I know that:
$$\begin{gathered}P\left(A\cap B\right)=P\left(A\right)P\left(B\right)\\ P\left(A^{C}\right)=1-P\left(A\right)\\ P\left(B^{C}\right)=1-P\left(B\right) \end{gathered} $$

My proof so far:
$$\begin{gathered}P\left(A^{C}\cap B^{C}\right)=\left(1-P\left(A\right)\right)\left(1-P\left(B\right)\right)=\\ 1-P\left(B\right)-P\left(A\right)+P\left(A\right)P\left(B\right)=1-P\left(B\right)-P\left(A\right)+P\left(A\cap B\right) \end{gathered} $$

After that, I'm stuck. Any help would be appreciated.

  • 8
    $\begingroup$ $=1-P(A \cup B)$... Also your first line after "What I have so far:" is what you're trying to prove, not what you have. $\endgroup$ Oct 10, 2016 at 19:14

4 Answers 4


Assume $A$ and $B$ are independent. Then \begin{align} P(A^c \cap B^c) &= P( (A \cup B)^c) \quad \tag{by De Morgan's Law}\\ &= 1 - P(A \cup B) \\ &= 1 - P(A) - P(B) + P(A \cap B) \\ &= 1 - P(A) - P(B) + P(A)P(B) \\ &= (1-P(A))(1-P(B)) \\ &= P(A^c)P(B^c). \end{align}

  • 1
    $\begingroup$ The addition law says that 𝑃(𝐴∪𝐵) = 𝑃(𝐴)+𝑃(𝐵)-𝑃(𝐴∩𝐵). Why are you doing the opposite in the second line? $\endgroup$ Sep 13, 2022 at 7:33
  • $\begingroup$ Maybe we should also mention DeMorgans law as a good way to understand why the first line is true. $\endgroup$
    – hxlaclhemy
    Oct 14, 2022 at 21:35

gradient23's proof is great, in my opinion, but I would like to show another proof that seems more intuitive to me, though much less rigorous.

The proof is based on a verbal definition of independence from wikipedia:

two events are independent [...] if the occurrence of one does not affect the probability of occurrence of the other

In addition, we use the fact that independence is symmetric.

The (non-rigorous) proof:

  • We assume that $A$ and $B$ are independent.
  • By definition, the occurrence of $A$ doesn't affect the probability of $B$.
  • Thus, the occurrence of $A$ also doesn't affect the probability of $B^C$.
  • So by definition, $A$ and $B^C$ are also independent, which by definition again means that the occurrence of $B^C$ doesn't affect the probability of $A$.
    (Here we used the symmetry of independence.)
  • Therefore, the occurrence of $B^C$ also doesn't affect the probability of $A^C$.
  • So by definition, $B^C$ and $A^C$ are also independent.

One could convert the proof to the language of math: $$\begin{gathered}A\text{ and }B\,\text{are independent}\\ \downarrow\\ P\left(B|A\right)=P\left(B\right)\\ \downarrow\\ 1-P\left(B^{C}|A\right)=1-P\left(B^{C}\right)\\ \downarrow\\ P\left(B^{C}|A\right)=P\left(B^{C}\right)\\ \downarrow\\ A\text{ and }B^{C}\,\text{are independent}\\ \downarrow\\ P\left(A|B^{C}\right)=P\left(A\right)\\ \downarrow\\ 1-P\left(A^{C}|B^{C}\right)=1-P\left(A^{C}\right)\\ \downarrow\\ P\left(A^{C}|B^{C}\right)=P\left(A^{C}\right)\\ \downarrow\\ B^{C}\text{ and }A^{C}\,\text{are independent} \end{gathered} $$

But now we used conditional probabilities, which might be a problem in case $P(A)=0$ or $P(B^C)=0$ (here is a discussion about this problem).


As you have found :

$$P(A') = 1-P(A)$$

$$P(B') = 1- P(B)$$

Now clearly $P(A')P(B') =1 - [ P(A) + P(B)] + P(A \cap B)$

From set algebra we know that $P(A) + P(B) = P(A \cup B) + P(A\cap B)$ Substituting, we have $P(A')P(B') = P([A\cup B]')$

Now from De morgans law we know that:

$[A\cup B]' = [A' \cap B']$

Substituting, we have $P(A')P(B') = P(A' \cap B')$ , as required.


By definition if two events are independent then $P(A | B)=P(A)=P(A/B')$.

So, $P(A/B)=P(A \cap B)/P(B)=P(A)$ and hence by multiplying both sides by $P(B)$ we get $P(A \cap B)= P(A)P(B)$.

Hence proved.


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