Find the probability $P(A^c \cap B^c)$ when only given $P(A), P(B)$.

Question

Given the following question:

Given 2 events $A$, $B$ where:

$P(A) = 0.4$, $P(B^c) = 0.7$, $P((A \cup B)^c) = 0.3$

Does $A$ and $B$ dependent each other?

I infered that $P(B) =1 - P(B^c) = 1 - 0.7 = 0.3$.

According to DeMorgan laws I know that $P((A \cup B) ^c) = P(A^c \cap B^c) = 0.3$.

All I have to know is whether $P(A \cap B) = P(A)P(B) = 0.3^2 = 0.09$.

Using all I have, how can I infer $P(A \cap B)$?

Thanks in advance.

Answer 1

$$P(A\cap B)=P(A)+P(B)-P(A\cup B) $$ $$=P(A)+(1-P(B^c))-(1-P((A\cup B)^c))$$ $$=P(A)-P(B^c)+P((A\cup B)^c)$$ This final thing is equal to zero, so they are not independent.

Answer 2

$P(A^c \cap B^c) = P(A^c) + P(B^c) - P(A^c \cup B^c) = 1 - P(A) + 1 - P(B) - P(A^c \cup B^c)$.