- #1

Luscinia

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## Homework Statement

Among a large group of patients recovering from shoulder injuries, it is found that 22% visit both a physical therapist and a chiropractor, whereas 12% visit neither of these. The probability that a patient visits a chiropractor exceeds by 0.14 the probability that a patient visits a physical therapist. Determine the probability that a random chosen member of this group visits a physical therapist.

## Homework Equations

Pr(A∪B)=Pr(A)+Pr(B)-Pr(A∩B)

Pr(A)'=1-Pr(A)

Pr(A∩B)=Pr(A)*Pr(B) (?)

## The Attempt at a Solution

Let P=Physical Therapy and C=Chiropractor

I've managed to solve this once by finding Pr(P∪C)':

1-(P+(P+0.14)-Pr(P∩C))=0.12

1-(2P+0.14-0.22)= 0.12

1+0.08-0.12=2P

P=0.48

What I would like to know is why is it not possible to solve the problem using the equation "Pr(A∩B)=Pr(A)*Pr(B)"

0.22=P*(P+0.14)

P^2+0.14P-0.22=0

Using the quadratic formula and ruling out the negative result, P=0.40 ≠ 0.48

Is there a condition that I've missed that forbids the use of "Pr(A∩B)=Pr(A)*Pr(B)"?