[tex]P(A)=0.5[/tex], [tex]P(B)=0.6[/tex], [tex]P(A\cup B)=0.8[/tex]
We find the probability of intersection using the inclusion/exclusion principle:
[tex]P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.3[/tex]
By definition of conditional probability,
[tex]P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.3}{0.6}=0.5[/tex]
For [tex]A[/tex] and [tex]B[/tex] to be independent, we must have
[tex]P(A\cap B)=P(A)\cdot P(B)[/tex]
in which case we have [tex]0.3=0.5\cdot0.6[/tex], which is true, so [tex]A[/tex] and [tex]B[/tex] are indeed independent.
Or, to establish independence another way, in terms of conditional probability, we must have
[tex]P(A\mid B)\cdot P(B)=P(A)\cdot P(B)\implies P(A\mid B)=P(A)[/tex]
which is also true.
