let’s look at the two cases, with $\begin{align*}P(A)\end{align*}$ representing the probability with each choice coming out of a full deck of 52 cards, and $\begin{align*}P(B)\end{align*}$ representing the probability when the deck gets smaller each pull:
::概率是不同的,但也许与你想象的不尽相同,至少以百分比表示。让我们看看这两个案例,P(A)代表着每个选择的概率,从52张牌的全甲板上出来,P(B)代表着每张牌变小时的概率:

$\begin{aligned} P (A) & = \frac{1 K i n g o f H e a r t s}{52 c a r d s} \times \frac{1 J a c k o f D i a m o n d s}{52 c a r d s} \times \frac{4 a c e s}{52 c a r d s} = \frac{1}{52} \times \frac{1}{52} \times \frac{1}{13} = \frac{1}{35152} \\ P (B) & = \frac{1 K i n g o f H e a r t s}{52 c a r d s} \times \frac{1 J a c k o f D i a m o n d s}{51 C a r d s} \times \frac{4 a c e s}{50 c a r d s} = \frac{1}{52} \times \frac{1}{51} \times \frac{4}{50} = \frac{4}{132600} = \frac{1}{33150} \end{aligned}$
$\begin{align*}P(A)&=\frac{1 \ King \ of \ Hearts}{52 \ cards}\times \frac{1 \ Jack \ of \ Diamonds}{52 \ cards}\times \frac{4 \ aces}{52 \ cards}=\frac{1}{52}\times \frac{1}{52}\times \frac{1}{13}=\frac{1}{35152} \\ P(B)&=\frac{1 \ King \ of \ Hearts}{52 \ cards}\times\frac{1 \ Jack \ of \ Diamonds}{51 \ Cards}\times\frac{4 \ aces}{50 \ cards}=\frac{1}{52}\times\frac{1}{51}\times\frac{4}{50}=\frac{4}{132600}=\frac{1}{33150}\end{align*}$
::P(A) = 1 红心王52卡片×1 钻石王52卡片×4 Aces52卡片=152×152×113=135152P(B) = 1 红心王52卡片×1 钻石王51卡片×4 彩卡50卡=152×151×450=4132600=133150

The difference in probability is approximately $\begin{align*}\frac{1}{2000}\end{align*}$ or $\begin{align*}\frac{5}{100}\end{align*}$ of 1%, pretty small difference!
::概率差约为12000或5100 %, 相差很小!

Examples
::实例

Example 1
::例1

What would be the probability of the coin landing heads on your first flip>
::你第一个翻转的硬币着陆头的概率是多少?

To calculate the probability of flipping heads, we need to calculate the union of 50% of the probability of flipping heads on each coin. (Why 50% of each probability? There are two coins, so the chance that you will pull either one is 50%)
::要计算翻转头的概率, 我们需要计算每个硬币翻转头的概率的50%的组合。 (为什么每个概率的50%? 有两个硬币, 所以您拉一个的概率是50% )

$\begin{aligned} P (h e a d s | e i t h e r c o i n) & = 50 % \times P (h e a d s | f a i r c o i n) + 50 % \times P (h e a d s | u n f a i r c o i n) \\ P (h e a d s | e i t h e r c o i n) & = 50 % (50 %) + 50 % (75 %) \\ P (h e a d s | e i t h e r c o i n) & = 25 % + 37.5 % \\ P (h e a d s | e i t h e r c o i n) & = 62.5 % \end{aligned}$
$\begin{align*}P(heads|either \ coin)&=50 \%\times P(heads|fair \ coin)+50 \%\times P(heads|unfair \ coin) \\ P(heads|either \ coin)&=50 \%(50 \%)+50 \%(75 \%) \\ P(heads|either \ coin)&=25 \%+37.5 \% \\ P(heads|either \ coin)&=62.5 \% \end{align*}$
::P(头或硬币)=50P(头或公平硬币)+50P(头或不公平硬币)P(头或两种硬币)=50%(50%)+50%(75%)P(头或两种硬币)=2537.5%(头或两种硬币)=62.5%

Example 2
::例2

What would be the probability of flipping tails four times in a row?
::连续四次翻翻尾巴的概率是多少?

To calculate the probability of flipping four tails in a row, we calculate the union of 50% of the probability of flipping four tails in a row with each coin, much like in question 1.
::为了计算一排翻四尾的概率, 我们计算出每一枚硬币翻四尾的概率的50%, 大致上与问题1相似。

$\begin{aligned} P (4 t a i l s | e i t h e r c o i n) & = 50 % \times P (4 t a i l s | u n f a i r c o i n) + 50 % \times P (4 t a i l s | f a i r c o i n) \\ P (4 t a i l s | e i t h e r c o i n) & = 50 % (25 % \times 25 % \times 25 % \times 25 %) + 50 % (50 % \times 50 % \times 50 % \times 50 %) \\ P (4 t a i l s | e i t h e r c o i n) & = 0.1953 % + 3.125 % \\ P (4 t a i l s | e i t h e r c o i n) & = 3.32 % \end{aligned}$
$\begin{align*} P(4 \ tails|either \ coin)&=50 \%\times P(4 \ tails|unfair \ coin)+50 \%\times P(4 \ tails|fair \ coin) \\ P(4 \ tails|either \ coin)&=50 \%(25 \%\times 25 \%\times 25 \%\times 25 \%)+50 \%(50 \%\times 50 \% \times 50 \%\times 50 \%) \\ P(4 \ tails|either \ coin)&=0.1953 \%+3.125 \% \\ P(4 \ tails|either \ coin)&=3.32\%\end{align*}$
::P( 4 尾巴) = 50 P( 4 尾巴) = 50 P( 4 尾巴) +50 P( 4 尾巴) +50 ( 4 尾巴) = 50% ( 25 25 25 25%) +50% ( 50 50 50 50%) P( 4 尾巴) = 0. 1953 3. 125 P( 4 尾巴) = 3.32%

Example 3
::例3

What would be the probability of flipping heads five times in a row?
::连续五次翻头的概率是多少?

Just like question 2, only this time the probability will end up greater, since the unfair coin has a large chance of heads:
::就如问题2一样, 但这次的概率会更大, 因为不公平的硬币有很大的正面机会:

$\begin{aligned} P (4 h e a d s | e i t h e r c o i n) & = 50 % \times P (4 h e a d s | u n f a i r c o i n) + 50 % \times P (4 h e a d s | f a i r c o i n) \\ P (4 h e a d s | e i t h e r c o i n) & = .50 \times (.75 \times .75 \times .75 \times .75)^{4} + .50 \times (.25 \times .25 \times .25 \times .25)^{4} \\ P (4 h e a d s | e i t h e r c o i n) & = .1582 + .0020 \\ P (4 h e a d s | e i t h e r c o i n) & = 16.02 % \end{aligned}$
$\begin{align*}P(4 \ heads|either \ coin)&=50 \% \times P(4 \ heads|unfair \ coin)+50 \% \times P(4 \ heads|fair \ coin) \\ P(4 \ heads|either \ coin)&=.50\times(.75\times .75\times.75\times .75)^4+.50\times (.25\times .25\times .25\times .25)^4 \\ P(4 \ heads|either \ coin)&=.1582+.0020 \\ P(4 \ heads|either \ coin)&=16.02\% \end{align*}$
::P( 4 头双方硬币) = 50P( 4 头双方硬币) +50P( 4 头双方硬币) P( 4 头双方硬币) = 50x( 75x. 75x. 75x. 75) 4+50x( 25x.25x.25x. 25x. 4) P( 4 头双方硬币) = 1582+. 0020P( 4头双方硬币) = 16.02%

Example 4
::例4

Assume you are using a limited portion of cards that only includes face cards (no number cards). Assume also that you each pull a card, you keep it until the end of the experiment . What would be the probability of pulling three kings in a row?
::假设您使用的牌数有限, 仅包括面牌( 无牌号) 。假设您每人拉一张牌, 直至实验结束。连续拉三个国王的概率是多少 ?

The key here is to note that you do not replace the card between pulls. That means that the probability changes with each trial. Let’s look at the situation for each trial individuality:
::关键是注意您不会在拉拉之间替换卡片。这意味着每次审判的概率都会发生变化。让我们看看每个审判的个别性:

$\begin{align*}(\text{T}1)\end{align*}$ : Since we are only dealing with face cards, our first trial will have 12 possible outcomes , four for each of three face cards. Four of the outcomes are favorable, since there are four kings.
:T1):因为我们只处理面牌,我们的第一次试验将产生12个可能的结果,每3张面卡中就有4个。其中4个结果是有利的,因为有4个国王。

$\begin{align*}(\text{T}2)\end{align*}$ Our second trial will have only 11 outcomes, since we are keeping the first card. There are only three favorable outcomes this time, since we “used up” a king if $\begin{align*}(\text{T}1)\end{align*}$ was favorable.
:T2) 我们的第二次审判将只有11个结果,因为我们保留着第一张牌。这次只有3个有利结果,因为我们“放弃了”国王,如果(T1)是有利的。

The third pull $\begin{align*}(\text{T}3)\end{align*}$ only has 10 outcomes, since we will already have the other two cards. Two of the outcomes are favorable, since there would be only two kings left.
::第三局(T3)只有10个结果,因为我们已经有另外两张牌了。有两个结果是有利的,因为只剩下两个国王了。

$\begin{aligned} P (t h r e e k i n g s | f a c e c a r d) & = P (T 1) \times P (T 2) \times P (T 3) \\ P (t h r e e k i n g s | f a c e c a r d) & = \frac{4 k i n g s}{12 f a c e c a r d s} \times \frac{3 k i n g s}{11 f a c e c a r d s} \times \frac{2 k i n g s}{10 f a c e c a r d s} \\ P (t h r e e k i n g s | f a c e c a r d) & = .3 \bar{3} \times .27 \bar{27} \times .2 \\ P (t h r e e k i n g s | f a c e c a r d) & = 0.0182 o r 1.82 % \end{aligned}$
$\begin{align*}P(three \ kings|face \ card)&=P(T1)\times P(T2)\times P(T3) \\ P(three \ kings|face \ card)&=\frac{4 \ kings}{12 \ face \ cards}\times \frac{3 \ kings}{11 \ face \ cards}\times \frac{2 \ kings}{10 \ face \ cards} \\ P(three \ kings|face \ card)&=.3 \overline{3}\times .27 \overline{27}\times .2 \\ P(three \ kings|face \ card)&=0.0182 \ or \ 1.82\%\end{align*}$
::P( 3 Kntsface card) = P( T1) xP( T2) xP( T3) P( 3 Knsface card) = 4 Kns12 脸卡片×3 Kns11 脸卡片×2 Kns10 脸卡片P( 3 Knts脸卡) = 33 2727 2P( 3 Knsfface card) = 0.0182 或 1. 82%

Example 5
::例5

What would be the probability of rolling a 5, given that you know you rolled an odd number?
::滚动5号的概率是多少? 因为你知道你滚动了一个奇数?

This is a ‘given that’ problem, so we can use the conditional probability formula:
::这是一个“给定的”问题, 所以我们可以使用有条件的概率公式:

$\begin{aligned} P (r o l l 5 | r o l l o d d) & = \frac{P (r o l l 5) \cap P (r o l l o d d)}{P (r o l l o d d)} \\ P (r o l l 5 | r o l l o d d) & = \frac{\frac{1}{6}}{\frac{1}{2}} \\ P (r o l l 5 | r o l l o d d) & = \frac{2}{6} o r \frac{1}{3} o r 33.33 % \end{aligned}$
$\begin{align*}P(roll \ 5 | roll \ odd)&=\frac{P(roll \ 5) \cap P(roll \ odd)}{P(roll \ odd)} \\ P(roll \ 5 | roll \ odd)&=\frac{\frac{1}{6}}{\frac{1}{2}} \\ P(roll \ 5 | roll \ odd)&=\frac{2}{6} \ or \ \frac{1}{3} \ or \ 33.33\%\end{align*}$
::P( 滚动 5 ) = P( 滚动 5 ) = 1612P( 滚动 5 ) = 26 或 13 或 33.33%

Review
::回顾

What is the probability that you roll two standard dice, and get 4’s on both, given that you know that you have already rolled a 4 on one of them?
::既然你知道你已经在其中之一上打过4个骰子,那么你滚过2个标准骰子并同时打4个骰子的概率是多少?

Assuming you are using a standard deck, what is the probability of drawing two cards in a row, without replacement, that are the same suit?
::假设你用的是标准甲板连排画两张牌的概率是多少而没有换换那一套西装呢?

What is the probability that a single roll of two standard dice will result in a sum greater than 8, given that one of the dice is a 6?
::鉴于一个骰子是6,一个两张标准骰子的单卷将产生超过8美元的总和的概率有多大?

Assuming a standard deck, what is the probability of drawing 3 queens in a row, given that the first card is a queen?
::假设是标准甲板,如果第一张牌是皇后,那么连续画3 Q的概率是多少?

There are 130 students in your class, 50 have laptops, and 80 have tablets. 20 of those students have both a laptop and a tablet. What is the probability that a randomly chosen student has a tablet, given that she has a laptop?
::你们班级有130名学生,50名学生有笔记本电脑,80名学生有平板电脑。其中20名学生有笔记本电脑和平板电脑。随机选择的学生有平板电脑的可能性有多大,因为她有笔记本电脑?

Thirty percent of your friends like both Twilight and The Hobbit, and half of your friends like The Hobbit. What percentage of your friends who like the Hobbit also like Twilight?
::30%的你的朋友喜欢暮暮和霍比特人一半的朋友喜欢霍比特人

Assume you pull and keep two candies from a jar containing sweet candies and sour candies. If the probability of selecting one sour candy and one sweet candy is 39%, and the probability of selecting a sweet candy first is 52%, what is the probability that you will pull a sour candy on your second pull, given that you pulled a sweet candy on your first pull?
::假设你从含有甜甜糖果和酸糖果的罐子中拉出并保留两个糖果。如果选择一种酸糖果和一种甜糖果的概率是39%,而选择一种甜甜糖果的概率是52%,那么你第二次拉出甜甜糖果的概率是52%,那么,考虑到你第一次拉出甜甜甜糖果的概率是多少?

The probability that a student has called in sick and that it is Monday is 12%. The probability that it is Monday and not another day of the school week is 20% (there are only five days in the school week). What is the probability that a student has called in sick, given that it is Monday?
::学生请病假和请病假的概率是12 % 。周一是周一而不是学校周的另一天的概率是20 % ( 每周只有5天 ) 。学生请病假的概率是多少? 因为是周一,请病假的概率是多少?

A neighborhood wanted to improve its parks so it surveyed kids to find out whether or not they rode bikes or skateboards. Out of 2300 children in the neighborhood that ride something, 1800 rode bikes, and 500 rode skateboards, while 200 of those ride both a bike and skateboard. What is the probability that a student rides a skateboard, given that he or she rides a bike?
::一个街区想改善公园,因此它调查了孩子是否骑过自行车或滑板。在附近骑过车的2300名儿童中,有1800人骑过自行车,500人骑过滑板,其中200人骑过自行车和滑板。如果学生骑过自行车,他们骑过滑板的可能性有多大?

A movie theatre is curious about how many of its patrons buy food, how many buy a drink, and how many buy both. They track 300 people through the concessions stand one evening, out of the 300, 78 buy food only, 113 buy a drink only and the remainder buy both. What is the probability that a patron buys a drink if they have already bought food?
::电影院对有多少人购买食物、多少人购买饮料和多少人购买两者都很好奇。他们有一晚通过特许摊位追踪300人,在300人中,78人只购买食物,113人只购买饮料,其余人两者都购买。如果已经购买了食物,那么赞助者购买饮料的可能性是多少?

A sporting goods store want to know if it would be wise to place sports socks right next to the athletic shoes. First they keep the socks and shoes in separate areas of the store. They track purchases for one day, Saturday, their busiest day. There were a total of 147 people who bought socks, shoes, or both in one given day. Of those 45 bought only socks, 72 bought only shoes and the remainder bought both. What is the probability that a person bought shoes, if they purchased socks?
::体育用品商店想知道把运动袜子放在运动鞋旁边是否明智。首先,它们把袜子和鞋放在商店的单独地区。它们跟踪购买时间一天,星期六,最忙的一天。在一天里,共有147人购买袜子、鞋子,或两者兼而有之。在这45人中,只有袜子买,72人只买鞋子,其余人买。一个人买鞋,如果买袜子的可能性有多大?

The following week they put socks right next to the shoes to see how it would affect Saturday sales. The results were as follows; a total of 163 people bought socks, shoes, or both. Of those 52 bought only socks, 76 bought only shoes and the remainder bought both. What is the probability that a person bought socks, if they purchased shoes?
::接下来的一周,他们把袜子放在鞋子旁边,看看它会如何影响星期六的销售。结果如下:共有163人购买袜子、鞋子或两者兼而有之。在这52人中,只有买袜子,76人只买鞋子,其余人买袜子,如果买鞋,其可能性有多大?

A florist wanted to know how many roses and daisies to order for the upcoming valentines rush. She used last year’s statistics to determine how many to buy. Last year she sold 52 arrangements with roses only, 15 arrangements with daises only, and 36 arrangements with a mixture of roses and daises. What is the probability that an arrangement has at least one daisy, given that it has at least one rose?
::一个花店家想知道,为即将到来的情人节潮流订购了多少玫瑰和花朵。她用去年的统计数据来决定要买多少。去年,她只卖了52个玫瑰协议,15个玫瑰协议,36个玫瑰协议,混合了玫瑰和花环。一项协议至少有1个乳房的可能性是多少,因为它至少有1个玫瑰?

Review (Answers)
::回顾(答复)

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::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。

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