# Finding Probabilities

1. Aug 15, 2014

### Luscinia

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Pr(A∪B)=Pr(A)+Pr(B)-Pr(A∩B)
Pr(A)'=1-Pr(A)

Pr(A∩B)=Pr(A)*Pr(B) (?)

3. 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)"?

2. Aug 15, 2014

### Orodruin

Staff Emeritus
Pr(A∩B)=Pr(A)*Pr(B) is only valid if A and B are independent.

3. Aug 15, 2014

### HallsofIvy

I've never liked using formulas for problems like this. Instead:

Suppose there are 10000 patients. 22%, or 2200, visit both a physical therapist and a chiropractor and 12%, or 1200 visit nether. The remaining 66%, or 6600 visit one or the other but not both.

The probability that a patient visits a chiropractor exceeds by 0.14 the probability that a patient visits a physical therapist. That is, if "n" is the number of people who go to a physical therapist only, then, including the 2200 who go to both, there are n+ 2200 who go to a physical therapist and so 1.14(n+ 2200)= 1.14n+ 2508 who go to a chiropractor and so 1.14n+ 2508- 2200= 1.14n+ 308 who go to a chiropractor only. We must have that the number of people who go to both, plus the number of people who go to the physical therapist only, plus the number of people who go to a chiropractor only, plus the number of people who go to neither, equal to 10000: 2200+ n+ 1.14n+ 308+ 1200= 10000.

Solve that equation for n. Then n+ 2200 go to a physical therapist. Divide that by 1000 to find the probability that one goes to a physical therapist.

4. Aug 15, 2014

### Orodruin

Staff Emeritus
I do not agree with this statement. The problem statement was:
In fact, I believe this was the problem writer's reason for not writing 14%. The statement is simply saying that P(chiropractor) = P(therapist)+0.14. Or in your terms:

N(chiropractor) = N(therapist) + 0.14*10000 = n + 2200 + 1400 = n + 3600

Other than that, I really do not see any advantage of multiplying by a number of people. All it is doing is multiplying all of the equations by that number - the logic is still the same.

5. Aug 15, 2014

### HallsofIvy

That makes no sense at all! You are asserting that if 10 people go a physical therapist, 10.14 will go to a chiropractor! What would be meant by .14 person?

6. Aug 15, 2014

### Orodruin

Staff Emeritus
a) No, that is not what I am saying and not what the problem is saying. It is saying that the probability of a person going to the chiropractor is 0.14 greater than the probability of a person going to the therapist. This has no direct link to the actual number or quotient between them. The math was in my previous post.

b) You have no way of knowing the number of participants in the study. Your choice of 10000 is completely arbitrary and another choice would have left you with a fractional number of people.

c) The problem deals with probabilities, not number of people. It may help to understand the procedure to think of a number, but it is perfectly fine to do without.

d) Even a discrete probability distribution may have an expectation value which is not an integer.

7. Aug 15, 2014

### Luscinia

Oh wow, that was much simpler than I thought. Thanks!
I guess I jump to equations too quickly and automatically assumed that it was independent events by default.

8. Aug 15, 2014

### Ray Vickson

It would not matter if the events were independent or not; you were GIVEN $P(C \cap T)$ as numerical input data and you could get $P(C \cup T)$ from other input data. That's all you need.

9. Aug 15, 2014

### Luscinia

Even if the data was given to me, wouldn't I still be able to somehow get the same number by doing calculations though? I don't really care about the right or wrong answer at this point, I just want to know if there is more than 1 way of solving this problem.

10. Aug 15, 2014

### Ray Vickson

I think there is essentially just one way to solve this particular problem. We have
$$P(C \cup T) = 1 - 0.12 = 0.88 \; (\text{essentially, given})\\ P(C \cap T) = 0.22 \; (\text{given})\\ P(C \cup T) = P(C) + P(T) - P(C \cap T) \: \Longrightarrow P(C) + P(T) = P(C \cup T) + P(C \cap T) = 0.88 + 0.12 = 1.10\\ P(C) - P(T) = 0.14 \; (\text{given})$$
Solving the two equations for $P(C), P(T)$ gives $P(C) = 0.62, P(T) = 0.48$.

Note that $P(C \cap T) = 0.22 \neq P(C) \cdot P(T) = 0.2976$. Therefore, the two events $C$ and $T$ are not independent. Trying to pretend that they are would lead to the wrong answer.