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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 a visiting a physical therapist exceeds the probability of of visiting a chiropractor by 16%. What is the probability of a randomly selected person from the group visiting a physical therapist?

My attempt:

GIVEN
A = Visiting a physical therapist
B = Visiting a chiroprator

Pr(A union B)' = .08
Pr(A union B) = 1 - .08 = .92
Pr(A intersect B) = .28
Pr(A) = Pr(B) + .16

SOLUTION
Pr(A union B) = Pr(A) + Pr(B) - Pr(A intersect B)
.92 = [Pr(B) + .16] + Pr(B) - .28
.92 = 2Pr(B) - .12
1.04 = 2 Pr(B)
.52 = Pr(B)

Since, Pr(A) = Pr(B) + .16 then P(A) = .52 + .16 = .68.

Is this solution correct?

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  • $\begingroup$ The solution is correct. $\endgroup$ – André Nicolas Sep 21 '13 at 3:48

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