# How is $P(A^c \cap B^c)$ the same as $1-P(A \cup B)$?

I don't understand how $$P(A^c \cap B^c) = 1-P(A \cup B)$$ is the same?

If I draw $$P(A^c \cap B^c)$$ as a Venn diagram: If I draw $$P(A \cup B)$$ as a Venn diagram: So if I subtract $$P(A \cup B)$$ from 1, wouldn't that mean that I subtract $$P(A \cup B)$$ from the universe $$\Omega$$, which would result int this: However, that would mean $$P(A^c \cap B^c) \neq 1-P(A \cup B)$$

Edit:

As pointed out by multiple people. My diagram for $$P(A^c \cap B^c)$$ should look like the following and therefore the assumption of $$P(A^c \cap B^c) = 1-P(A \cup B)$$ is valid: • Your first diagram is of $A^c \cup B^c$ not $A^c \cap B^c$. The yellow areas include both $A \cap B^c$ and $A^c \cap B$ – Henry Jan 24 '20 at 0:09
• your first diagram is wrong – Masacroso Jan 24 '20 at 0:11
• The third diagram is the correct one for the left side as well. – Berci Jan 24 '20 at 0:12
• So yellow should be only the universe? – Tom el Safadi Jan 24 '20 at 0:15
• I edited my question with the help of you guys. Thanks!! – Tom el Safadi Jan 24 '20 at 0:19

## 2 Answers

Your first diagram is incorrect. You seem to have drawn $$A^c\cup B^c$$ (as some other people have mentioned).

I would recommend drawing $$A^c$$ independently first, which consists of the rest of the universe and $$B-A$$. Then draw $$B^c$$ in a different color, noting again that you have the rest of the universe and $$A-B$$. The intersection of $$A-B$$ and $$B-A$$ is neither $$A$$ nor $$B$$, so you will end up without $$A$$ or $$B$$ in your final intersection. However, the rest of the universe is in both $$A^c$$ and $$B^c$$, therefore so is its intersection. Thus, you get that $$P(A^c\cap B^c)$$ is just the universe without $$A$$ or $$B$$, which is equivalent to $$1-P(A\cup B)$$.

According to the DeMorgan's laws, one has \begin{align*} \textbf{P}(\Omega) & = \textbf{P}((A\cup B)\cup(A\cup B)^{c})\\ & = \textbf{P}(A\cup B) + \textbf{P}((A\cup B)^{c})\\ & = \textbf{P}(A\cup B) + \textbf{P}(A^{c}\cap B^{c}) = 1 \end{align*} from whence the result follows immediately, since $$X\cup X^{c} = \Omega$$ and $$X\cap X^{c} = \varnothing$$ for every event $$X\subseteq\Omega$$.