已知条件概率求P(B)：备考考试遇求解循环困境

Hey there, let's break this down and get past that loop you're stuck in—you've already got two key equations, so we just need to tie them to the given $P(A)$ value using basic probability rules.

First, let's recap what we know:

• $P(A|B) = 0.4$
• $P(A'|B') = 0.4$
• $P(A) = 0.5$

You correctly derived these two equations:

• $0.6P(B) = P(A' \cap B)$ (since $P(A'|B) = 1 - P(A|B) = 0.6$, multiplying by $P(B)$ gives the joint probability)
• $0.6P(B') = P(A \cap B')$ (similarly, $P(A|B') = 1 - P(A'|B') = 0.6$, hence this joint probability)

Now here's the step that connects everything: the law of total probability tells us that $P(A)$ can be split into the probability of $A$ occurring with $B$, plus $A$ occurring without $B$:
$$P(A) = P(A \cap B) + P(A \cap B')$$

We can rewrite $P(A \cap B)$ by subtracting your first equation from $P(B)$:
$P(A \cap B) = P(B) - P(A' \cap B) = P(B) - 0.6P(B) = 0.4P(B)$

And we already have $P(A \cap B') = 0.6P(B')$ from your work. Since $P(B') = 1 - P(B)$, substitute that in, then plug everything into the total probability equation:
$$0.5 = 0.4P(B) + 0.6(1 - P(B))$$

Now solve for $P(B)$:

1. Expand the right-hand side: $0.4P(B) + 0.6 - 0.6P(B) = 0.5$
2. Combine like terms: $-0.2P(B) + 0.6 = 0.5$
3. Rearrange to isolate $P(B)$: $-0.2P(B) = 0.5 - 0.6 = -0.1$
4. Divide both sides by $-0.2$: $P(B) = \frac{-0.1}{-0.2} = 0.5$

That's it! The key was linking your derived joint probabilities back to the given $P(A)$ using total probability, which breaks the loop you were stuck in.

内容的提问来源于stack exchange，提问作者dembrownies