I just started on online probability and statistics class, and I feel like I'm missing some kind of principle rule here. On our homework, I missed (and can't figure out how to find) the following:
$P(A \cup B) = 0.85$, $P(A \cup B^c) = 0.73$. What is $P(A)$?
The answer given is $0.58$. Can someone explain how this is gotten?
On
Consider the probability of the complement event. The complement of $A\cup B$ is $A^c\cap B^c$, so $P(A^c\cap B^c)=1-P(A\cup B)=0.15$. Similarly, $P(A^c\cap B)=1-P(A\cup B^c)=0.27$. Since $A^c$ is the disjoint union of the events $A^c\cap B$ and $A^c\cap B^c$, deduce $$P(A^c)=P(A^c\cap B)+P(A^c\cap B^c)=0.15+ 0.27 = 0.42,$$ and therefore $P(A)=1-P(A^c)=0.58$.
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ $$P(A\cup B^c)=P(A)+P(B^c)-P(A\cap B^c)$$
Add those two equalities:
$$P(A\cup B)+P(A\cup B^c)=2P(A)+(P(B)+P(B^c))-(P(A\cap B)+P(A\cap B^c))$$
Recall $P(B)+P(B^c)=1$ and $P(A\cap B)+P(A\cap B^c)=P(A)$ because $A\cap B$ and $A\cap B^c$ are disjoint and their union is $A$. Replacing in the above equality, we get:
$$0.85+0.73=2P(A)+1-P(A)=P(A)+1$$
i.e. $P(A)=0.85+0.73-1=0.58$.
Update: Another way:
$$P(A\cup B)+P(A\cup B^c)=P((A\cup B)\cup(A\cup B^c))+P((A\cup B)\cap(A\cup B^c))=P(A\cup B\cup B^c)+P(A\cup(B\cap B^c))=1+P(A)$$
(because $B\cup B^c=X, B\cap B^c=\emptyset$), so $P(A)=P(A\cup B)+P(A\cup B^c)-1=0.85+0.73-1=0.58$.