Solving for P(A) in Probability of Events: A^c U B = 0.6 and A^c U B^c = 0.8

[notorious9000]

I'm having difficulty time of trying to figure out answer.

Question:
P(A^c U B) = 0.6
P((A^c U B^C) = 0.8
Find P(A) ?

Sol:
P(A^c U B) = P(A^c) + P(B) - P(A^c Intersection B)
P(A^c U B^c) = P(A^c) + P(B^c) - P(A^c Intersection B^c)
0.6 + 0.8 = 2*P(A^c) + 1 - P(A^c) <==== HOW ?
P(A^c)=0.4, P(A)=1-P(A^c)=0.6

PROBLEM:
I have the solution, but I don't get how he got "- P(A^c)"
I think my instructor skipped some steps.

 

[lanedance]

it should be clear it you write out the steps
$$P(A^c \cup B) = P(A^c) + P(B) - P(A^c \cap B)$$
$$P(A^c \cup B^c) = P(A^c) + P(B^c) - P(A^c \cap B^c)$$

now add them together
$$P(A^c \cup B^c) + P(A^c \cup B) = 2P(A^c) + (P(B^c)+P(B)) - (P(A^c \cap B^c) +P(A^c \cap B) )$$

now see if you can fill in the last bit