I was under the impression that 3 events are mutually independent when A and B, B and C & A and C are all (separately) independent.
Events are independent when:
$P(A \cap B) = P(A) \cdot P(B)$
or equivalently:
$\require{cancel} P(A|B) = P(A)$ (because of Bayes', $P(A|B)=\frac{P(A \cap B)}{P(B)} \to P(A|B) = \frac{P(A) \cdot \bcancel{P(B)}}{\bcancel{P(B)}} \to P(A|B) = P(A)$)
So given:
Events A, B and C, s.t. $P(A) = P(B) = 0.4, P(C) = 0.5$ and $P(A \cap B) = 0.16, P(A \cap C) = 0.2, P(B | C) = 0.4$
I calculated:
$P(A \cap B) = 0.16 = 0.4 \cdot 0.4 = P(A) \cdot P(B)$
$P(A \cap C) = 0.2 = 0.4 \cdot 0.5 = P(A) \cdot P(C)$
and
$P(B | C) = 0.4 = 0.4 = P(B)$
I thought this would be more than sufficient to tell that A, B and C are (mutually) independent.
However, I started to doubt when the solutions given for this exercise stated that A, B and C are dependent.
Please explain why my reasoning is incorrect, or verify that the solutions given were incorrect.