条件概率下事件交集概率的验证问询

Hi there! Let's break down your question and your current reasoning step by step.

First, the core issue here is: the equation $P(A \cap B|C) = P(A|C)P(B|C)$ only holds if events $A$ and $B$ are conditionally independent given $C$—and your problem statement doesn't mention this key condition! So the short answer is: no, it's not necessarily true that $P(A \cap B|C) = (0.05)^2$.

Let's clear up your reasoning gaps: #

• Your initial thought that you need to prove $P(A \cap B|C) = P(A|C)P(B|C)$ is correct for the conclusion to hold, but this equality isn't automatically true. It requires conditional independence, which isn't given here.
• Your assumption that $A, B, C$ are mutually exclusive is a mistake. Mutually exclusive events can't occur at the same time, but $P(A|C) = 0.05 > 0$ directly tells us that $A$ and $C$ can happen together—so they can't be mutually exclusive. That invalidates the rest of that derivation path.

A concrete counterexample to make this clear: #

Suppose $C$ represents "rolling a 100-sided die". Let $A$ be "rolling a number from 1 to 5", and $B$ be exactly the same event as $A$ (so $A = B$). Then:

• $P(A|C) = 5/100 = 0.05$, $P(B|C) = 0.05$ (matches your problem's given values)
• But $P(A \cap B|C) = P(A|C) = 0.05$, which is way larger than $(0.05)^2 = 0.0025$.

This shows that without the conditional independence assumption, the equality doesn't hold.

To fix your derivation approach: #

If you did have conditional independence of $A$ and $B$ given $C$, then by definition:
$$P(A \cap B|C) = P(A|C)P(B|C)$$
Which would give you $(0.05)^2$. But since that condition isn't stated in the problem, you can't conclude this.

Without conditional independence, the most you can say is that $P(A \cap B|C)$ ranges from $\max(0, P(A|C) + P(B|C) - 1)$ to $\min(P(A|C), P(B|C))$—so in this case, between 0 and 0.05.

备注：内容来源于stack exchange，提问作者Roma_Rayado