Question

Please helpppp it has to do with P(A U B)

Answer 1

$P(A)=0.5$, $P(B)=0.6$, $P(A\cup B)=0.8$

We find the probability of intersection using the inclusion/exclusion principle:


$P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.3$


By definition of conditional probability,


$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.3}{0.6}=0.5$


For $A$ and $B$ to be independent, we must have


$P(A\cap B)=P(A)\cdot P(B)$

in which case we have $0.3=0.5\cdot0.6$, which is true, so $A$ and $B$ are indeed independent.


Or, to establish independence another way, in terms of conditional probability, we must have

$P(A\mid B)\cdot P(B)=P(A)\cdot P(B)\implies P(A\mid B)=P(A)$


which is also true.