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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 $0.68$

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    You should have written $ P (A' \cap B') = 0.08 $. $ P(A' \cup B') $ is unknown. The question states that 0.08 of the patients do not see a physiotherapist or a chiropractor, which means that they belong to both $ A' $ and $ B' $. – 2017-01-05
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    @MichaelR I've made the edit, I will take a look over again and see if I understand it. – 2017-01-05

2 Answers 2

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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 $.

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    how do you go about finding $P(A \cup B)$? – 2017-01-05
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    $ P (A \cup B) = 1 - P(A' \cap B') = 1 - 0.08 = 0.92 $. – 2017-01-05
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Since $P(A \cap B) + P(A' \cup B') = 1$, we have $$ P(A \setminus B) + P(B \setminus A) = .64. $$

Assuming that $A=chiropractor$ and $B=therapist$, the remaining info says that $$ P(B) = P(A)+.16. $$

Can you take it from there?