In general, how do you approach proving a statement?

Question

I struggle to even start when it comes to proving statements, so I'm reading a book on stochastic processes and at the very beginning they provide some basic properties of Probabilities and one of them is

$P(A^c) = 1 - P(A)$

and next they ask me to prove it as an exercise, however simple this may seem, I just don't know how to approach even proving such a statement?

Any guidance?

Answer 1

If $A$ denotes one event then $A^c$ denotes that $A$ does not happened. So $A \cup A^c=S$ where S is the probability space. We know from axiom of probability that $P(S)=1$ and $P(A_1 \cup A_2 \cup... \cup A_n\cup ...)=P(A_1)+P(A_2)+...+P(A_n)+...$ for mutually exclusive events $A_i$. Using these two axioms we have, $$P(A)+P(A^c)=P(S)=1 \implies P(A^c)=1-P(A). \blacksquare $$

Answer 2

Let us reinforce the statements of the Kolmogorov's axioms for convenience.

• For every $A\subseteq\Omega$, $0\leq\textbf{P}(A)\leq 1$.

• Given the sample space $\Omega$, one has that $\textbf{P}(\Omega) = 1$.

• If $A_{i}\cap A_{j} = \varnothing$ for every $i\neq j$, then $\textbf{P}\left(\bigcup_{k=1}^{\infty}A_{k}\right) = \sum_{k=1}^{\infty}\textbf{P}(A_{k})$

Based on such considerations, since $A\cap A^{c} = \varnothing$ and $A\cup A^{c} = \Omega$, one has that \begin{align*} \textbf{P}(\Omega) = \textbf{P}(A\cup A^{c}) = \textbf{P}(A) + \textbf{P}(A^{c}) = 1 \Longrightarrow \textbf{P}(A^{c}) = 1 - \textbf{P}(A). \end{align*}

Hopefully this helps.