Bu kişinin kadın olma olasılığı nedir?

Bir perdenin arkasında bir kişi var - o kişinin kadın mı yoksa erkek mi olduğunu bilmiyorum.

Kişinin uzun saçlı olduğunu ve uzun saçlı insanların% 90'ının kadın olduğunu biliyorum.

Kişinin nadir kan grubu AX3 olduğunu ve bu kan grubuna sahip bütün kişilerin% 80'inin kadın olduğunu biliyorum.

İnsanın kadın olma olasılığı nedir?

NOT: Bu orijinal formülasyon iki başka varsayımla genişletilmiştir: 1. Kan grubu ve saç uzunluğu bağımsızdır 2. Erkek: Kadın nüfusu en fazla 50:50

(Buradaki özel senaryo o kadar önemli değil - daha doğrusu, bunu yanıtlamak için doğru yaklaşımı aklımdan almamı gerektiren acil bir projem var. farklı istatistiksel teorilere göre çok sayıda tartışılabilir cevabı olan bir şeyden daha fazlası.)

Birçok insan “popülasyon”, içindeki alt gruplar ve oranlar (olasılıklar yerine) açısından düşünmeyi faydalı bulmaktadır . Bu, kendisini görsel akıl yürütmeye borç verir.

Rakamları ayrıntılı olarak açıklayacağım, ancak amaç, iki rakamın hızlı bir şekilde karşılaştırılmasının derhal ve ikna edici bir şekilde, soruya nasıl ve neden belirli bir cevabın verilemeyeceğini göstermesi gerektiğidir. Biraz daha uzun bir inceleme, bir cevabı belirlemek ya da en azından cevapları sınırlamak için hangi ek bilgilerin faydalı olacağına işaret edecektir.

efsane

Çapraz tarama : dişi / Katı arka plan : erkek.

Üst : uzun saçlı / Alt : kısa saçlı.

Sağ (ve renkli) : AX3 / Sol (renksiz) : AX3 olmayan.

Veri

Üst tarama, üst dikdörtgenin% 90'ıdır ("uzun saçlı insanların% 90'ı kadındır").

Sağ renkli dikdörtgendeki toplam çapraz tarama, bu dikdörtgenin% 80'idir ("Bu kan grubuna sahip tüm kişilerin% 80'i kadındır").

açıklama

Bu şema, popülasyonun (incelenen tüm kadın ve kadın olmayan kadınların) eşzamanlı olarak kadın / kadın olmayan, AX3 / AX3 olmayan ve uzun saçlı / uzun saçlı ("kısa") olarak nasıl bölündüğünü göstermektedir. En azından yaklaşık olarak oranları temsil etmek için alan kullanır (resmi daha net hale getirmek için biraz abartı vardır).

Bu üç ikili sınıflamanın sekiz olası grup oluşturduğu açıktır. Her grup burada belirir.

Verilen bilgiler, çapraz çizgili üst dikdörtgenin (uzun saçlı dişiler) üst dikdörtgenin% 90'ını (uzun saçlı insanların tümü) oluşturduğunu belirtir. Ayrıca, renkli dikdörtgenlerin (AX3 ile uzun saçlı kadınlar ve AX3 ile kısa saçlı kadınlar) birleştirilmiş çapraz taranmış bölümlerinin sağdaki renkli bölgenin% 80'ini (AX3'lü tüm insanlar) oluşturduğunu belirtir. Birinin sağ üst köşesinde (ok) yalan söylediği söylenir: AX3'lü uzun saçlı insanlar. Bu dikdörtgenin yüzde kaçı çapraz çizgilidir (dişi)?

Ayrıca (dolaylı olarak) kan grubunun ve saç uzunluğunun bağımsız olduğunu varsaydım. grubunun : Renkli olan üst dikdörtgenin (uzun saç) oranı (AX3) renkli alt dikdörtgenin (kısa saç) oranına eşittir (AX3). Bağımsızlığın anlamı budur. Bu gibi soruları ele alırken yapmak adil ve doğal bir varsayımdır, ancak elbette belirtilmesi gerekir.

Çapraz çizgili üstteki dikdörtgenin (uzun saçlı dişiler) konumu bilinmemektedir. Çapraz çizgili taranmış dikdörtgeni yan yana kaydırmayı ve alt taranmış taranmış dikdörtgeni yana kaydırmayı ve genişliğini değiştirmeyi hayal edebiliyoruz. Bunu yaparsak, renkli dikdörtgenin% 80'i çapraz çizgili kalır, böyle bir değişiklik belirtilen bilgilerin hiçbirini değiştirmez ancak yine de sağ üst dikdörtgenin içindeki kadınların oranını değiştirebilir. Açıkçası, oran% 0 ile% 100 arasında herhangi bir yerde olabilir ve yine de bu resimde olduğu gibi verilen bilgilerle tutarlı olabilir:

---

Bu yöntemin bir gücü , soruya birden fazla cevabın varlığını tespit etmesidir. Biri tüm bunları cebirsel olarak çevirebilir ve olasılıkları belirleyerek, olası örnekler olarak spesifik durumlar sunabilir, ancak bu durumda bu örneklerin verilerle gerçekten tutarlı olup olmadığı sorusu ortaya çıkacaktır. Örneğin, birileri uzun saçlı insanların belki de% 50'sinin AX3 olduğunu söylerse, başlangıçta, mevcut tüm bilgiler verildiğinde bunun mümkün olduğu açık değildir. Nüfusun ve alt gruplarının bu (Venn) şemaları bu gibi şeyleri açıkça ortaya koymaktadır.

Bu şartlı bir olasılık sorunudur. Kişinin uzun saçlı ve kanlı Ax3 tipi olduğunu biliyorsun. Let A = { 'kişi uzun saçlı' } Demek P ( C | A ve B ) . P ( C | A ) = 0.9 ve P ( C | B ) = 0.8 olduğunu biliyorsunuz. Bu P ( C | A ve B ) ' yi hesaplamak için yeterlimi? Diyelim ki P ( A ve B ve 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)$$P(A\ \text{and}\ B\ \text{and}\ C)=0.7$. Then $$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).$$ Suppose $P(A\ \text{and}\ B)=0.8$. Then, by the above, $P(C|A\ \text{and}\ B)=0.875$$P(A\ \text{and}\ B)=0.9$$P(C|A\ \text{and}\ B)$

Şimdi her ikisi de olduğunda mümkündür$P(C|A)=0.9$ and $P(C|B)=0.8$. So we can't tell for sure what $P(C|A\ \text{and}\ B)$ is.

Fascinating discussion ! I am wondering if we specified P(A) and P(B) as well whether the ranges of P(C| A,B) will not be much narrower than the full interval [0,1], simply because of the many constraints we have.

Sticking to the notation introduced above:

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)

it does not seem possible to assume that events A and B are conditionally independent given C ! That leads directly to a contradiction: if $P(A \wedge B | C) = P(A| C) \cdot P(B| C) = P(C| A) \frac{P(A)}{P(C)} \cdot P(C| B) \frac{P(B)}{P(C)}$

then

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

If we now assume that A and B are independent as well: $P(A \wedge B) = P(A) P(B)$ most terms cancel and we end up with

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

Following up on whuber's wonderful geometric representation of the problem: While it is true that generally speaking $P(C | A \wedge B)$ can assume any value in the interval $[0,1]$ the geometric constraints do narrow the range of possible values significantly for values of $P(A)$ and $P(B)$ that are not "too small". (Though we can also upper bound the marginals: $P(A)$ and $P(B)$)

Let us compute the {\bf smallest possible value} for $P(C | A \wedge B)$ under the following geometric constraints:

1. The fraction of the upper area (A TRUE) covered by the upper rectangle must be equal to $P(C|A)=0.9$

2. The sum of the areas of the two rectangles must be equal to $P(C)=0.5$

3. The sum of the fraction of the areas of the two colored rectangles (i.e. their overlap with event B) must be equal to $P(C|B)=0.8$

4. (trivial) The upper rectangle cannot be moved beyond the left boundary and should not be moved beyond its minimum overlap to the left.

5. (trivial) The lower rectangle cannot be moved beyond the right boundary and should not be moved beyond its maximum overlap to the right.

These constraints limit how freely we can slide the hashed rectangles and in turn generate lower bounds for $P(C | A \wedge B)$. The figure below (created with this R script ) shows two examples

Running through a range of possible values for P(A) and P(B) (R script) generates this graph

In conclusion, we can lower bound the conditional probability P(c|A,B) for given P(A), P(B)

Make the hypotheses is that the person behind a curtain is a woman.

We area given 2 pieces of evidence, namely:

Evidence 1: We know the person has long hair (and we're told that 90% of all people with long hair are female)

Evidence 2: We know the person has a rare blood type AX3 (and we're told that 80% of all people with this blood type are female)

Given just Evidence 1, we can state that the person behind a curtain has a 0.9 probability value of being a woman (assuming 50:50 split between men and women).

Regarding the question posed earlier in the thread, namely "Would you agree that the answer must be GREATER than 0.9?", without doing any Math, I would say intuitively, the answer must be "yes" (it is GREATER than 0.9). The logic is that Evidence 2 is supporting evidence (again, assuming a 50:50 split for the number of men and women in the world). If we were told that 50% of all people with AX3 type blood were female, then Evidence 2 would be neutral and have no bearing. But since we're told that 80% of all people with this blood type are female, Evidence 2 is supporting evidence and logically should push the final probability of a woman above 0.9.

To calculate a specific probability, we can apply Bayes' rule for Evidence 1 and then use Bayesian updating to apply Evidence 2 to the new hypothesis.

Suppose:

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 (assume 50%)

Applying Bayes rule to Evidence 1:

P(C|A) = (P(A|C) * P(C)) / P(A)

In this case, again if we assume 50:50 split between men and women:

P(A) = (0.5 * 0.9) + (0.5 * 0.1) = 0.5

So, P(C|A) = (0.9 * 0.5) / 0.5 = 0.9 (Not surprising, but it would be different if we didn't have 50:50 split between men and women)

Using Bayesian updating to apply Evidence 2 and plugging in 0.9 as the new prior probability, we have:

P(C|A AND B) = (P(B|C) * 0.9) / P(E)

Here, P(E) is the probability of Evidence 2, given the hypotheses that the person already has a 90% chance of being female.

P(E) = (0.9 * 0.8) + (0.1 * 0.2) [this is law of total probability: (P(woman)*P(AX3|woman) + P(man)*P(AX3|man)] So, P(E) = 0.74

So, 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