这个人是女性的几率是多少？

窗帘后面有一个人-我不知道这个人是女性还是男性。

我知道这个人长发，而且所有长发中有90％是女性

我知道该人患有罕见的AX3血型，并且所有这种血型的人中有80％是女性。

这个人是女性的几率是多少？

注意：最初的配方在两个假设的基础上进行了扩展：1.血型和头发长度是独立的。2.总体人口中男性与女性的比例为50:50

（这里的具体情况不是那么重要-而是，我有一个紧急项目，要求我有正确的方法来回答这个问题。我的直觉是这是一个简单的概率问题，一个简单的确定性答案，而不是而不是根据不同的统计理论有多个值得商answers答案的事物。）

许多人发现根据“人口”，其中的子组和比例（而不是概率）进行思考很有帮助。 这有助于视觉推理。

我将详细解释这些数字，但目的是对这两个数字进行快速比较，应立即并令人信服地指出如何以及为什么无法对该问题给出具体答案。稍长的检查将提示哪些附加信息对确定答案或至少获得答案的界线有用。

传说

阴影线：女性/ 底色：男性。

顶：长毛/ 底：短毛。

右（有色）：AX3 / 左（无色）：非AX3。

数据

顶部阴影线是顶部矩形的90％（“长发的所有人中90％是女性”）。

右侧彩色矩形中的总阴影线是该矩形的80％（“所有这种血型的人中有80％是女性。”）

说明

该图示意性地示出了如何将（所考虑的所有女性和非女性中的）人口同时划分为女性/非女性，AX3 /非AX3和长发/非长发（“短”）。它使用面积（至少近似地）来表示比例（有些夸张以使图片更清晰）。

显然，这三个二进制分类创建了八个可能的组。每个组出现在这里。

给出的信息表明，上面的阴影线矩形（长发女性）占上面矩形的90％（所有长发人）。它还指出，彩色矩形的组合阴影线部分（带有AX3的长发女性和带有AX3的短发女性）占右侧有色区域的80％（所有带有AX3的人）。 我们被告知有人躺在右上角（箭头）：AX3长发的人。斜线阴影（女性）占矩形的比例是多少？

我还（隐式地）假设血型和头发长度是独立的：上色（AX3）的上矩形（长发）的比例等于上色（AX3）的下矩形（短发）的比例。这就是独立的意义。在解决此类问题时，这是一个公平自然的假设，但是当然需要说明。

上面的阴影线矩形（长发雌虫）的位置未知。 我们可以想象左右滑动顶部阴影线的矩形，并左右滑动底部阴影线的矩形，并可能改变其宽度。 如果我们这样做是为了使80％的彩色矩形保持交叉阴影线，则这种更改将不会更改任何声明的信息，但是会更改右上矩形中的女性比例。显然，该比例可能在0％到100％之间，并且仍然与给出的信息一致，如下图所示：

---

这种方法的优点之一是可以建立问题的多个答案。一个人可以用代数的方式翻译所有这些内容，并通过规定概率来提供特定的情况作为可能的例子，但是随后出现的问题是，这些例子是否真的与数据一致。例如，如果有人建议也许50％的长发人是AX3，那么从所有可用信息开始，甚至还不可能做到这一点。这些（Venn）总体图及其子组图清楚地表明了这一点。

这是条件概率的问题。您知道该人的头发和血液类型均为Ax3。令A = { “人长发” } 因此，您寻求 P （C | A 和B ）。您知道 P （C | A ）= 0.9且 P （C | B ）= 0.8。 这足以计算 P （C | A 和B ）吗？假设 P （A 和B 和C ）=

$$\ \ \ \ \ A =\{\text{'The person has long hair'}\}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B = \{\text{'The person has blood type Ax3'}\} \\ C =\{\text{'The person is female'}\}.$$
$P(C|A\ \text{and}\ B)$$P(C|A)=0.9$$P(C|B)=0.8$
$P(C|A\ \text{and}\ B)$$P(A\ \text{and}\ B\ \text{and}\ C)=0.7$。那么 假设P （A 和B ）= 0.8。然后，通过上述，P （C | A 和B ）= $$P(C|A\ \text{and}\ B)=P(A\ \text{and}\ B\ \text{and}\ C)/ P(A\ \text{and}\ B)=0.7/P(A\ \text{and}\ B).$$ $P(A\ \text{and}\ B)=0.8$$P(C|A\ \text{and}\ B)=0.875$。另一方面，如果，则P （C | A 和B ） = 0.78。$P(A\ \text{and}\ B)=0.9$$P(C|A\ \text{and}\ B)$

当和P （C | B ）= 0.8时，两者都是可能的。因此，我们无法确定P （C | A 和B ）是什么。$P(C|A)=0.9$$P(C|B)=0.8$$P(C|A\ \text{and}\ B)$

有趣的讨论！我想知道我们是否还指定了P（​​A）和P（B），是否仅仅因为有很多限制，P（C | A，B）的范围是否不会比整个间隔[0,1]窄很多？我们有。

坚持上面介绍的符号：

A =那个人长发的事件

B =人的血液类型为AX3

C =人是女性的事件

P（C | A）= 0.9

P（C | B）= 0.8

P（C）= 0.5（即，假设总体人口中男女比例相等）

在给定C的情况下，似乎不可能假设事件A和B是条件独立的！直接通向一个矛盾：如果$P(A \wedge B | C) = P(A| C) \cdot P(B| C) = P(C| A) \frac{P(A)}{P(C)} \cdot P(C| B) \frac{P(B)}{P(C)}$

然后

$P(C| A \wedge B ) = P(A \wedge B | C) \cdot \left( \frac{P(C)}{P(A \wedge B)} \right) = P(C| A) \frac{P(A)}{P(C)} \cdot P(C| B) \frac{P(B)}{P(C)} \cdot \left( \frac{P(C)}{P(A \wedge B)} \right)$

如果我们现在假定A和B是独立，以及：的大多数术语取消和我们结了$P(A \wedge B) = P(A) P(B)$

$P(C| A \wedge B ) = \frac{P(C| A) \cdot P(C| B)}{P(C)} = \frac{0.9 \cdot 0.8}{0.5} > 1$

对这个问题的whuber的精彩几何表示后续：尽管这是事实，一般来说可以在区间承担任何值[ 0 ，1 ]的几何约束做显著缩小的可能值的范围为的值P （甲）和P （乙）不属于“太小”。（尽管我们也可以将边际上界：P （A ）和P （B ））$P(C | A \wedge B)$$[0,1]$$P(A)$$P(B)$$P(A)$$P(B)$

让我们计算{\ BF最小可能值}对在下列几何约束：$P(C | A \wedge B)$

1.上方矩形所覆盖的上方区域的分数（A TRUE）必须等于$P(C|A)=0.9$

2.两个矩形的面积之和必须等于$P(C)=0.5$

3.两个彩色矩形的面积之和（即它们与事件B的重叠）之和必须等于$P(C|B)=0.8$

4.（平凡的）上方的矩形不能移出左侧边界，也不能移至左侧的最小交叠处。

5.（平凡的）下部矩形不能移到右边界之外，也不能移到右部最大重叠之外。

这些约束条件限制如何自由地我们可以滑动散列矩形和反过来产生用于下界。下图（使用此R脚本创建）显示了两个示例 $P(C | A \wedge B)$

在P（A）和P（B）的一系列可能值中运行（R脚本）会生成此图

总之，对于给定的P（A），P（B），我们可以下限条件概率P（c | A，B）

假设是窗帘后面的人是女人。

我们给出了2条证据，即：

证据1：我们知道这个人长发（据告知，所有长发人中有90％是女性）

证据2：我们知道该人患有罕见的AX3血型（我们被告知这种血型的所有人中有80％是女性）

仅凭证据1，我们可以说窗帘后面的人有一个成为女人的概率值为0.9（假设男人和女人之间的比例为50:50）。

关于线程前面提出的问题，即“您是否同意答案必须大于0.9？”，我不做任何数学运算，直觉地说，答案必须是“是”（大于0.9）。逻辑是证据2支持证据（同样，假设世界上男女人数以50:50的比例分配）。如果我们被告知所有AX3型血液患者中有50％是女性，那么证据2将是中性的，没有任何影响。但是由于我们被告知所有这种血型的人中有80％是女性，因此证据2支持了证据，因此从逻辑上讲应该将女性的最终概率提高到0.9以上。

要计算特定概率，我们可以对证据1应用贝叶斯规则，然后使用贝叶斯更新将证据2应用于新假设。

假设：

A =那个人长发的事件

B =人的血液类型为AX3

C =人是女性的事件（假设50％）

将贝叶斯规则应用于证据1：

P（C | A）=（P（A | C）* P（C））/ P（A）

在这种情况下，再次假设男人和女人之间的比例为50:50：

P（A）=（0.5 * 0.9）+（0.5 * 0.1）= 0.5

因此，P（C | A）=（0.9 * 0.5）/ 0.5 = 0.9（不足为奇，但是如果我们不按50:50的比例分配性别，则情况会有所不同）

使用贝叶斯更新应用证据2并插入0.9作为新的先验概率，我们具有：

P（C | A AND B）=（P（B | C）* 0.9）/ P（E）

在此假设给定某人已经有90％的机会成为女性的假设，P（E）是证据2的概率。

P（E）=（0.9 * 0.8）+（0.1 * 0.2）[这是总概率定律：（P（女性）* P（AX3 |女性）+ P（男性）* P（AX3 |男性）] ，P（E）= 0.74

因此，P（C | A AND B）=（0.8 * 0.9）/ 0.74 = 0.97297

Question Restatement and Generalisation

$A$, $B$, and $C$ are binary unknowns whose possible values are $0$ and $1$. Let $Z_i$ stand for the proposition, "The value of $Z$ is $i$". Also let $(X | Y)$ stand for "The probability that $X$, given that $Y$". What is $(A_a | B_b C_c I)$, given that

1. $(A_{a_1} | B_{b_1} I) = u_1$ and $(A_{a_2} | C_{c_2} I) = u_2$
2. $(A_{a_1} | B_{b_1} I) = u_1$ and $(A_{a_2} | C_{c_2} I) = u_2$ and $(B C | I) = (B | I)(C | I)$
3. $(A_{a_1} | B_{b_1} I) = u_1$ and $(A_{a_2} | C_{c_2} I) = u_2$ and $(A_0 | I) = \frac{1}{2}$
4. $(A_{a_1} | B_{b_1} I) = u_1$ and $(A_{a_2} | C_{c_2} I) = u_2$ and $(A_0 | I) = \frac{1}{2}$ and $(B C | I) = (B | I)(C | I)$

and that $I$ contains no relevant information besides what is implicit in the assignments? The last conjunct of conditions 2 and 4 is shorthand for the independence statement

$$(B_j C_k | I) = (B_j | I)(C_k | I) \quad , \quad j = 0, 1 \quad k = 0,1$$ Treat each of the four cases in turn.

Answers

Case 1

We have to specify the distribution $(ABC | I)$. The problem is underdetermined, because $(ABC | I)$ requires eight numbers, but we have only three equations---the two given conditions and the normalisation condition.

It has been shown by various esoteric means that the distribution to assign when the information doesn't otherwise determine a solution is the one that, of all distributions consistent with the known information, has the greatest entropy. Any other distribution implies that we know more than the known information, which of course is a contradiction.

All we need to do, therefore, is assign the maximum entropy distribution. This is more easily said than done, and I have not found a general closed-form solution. But particular solutions can be found using a numerical optimiser. We maximise

$$- \sum_{i,j,k} (A_i B_j C_k | I) \ln (A_i B_j C_k | I)$$ subject to the constraints $$\sum_{i,j,k} (A_i B_j C_k | I) = 1$$ and $$(A_{a_1} | B_{b_1} I) = u_1 \quad\quad \text{i.e.} \quad \frac{\sum\limits_k (A_{a_1} B_{b_1} C_k | I )}{\sum\limits_{i,k} (A_i B_{b_1} C_k | I)} = u_1$$ and $$(A_{a_2} | C_{c_2} I) = u_2 \quad\quad \text{i.e.} \quad \frac{\sum\limits_j (A_{a_2} B_j C_{c_2} | I)}{\sum\limits_{i,j} (A_i B_j C_{c_2} | I)} = u_2$$ Now let's apply this to the question. If we have

1. "The person is female" $\longleftrightarrow A_1$
2. "The person has long hair" $\longleftrightarrow B_1$
3. "The person has blood type AX3" $\longleftrightarrow C_1$

then $a = 1$, $b = 1$, $c = 1$, $a_1 = 1$, $b_1 = 1$, $a_2 = 1$, $c_2 = 1$, $u_1 = 0.9$, $u_2 = 0.8$, and we find that for the maximum entropy solution, $(A_1 | B_1 C_1 I) \simeq 0.932$. Therefore the probability that the person behind the curtain is female, given that he/she has long hair and blood type AX3, is 0.932.

Case 2

Now we repeat the exercise with the extra constraint that for a given person, knowing the value of $B$ (the hair state) does not affect our estimate of the value of $C$ (the blood type state), and vice versa. Everything is the same as in Case 1, except there are two extra constraints in the optimisation, namely:

$$\begin{align*} (B_0 | C_l I) &= (B_0 | I) \quad , \quad l = 0, 1 \\ \end{align*}$$ i.e. $$\begin{align*} \frac{\sum\limits_i (A_i B_0 C_l | I)}{\sum\limits_{i,j} (A_i B_j C_l | I)} &= \sum_{i,k} (A_i B_0 C_k | I) \quad , \quad l = 0, 1 \end{align*}$$ This gives $(A_1 | B_1 C_1 I) \simeq 0.936$, so the probability that the person behind the curtain is female, given that he/she has long hair and blood type AX3, is 0.936.

Case 3

Now we remove the independence condition and replace it with the prior condition that there is an equal chance that a given person is male or female:

$$(A_0 | I) = \frac{1}{2} \quad \quad \text{i.e.} \quad \sum_{j,k} (A_0 B_j C_k | I) = \frac{1}{2}$$ This time $(A_1 | B_1 C_1 I) \simeq 0.973$, so the probability that the person behind the curtain is female, given that he/she has long hair and blood type AX3, is 0.973.

Case 4

Finally we reintroduce the independence constraints of Case 2, and find that $(A_1 | B_1 C_1 I) \simeq 0.989$. Therefore the probability that the person behind the curtain is female, given that he/she has long hair and blood type AX3, is 0.989.

I believe now that, if we assume a ratio of men and women in the population at large, then there is a single indisputable answer.

A = the event that the person has long hair

B = the event that the person has blood type AX3

C = the event that person is female

P(C|A) = 0.9

P(C|B) = 0.8

P(C) = 0.5 (i.e. let's assume an equal ratio of men and women in the population at large)

Then P(C|A and B) = [P(C|A) x P(C|B) / P(C)] / [[P(C|A) x P(C|B) / P(C)] + [[1-P(C|A)] x [1-P(C|B)] / [1-P(C)]]]

in this case, P(C|A and B) = 0.972973

Note: In order to get a definitive answer, the below answers assume that the probability of a person, a long-haired man, and a long-haired women having AX3 are approximately the same. If more accuracy is desired, this should be verified.

You start out with the knowledge that the person has long hair, so at this point the odds are:

90:10

Note: The ratio of males to females in the general population does not matter to us once we find out the person has long hair. For example, if there were 1 female in a hundred in the general population, a randomly-selected long-haired person would still be a female 90% of the time. The ratio of females to males DOES matter! (see the update below for details)

Next, we learn that the person has AX3. Because AX3 is unrelated to long hair, the ratio of men to women is known to be 50:50, and because of our assumption of the probabilities being the same, we can simply multiply each side of the probability and normalize so that the sum of the sides of the probability equals 100:

(90:10) * (80:20)
==> 7200:200

    Normalize by dividing each side by (7200+200)/100 = 74

==> 7200/74:200/74
==> 97.297.. : 2.702..

Thus, the chance that the person behind the curtain is female is approximately 97.297%.

UPDATE

Here's a further exploration of the problem:

Definitions:

f - number of females
m - number of males
fl - number of females with long hair
ml - number of males with long hair
fx - number of females with AX3
mx - number of males with AX3
flx - number of females with long hair and AX3
mlx - number of males with long hair and AX3
pfl - probability that a female has long hair
pml - probability that a male has long hair
pfx - probability that a female has AX3
pmx - probability that a male has AX3

First, we are given that 90% of long-haired people are females, and 80% of people with AX3 are female, so:

fl = 9 * ml
pfl = fl / f
pml = ml / m 
    = fl / (9 * m)

fx = 4 * mx
pfx = fx / f
pmx = mx / m 
    = fx / (4 * m)

Because we assumed that the probability of AX3 is independent of gender and long hair, our calculated pfx will apply to women with long hair, and pmx will apply to men with long-hair to find the number of them that likely have AX3:

flx = fl * pfx 
    = fl * (fx / f) 
    = (fl * fx) / f
mlx = ml * pmx 
    = (fl / 9) * (fx / (4 * m)) 
    = (fl * fx) / (36 * m)

Thus, the likely ratio of the number of females with long-hair and AX3 to the number of males with long-hair and AX3 is:

flx             :   mlx
(fl * fx) / f   :   (fl * fx) / (36 * m)
1/f             :   1 / (36m)
36m             :   f

Because it is given that there is an equal number of 50:50, you can cancel both sides and end with 36 females to every male. Otherwise, there are 36*m/f females for every male in the specified subgroup. For example, if there were twice as many women as men, there would be 72 females to each male of those that have long-hair and AX3.

98% Female, simple interpolation. First premise 90% female, leaves 10%, second premise only leaves 2% of the existing 10%, hence 98% female