We're told that
[tex]P(A\cap B)=0.15[/tex]
[tex]P(A\cup B)^C=0.06\implies P(A\cup B)=0.94[/tex]
[tex]P(B\mid A)=P(B^C\mid A)=0.5[/tex]
where the last fact is due to the law of total probability:
[tex]P(A)=P(A\cap B)+P(A\cap B^C)[/tex]
[tex]\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)[/tex]
[tex]\implies 1=P(B\mid A)+P(B^C\mid A)[/tex]
so that [tex]B\mid A[/tex] and [tex]B^C\mid A[/tex] are complementary.
By definition of conditional probability, we have
[tex]P(B\mid A)=P(B^C\mid A)[/tex]
[tex]\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}[/tex]
[tex]\implies P(A\cap B)=P(A\cap B^C)[/tex]
We make use of the addition rule and complementary probabilities to rewrite this as
[tex]P(A\cap B)=P(A\cap B^C)[/tex]
[tex]\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)[/tex]
[tex]\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)[/tex]
[tex]\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)][/tex]
[tex]\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)][/tex]
[tex]\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C[/tex]
[tex]\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)[/tex]
By the law of total probability,
[tex]P(B)=P(A\cap B)+P(A^C\cap B)[/tex]
[tex]\implies P(A^C\cap B)=P(B)-P(A\cap B)[/tex]
and substituting this into [tex](*)[/tex] gives
[tex]2P(B)=P(A\cup B)+[P(B)-P(A\cap B)][/tex]
[tex]\implies P(B)=P(A\cup B)-P(A\cap B)[/tex]
[tex]\implies P(B)=0.94-0.15=\boxed{0.79}[/tex]
