Probability proof. Check my answer?

Question

Let $A$ and $B$ be events defined on a sample space $\Omega$. Prove that

$$\textbf{P}(A \cap B) \geq\ 1 - \textbf{P}(A^{c})- \textbf{P}(B^{c}) $$

Answer:

\begin{align*} \textbf{P}(A\cap B) & = \textbf{P}(A)\textbf{P}(B) \\ & = (1 - \textbf{P}(A^{c}))(1 - \textbf{P}(B^{c})) \\ & = 1 - \textbf{P}(A^{c}) - \textbf{P}(B^{c}) + \textbf{P}(A^{c})\textbf{P}(B^{c}) \geq 1 - \textbf{P}(A^{c}) - \textbf{P}(B^{c}) \end{align*}

Answer 1

Notice that

\begin{align*} &\textbf{P}(A\cup B) \leq 1 \Leftrightarrow \textbf{P}(A) + \textbf{P}(B) - \textbf{P}(A\cap B) \leq 1 \Leftrightarrow\\\\ & 1 - \textbf{P}(A^{c}) + 1 - \textbf{P}(B^{c}) - \textbf{P}(A\cap B) \leq 1 \Leftrightarrow\\\\ & 1 - \textbf{P}(A^{c}) - \textbf{P}(B^{c}) \leq \textbf{P}(A\cap B) \end{align*}

Where we have used the following properties \begin{cases} \textbf{P}(X) \leq 1\\\\ \textbf{P}(X) = 1 - \textbf{P}(X^{c})\\\\ \textbf{P}(X\cup Y) = \textbf{P}(X) + \textbf{P}(Y) - \textbf{P}(X\cap Y) \end{cases}

Hope this helps.

Answer 2

Note that $P(S^c)=1-P(S)$ and $P(A\cup B)\le 1$ therefore$$P(A)+P(B)-P(A\cap B)\le 1\\P(A)+P(B)-1\le P(A\cap B)\\P(A)+P(B)-1-1+1\le P(A\cap B)\\-P(A^c)-P(B^c)+1\le P(A\cap B)$$and finally by rearranging the terms $$P(A\cap B)\ge 1-P(A^c)-P(B^c)$$with the equality if and only if $$A\cup B=S$$

Answer 3

You wrote: $$ \textbf{P}(A\cap B) = \textbf{P}(A)\textbf{P}(B) $$ I claim this is wrong. Suppose that we are flipping a fair coin. Let $A$ be the event the coin comes up heads. Let $B$ be the event the coin comes up heads. This give us: \begin{align*} \textbf{P}(A) &= \frac{1}{2} \\ \textbf{P}(B) &= \frac{1}{2} \\ \textbf{P}(A\cap B) &= \frac{1}{2} \\ \textbf{P}(A)\textbf{P}(B) &= \frac{1}{4} \\ \end{align*} Hence their is a contradiction and your original statement is wrong.

I hope this helps.

Bob