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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: enter image description here

If I draw $P(A \cup B)$ as a Venn diagram:

enter image description here

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:

enter image description here

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:

enter image description here

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    $\begingroup$ 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 $A^c \cap B$ $\endgroup$ – Henry Jan 24 '20 at 0:09
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    $\begingroup$ your first diagram is wrong $\endgroup$ – Masacroso Jan 24 '20 at 0:11
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    $\begingroup$ The third diagram is the correct one for the left side as well. $\endgroup$ – Berci Jan 24 '20 at 0:12
  • $\begingroup$ So yellow should be only the universe? $\endgroup$ – Tom el Safadi Jan 24 '20 at 0:15
  • $\begingroup$ I edited my question with the help of you guys. Thanks!! $\endgroup$ – Tom el Safadi Jan 24 '20 at 0:19
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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 $1-P(A\cup B)$.

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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 $X\subseteq\Omega$.

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