1-1. Of a group of patients having injuries, 28% visit both a physical therapist and a chiropractor while...

Question

1.1-1. Of a group of patients having injuries, 28% visit both a physical therapist and a chiropractor while 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?

SnapAndSolve AI · Accepted Answer

Let P(T) be the probability of visiting a physical therapist and P(C) be the probability of visiting a chiropractor.

We know:

P(T and C) = 0.28
P(neither T nor C) = 0.08
P(T) = P(C) + 0.16

Using the formula for the union of two events:

P(T or C) = P(T) + P(C) - P(T and C)

Since P(neither T nor C) = 0.08, we have:

P(T or C) = 1 - 0.08 = 0.92

Substitute the known values:

0.92 = P(T) + P(C) - 0.28

0.92 = (P(C) + 0.16) + P(C) - 0.28

0.92 = 2P(C) + 0.16 - 0.28

0.92 = 2P(C) - 0.12

1.04 = 2P(C)

P(C) = 0.52

Therefore, P(T) = P(C) + 0.16 = 0.52 + 0.16 = 0.68

The probability of a randomly selected person visiting a physical therapist is 0.68 or 68%.

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1-1. Of a group of patients having injuries, 28% visit both a physical therapist and a chiropractor while...

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