There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that plant X is always idles. Event B is that at least two plants generate electricity. P(A∩B) is 15% and probability of complement of AUB is 6%. All outcomes in event A are equally likely. Compute the Probability of the event B?

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]