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25 May, 12:55

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 d le to 010.0 validedo a patient visits a physical therapist. oblodiog bemion-l dans h an n itidadong Determine the probability that a randomly chosen member of this group visits a physical therapist.

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  1. 25 May, 13:36
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    0.48

    Step-by-step explanation:

    First of all. let's recall probability formula needed for solving the problem.

    We know that, If the events A and B are not mutually exclusive, the probability is:

    probability of event (A or B) = probability of event A + probability of event B - p (A and B).

    This formula will be used in calculations of this problem.

    Let's consider that event A is patient visiting physical therapist; event B is patient visiting chiropractor.

    P (A∩B) = 0.22 (visiting both physical therapist and chiropractor) ⇔ p (A and B)

    P (B) = P (A) + 0.14 [ Probability patient visiting chiropractor is 0.14 more than probability visiting physical therapist]

    Patients visiting none of these is 12%

    Those who visit either therapist or chiropractor are amount to 1 - 0.12 = 0.88

    Now we should use the formula mentioned in the beginning of the text:

    0.88 = P (A) + P (B) - 0.22 ⇒ P (A) + P (B) = 1.1

    If we replace P (B) by P (A) + 0.14 ⇒ 2P (A) + 0.14 = 1.1 ⇒ 2P (A) = 1.1 - 0.14 = 0.96

    So, P (A) = 0.96/2 = 0.48 [Probability of patients visiting therapist]
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