ThmDex – An index of mathematical definitions, results, and conjectures.

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Two disjoint events are not necessarily independent

Formulation 0

Let

Then

(1)	$\begin{equation} \{ 0 \}, \{ 1 \} \in \mathcal{F} \end{equation}$
(2)	$\begin{equation} \{ 0 \} \cap \{ 1 \} = \emptyset \end{equation}$
(3)	$\begin{equation} \mathbb{P}(\{ 0 \} \cap \{ 1 \}) = \mathbb{P}(\emptyset) = 0 \neq \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} = \mathbb{P} \{ 0 \} \mathbb{P} \{ 1 \} \end{equation}$

Also known as

Counterexample: two disjoint events which are not independent

Proofs

Proof 0

Let

$P = (\{ 0 ,1 \}, \mathcal{F}, \mathbb{P})$ be the D5511: Standard single boolean trial probability space.

Clear.

$\square$