simple probability questions i) Of a group of patients having injuries, 28% visit both a physical therapist and a chiropractor and 8% visit neither. Say that the probability of visiting a physical therapist exceeds the probability of visiting a chiropractor by 16%. What is the probability of a randomly selected person from this group visiting a physical therapist?
ii)given that P(A U B) = 0.76 and P(A U B')=0.87, find P(A)
iii)For each positive integer n, let p({n})=(1/2)^2. Consider the even A={n:1<=n<=10}, B={n:1<=n<=20}, C={n:11<=n<=20} find P(A)
out of like 13 questions in the book these were the 3 I didnt manage to solve, so anyone know how to go about these one? wouldnt suprise me at all if they are extremely easy, anyway thanks in advance for solutions/tips on any of the questions :)
 A: i) Let us denote $A$ to be the physical therapist and $B$ to be chiropractor. We know that $\mathbb{P}(A\cup B)=.28$ and $\mathbb{P}(A\cap B)=.08$. That means that the rest, i.e. $\mathbb{P}(A\cap B')$ and $\mathbb{P}(A'\cap B)$ has to have $1-0.08-0.28=0.64$. Moreover, we know that $\mathbb{P}(A)=\mathbb{P}(B)+0.16$. This can be rewritten as $\mathbb{P}(A\cap B)+\mathbb{P}(A,B')=\mathbb{P}(A\cap B)+\mathbb{P}(A'\cap B)+0.16$. This can be simplified to $$\mathbb{P}(A\cap B')=\mathbb{P}(A'\cap B)+0.16$$. In combination with $$\mathbb{P}(A,B')+\mathbb{P}(A'\cap B)=0.64$$, you obtain
$$\mathbb{P}(A\cap B')+\mathbb{P}(A\cap B')-0.16=0.64$$
$$\mathbb{P}(A\cap B')+\mathbb{P}(A,B')=0.8$$
$$2\mathbb{P}(A\cap B')=0.8$$
$$\mathbb{P}(A\cap B')=0.4$$
and finally $\mathbb{P}(A)=\mathbb{P}(A,B')+\mathbb{P}(A\cap B)=0.4+0.28=0.68$.
ii) From $\mathbb{P}(A\cup B)=.76$, you know that $\mathbb{P}(A' \cap B')=.24$. Similarly, you can infer from $\mathbb{P}(A\cup B')=.87$ that $\mathbb{P}(A'\cap B)=.13$ Thus, you know that $\mathbb{P}(A')=\mathbb{P}(A'\cap B) + \mathbb{P}(A' \cap B')=.13+.24=.37$. From this, you obtain $\mathbb{P}(A)=.63$. 
iii) If I understand correctly, there is a probability for each $n\in\mathbb{N}$. The probability can be hardly (1/2)^2$=\frac{1}{4}$ - in that case the sum would not be one. I have the following interpretation: originaly there was (1/2)^n=$\left(\frac{1}{2}\right)^n$. In that case, the sum is one. You get for $A$:
$$\sum_{n=1}^{10}\left(\frac{1}{2}\right)^n=1-\left(\frac{1}{2}\right)^{10}$$ 
Please check your formulation and tell me whether it fits.
