Por que

eu suponho que

$$P(A|B) = P(A | B,C) * P(C) + P(A|B,\neg C) * P(\neg C)$$

está correto, enquanto

$$P(A|B) = P(A | B,C) + P(A|B,\neg C)$$

está incorreto.

No entanto, tenho uma "intuição" sobre a mais recente, ou seja, você considera a probabilidade P (A | B) dividindo dois casos (C ou Não C). Por que essa intuição está errada?

Suponha-se, como um exemplo contador fácil, que a probabilidade $P(A)$ de $A$ é $1$ , independentemente do valor de $C$ . Então, se tomarmos a equação incorreta , obtemos:

$P(A | B) = P(A | B, C) + P(A | B, \neg C) = 1 + 1 = 2$

Obviamente, isso não pode estar correto, provavelmente não pode ser maior que $1$ . Isso ajuda a criar a intuição de que você deve atribuir um peso a cada um dos dois casos proporcionalmente à probabilidade desse caso , o que resulta na primeira equação (correta). .

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Isso o aproxima da sua primeira equação, mas os pesos não estão completamente corretos. Veja o comentário de A. Rex para obter os pesos corretos.

A resposta de Dennis tem um ótimo contra-exemplo, refutando a equação errada. Esta resposta procura explicar por que a seguinte equação está correta:

$$P(A|B) = P(A|C,B) P(C|B) + P(A|\neg C,B) P(\neg C|B).$$

Como todo termo é condicionado em , podemos substituir todo o espaço de probabilidade por B e soltar o termo B. Isso nos dá:$B$$B$$B$

$$P(A) = P(A|C) P(C) + P(A|\neg C) P(\neg C).$$

Então você está perguntando por que essa equação tem os termos e P ( ¬ C ) .$P(C)$$P(\neg C)$

A razão é que é a porção de um em C e P ( A | ¬ C ) P ( ¬ C ) é a porção de uma em ¬ C e a dois adicionar-se a um . Veja o diagrama. Por outro lado, P ( A | C ) é a proporção de C contendo A e P ( $P(A|C) P(C)$$A$$C$$P(A|\neg C) P(\neg C)$$A$$\neg C$$A$$P(A|C)$$C$$A$ é a proporção de ¬ C contendo um - estas são as proporções de diferentes regiões, de forma que não têm denominadores comuns, de modo adicionando-lhes é insignificante.$P(A|\neg C)$$\neg C$$A$

I know you've already received two great answers to your question, but I just wanted to point out how you can turn the idea behind your intuition into the correct equation.

First, remember that $P(X \mid Y) = \dfrac{P(X \cap Y)}{P(Y)}$$P(X \cap Y) = P(X \mid Y)P(Y)$

$$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$$

We will keep rewriting the right-hand side until we get the desired equation.

The casework in your intuition expands the event $A$ into $(A \cap C) \cup (A \cap \neg C)$, resulting in

$$P(A \mid B) = \dfrac{P(((A \cap C) \cup (A \cap \neg C)) \cap B)}{P(B)}$$

As with sets, the intersection distributes over the union:

$$P(A \mid B) = \dfrac{P((A \cap B \cap C) \cup (A \cap B \cap \neg C))}{P(B)}$$

Since the two events being unioned in the numerator are mutually exclusive (since $C$ and $\neg C$ cannot both happen), we can use the sum rule:

$$P(A \mid B) = \dfrac{P(A \cap B \cap C)}{P(B)} + \dfrac{P(A \cap B \cap \neg C)}{P(B)}$$

We now see that $P(A \mid B) = P(A \cap C \mid B) + P(A \cap \neg C \mid B)$; thus, you can use the sum rule on the event on the event of interest (the "left" side of the conditional bar) if you keep the given event (the "right" side) the same. This can be used as a general rule for other equality proofs as well.

We re-introduce the desired conditionals using the second equation in the second paragraph:

$$P(A \cap (B \cap C)) = P(A \mid B \cap C)P(B \cap C)$$ and similarly for $\neg C$.

We plug this into our equation for $P(A \mid B)$ as:

$$P(A \mid B) = \dfrac{P(A \mid B \cap C)P(B \cap C)}{P(B)} + \dfrac{P(A \mid B \cap \neg C)P(B \cap \neg C)}{P(B)}$$

Noting that $\dfrac{P(B \cap C)}{P(B)} = P(C \mid B)$ (and similarly for $\neg C$), we finally get

$$P(A \mid B) = P(A \mid B \cap C)P(C \mid B) + P(A \mid B \cap \neg C)P(\neg C \mid B)$$

Which is the correct equation (albeit with slightly different notation), including the fix A. Rex pointed out.

Note that $P(A \cap C \mid B)$ turned into $P(A \mid B \cap C)P(C \mid B)$. This mirrors the equation $P(A \cap C) = P(A \mid C)P(C)$ by adding the $B$ condition to not only $P(A \cap C)$ and $P(A \mid C)$, but also $P(C)$ as well. I think if you are to use familiar rules on conditioned probabilities, you need to add the condition to all probabilities in the rule. And if there's any doubt whether that idea works for a particular situation, you can always expand out the conditionals to check, as I did for this answer.

Probabilities are ratios; the probability of A given B is how often A happens within the space of B. For instance, $P(\text{rain|March})$ is the number of rainy days in March divided by the number of total days in March. When dealing with fractions, it makes sense to split up numerators. For instance,

$$\begin{align} P(\text{rain or snow|March}) &= \frac{(\text{number of rainy or snowy days in March})}{(\text{total number of days in March})} \\[7pt] &= \frac{(\text{number of rainy days in March})}{(\text{total number of days in March)}} + \\[4pt] &\qquad \frac{\text{(number of snowy days in March)}}{(\text{total number of days in March)}} \\[7pt] &= P(\text{rain|March})+P(\text{snow|March}) \end{align}$$

This of course assumes that "snow" and "rain" are mutually exclusive. It does not, however, make sense to split up denominators. So if you have $P(\text{rain|February or March})$, that is equal to

$$\frac{(\text{number of rainy days in February and March})}{(\text{total number of days in February and March})}.$$

But that is not equal to

$$\frac{(\text{number of rainy days in February})}{(\text{total number of days in February})} + \frac{(\text{number of rainy days in March)}}{(\text{total number of days in March)}}.$$

If you're having trouble seeing that, you can try out some numbers. Suppose there are 10 rainy days in February and 8 in March. Then we have

$$\frac{(\text{number of rainy days in February and March})}{(\text{total number of days in February and March)}} = (10+8)/(28+31) = 29.5\%$$

and

$$\begin{align} \frac{(\text{number of rainy days in February})}{(\text{total number of days in February)}} + \frac{(\text{number of rainy days in March)}}{(\text{total number of days in March)}} &= (10/28)+(8/31) \\ &= 35.7\% + 25.8\% \\ &= 61.5\% \end{align}$$

The first number, 29.5%, is the average of 35.7% and 25.8% (with the second number weighted slightly more because there is are more days in March). When you say $P(A|B)=P(A|B,C)+P(A|B,¬C)$ you're saying that $\frac{x_1+x_2}{y_1+y_2} = \frac{x_1}{y_1}+\frac{x_2}{y_2}$, which is false.

If I go to Spain, I can get sunburnt.

$$P(sunburnt | Spain)=0.2$$ This tells me nothing about getting sunburnt if not going to Spain, let's say $$P(sunburnt|\neg Spain)=0.1$$ This year I'm going to Spain, so $$P(sunburnt)=0.2$$ Letting $B=\Omega$, this is, $P(B)=1$, your intuition would imply $$P(A)=P(A|C)+P(A|\neg C)$$ which by the previous argument, isn't neccesarily true.