To solve this problem, we'll use the principles of probability and the relationships between different events. Let's define the following:

• $P(P)$: The probability that a patient visits a physical therapist.
• $P(C)$: The probability that a patient visits a chiropractor.
• $P(P \cap C)$: The probability that a patient visits both a physical therapist and a chiropractor. This is given as 22%, or $P(P \cap C) = 0.22$.
• $P(\text{neither})$: The probability that a patient visits neither a physical therapist nor a chiropractor. This is given as 12%, or $P(\text{neither}) = 0.12$.
• $P(P \cup C)$: The probability that a patient visits either a physical therapist, a chiropractor, or both.

Using the fact that the probability of the union of two events $P(P \cup C)$ is related to the probability of each event and their intersection, we have:

$P(P \cup C) = P(P) + P(C) - P(P \cap C)$

Since $P(\text{neither}) = 1 - P(P \cup C)$, we can write:

$P(\text{neither}) = 1 - (P(P) + P(C) - P(P \cap C))$

Substituting the given values:

$0.12 = 1 - (P(P) + P(C) - 0.22)$

This simplifies to:

$P(P) + P(C) = 1 - 0.12 + 0.22 = 1.10$

We are also given that the probability of visiting a chiropractor exceeds the probability of visiting a physical therapist by 0.14:

$P(C) = P(P) + 0.14$

Substituting this into the equation $P(P) + P(C) = 1.10$:

$P(P) + (P(P) + 0.14) = 1.10$

This simplifies to:

$2P(P) + 0.14 = 1.10$

Subtracting 0.14 from both sides:

$2P(P) = 0.96$

Dividing by 2:

$P(P) = 0.48$

So, the probability that a randomly chosen patient visits a physical therapist is $\boxed{0.48}$.

Would you like further details or have any questions?

Here are 5 related questions you might consider:

1. What is the probability that a patient visits only a chiropractor?
2. How can you calculate the probability that a patient visits only a physical therapist?
3. What is the total probability that a patient visits at least one of the two specialists?
4. How would the solution change if the probabilities of visiting a physical therapist and a chiropractor were equal?
5. How would the result change if the probability of visiting neither specialist was different?

Tip: In probability problems involving two events, always consider both their union and intersection to fully understand the relationships between them.