Question

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?

Answer 1

We're told that

`P(A\cap B)=0.15`

`P(A\cup B)^C=0.06\implies P(A\cup B)=0.94`

`P(B\mid A)=P(B^C\mid A)=0.5`

where the last fact is due to the law of total probability:

`P(A)=P(A\cap B)+P(A\cap B^C)`

`\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)`

`\implies 1=P(B\mid A)+P(B^C\mid A)`

so that
`B\mid A` and
`B^C\mid A` are complementary.

By definition of conditional probability, we have

`P(B\mid A)=P(B^C\mid A)`

`\implies(P(A\cap B))/(P(A))=(P(A\cap B^C))/(P(A))`

`\implies P(A\cap B)=P(A\cap B^C)`

We make use of the addition rule and complementary probabilities to rewrite this as

`P(A\cap B)=P(A\cap B^C)`

`\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)`

`\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)`

`\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]`

`\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]`

`\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C`

`\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)`

By the law of total probability,

`P(B)=P(A\cap B)+P(A^C\cap B)`

`\implies P(A^C\cap B)=P(B)-P(A\cap B)`

and substituting this into
`(*)` gives

`2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]`

`\implies P(B)=P(A\cup B)-P(A\cap B)`

`\implies P(B)=0.94-0.15=\boxed{0.79}`