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.

Calculate the probability that a randomly chosen member of this group visits a physical therapist.

• 0.26
• 0.38
• 0.40
• 0.48
• 0.62

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: D

Let

[math]C[/math] = event that patient visits a chiropractor

[math]T[/math] = event that patient visits a physical therapist

We are given that

[[math]] \operatorname{P}[C ] = \operatorname{P}[T ] + 0.14 \operatorname{P}( C ∩ T ) = 0.22 \operatorname{P}( C^c ∩ T^c ) = 0.12 [[/math]]

Therefore,

[[math]] \begin{align*} 0.88 = 1 − \operatorname{P}[C^c ∩ T^c ] = \operatorname{P}[C ∪ T ] &= \operatorname{P}[C ] + \operatorname{P}[T ] − \operatorname{P}[C ∩ T ] \\ &= \operatorname{P}[T ] + 0.14 + \operatorname{P}[T ] − 0.22 \\ &= 2 \operatorname{P}[T ] − 0.08 \end{align*} [[/math]]

or

[[math]] \operatorname{P}[T ] = ( 0.88 + 0.08 ) 2 = 0.48 [[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.