probability problem with the intersection of 2 events given The question is: 1.) 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? What I gathered while trying to find the answer is: $A\cap B\:=.28$ $A'\cap B'\:=.08$ $1.0 - (0.28 + 0.08) = .64$ The confusion comes from figuring out what $A$ and $B$ is with the remaining info: > **physical therapist exceeds the probability of visiting a chiropractor by 16%.** venn diagram open for editing here. ![enter image description here]( _note:_ the answer for the question is $0.68$

Let $ A $ denote a physiotherapist visit and $ B $ denote a chiropractor visit.

What is the probability of visiting a physiotherapist? $ P(A) $

What is the probability of visiting a chiropractor? $ P(B) $

We are told that the probability of visiting a physiotherapist exceeds the probability of visiting a chiropractor by 16%, which means that:

$$ 0.16 = P(A) - P(B) $$

Recall the formula for the intersection of two events: $ P(A \cap B) = P(A) + P(B) - P(A \cup B) $

Let $ 0.16 = P(A) - P(B) $ be equation 1 and let $ P(A \cap B) = P(A) + P(B) - P(A \cup B) $ be equation 2.

Then add them together to get:

$$ 0.16 + P(A \cap B) = P(A) - P(B) + P(A) + P(B) - P(A \cup B) $$

You can now solve for $ P(A) $, i.e. $ P(A) = \frac{1}{2} \big(0.16 + P(A \cap B) + P(A \cup B) \big) $

Substituting $ P(A \cap B) = 0.28 $ and $ P(A \cup B) = 1 - P(A' \cap B') = 1 - 0.08 = 0.92 $ into this equation, we find that $ P(A) = 0.68 $.