Let $A,B,C$ be three events, what would be an example that $P(A|C)=P(A)$ and $P(B|C)=P(B)$ do not imply $P(A\cap B|C)= P(A\cap B)$?
Any counter example for $P(A|C)=P(A),P(B|C)=P(B)$ but $P(A\cap B|C)\neq P(A\cap B)$?
5
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probability
examples-counterexamples
2 Answers
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Consider a dice with 8 faces with numbers from $1$ to $8$ on them respectively. We roll that dice. Denote $X$ as the resulting face.
Let $A=[X=1,X=2]$, $B = [X=2,X = 3]$, $C=[X~\text{is even}]$. Then we definitely have $$P(A|C)=P(X=2\vert C)=\frac14=P(A)$$ and $$P(B|C)=P(X=2\vert C)=\frac14=P(B).$$
But $$P(A\cap B|C) = P(X = 2\vert C) = \frac14\ne \frac18 = P(X = 2) = P(A\cap B) $$
3
Let $\Omega = \{1,2,3,4\}$ be the sample space, events $A = \{1,2\}$, $B = \{1,3\}$, $C = \{1,4\}$.
\begin{align} P(A|C) &= P(\text{choose 1 or 2 | 1 or 4 chosen}) = \frac12 \\ P(A) &= P(\text{choose 1 or 2}) = \frac12 = P(A|C) \\ P(B|C) &= P(B) = \frac12 \quad \text{similarly} \\ P(A\cap B | C) &= P(\text{choose 1 | 1 or 4 chosen}) = \frac12, \text{ but} \\ P(A\cap B) &= P(\text{choose 1}) = \frac14 \ne P(A\cap B | C). \end{align}