Apr 28'23

Exercise

The probability that a visit to a primary care physician’s (PCP) office results in neither lab work nor referral to a specialist is 35%. Of those coming to a PCP’s office, 30% are referred to specialists and 40% require lab work. Calculate the probability that a visit to a PCP’s office results in both lab work and referral to a specialist.

  • 0.05
  • 0.12
  • 0.18
  • 0.25
  • 0.35

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

2 Answers
Oct 23'25
Step 1: Define Events and Given Probabilities

Define the events involved in the problem and list the probabilities provided.

  • Let [math]R[/math] be the event that a visit to a primary care physician's (PCP) office results in a referral to a specialist.
  • Let [math]L[/math] be the event that a visit to a PCP's office requires lab work.

The problem provides the following probabilities:

Given Probabilities
Event Description Probability
Neither lab work nor referral to a specialist [math]P(R^c \cap L^c) = 0.35[/math]
Referred to a specialist [math]P(R) = 0.30[/math]
Requires lab work [math]P(L) = 0.40[/math]
Step 2: Utilize De Morgan's Law and the Complement Rule

The event "neither lab work nor referral to a specialist" means that the visit is not in event [math]R[/math] AND not in event [math]L[/math]. This can be expressed using De Morgan's Law, which states that the complement of the union of two events is the intersection of their complements:

[[math]]R^c \cap L^c = (R \cup L)^c[[/math]]
Given [math]P(R^c \cap L^c) = 0.35[/math], we can write:
[[math]]P((R \cup L)^c) = 0.35[[/math]]
Using the complement rule of probability, [math]P(E) = 1 - P(E^c)[/math], we can find the probability of the union of events [math]R[/math] and [math]L[/math]:
[[math]]P(R \cup L) = 1 - P((R \cup L)^c)[[/math]]
[[math]]P(R \cup L) = 1 - 0.35[[/math]]
[[math]]P(R \cup L) = 0.65[[/math]]

Step 3: Apply the Principle of Inclusion-Exclusion

The Principle of Inclusion-Exclusion for two events [math]R[/math] and [math]L[/math] states that the probability of their union is:

[[math]]P(R \cup L) = P(R) + P(L) - P(R \cap L)[[/math]]
We are asked to calculate the probability that a visit results in both lab work and referral to a specialist, which is [math]P(R \cap L)[/math]. We can rearrange the formula above to solve for [math]P(R \cap L)[/math]:
[[math]]P(R \cap L) = P(R) + P(L) - P(R \cup L)[[/math]]

Step 4: Calculate the Probability of Both Events Occurring

Now, substitute the known probabilities from Step 1 and Step 2 into the rearranged formula from Step 3:

  • [math]P(R) = 0.30[/math]
  • [math]P(L) = 0.40[/math]
  • [math]P(R \cup L) = 0.65[/math]

[[math]]P(R \cap L) = 0.30 + 0.40 - 0.65[[/math]]
[[math]]P(R \cap L) = 0.70 - 0.65[[/math]]
[[math]]P(R \cap L) = 0.05[[/math]]
Thus, the probability that a visit to a PCP’s office results in both lab work and referral to a specialist is 0.05.

Key Insights
  • Defining clear notation for events ([math]R[/math] for referral, [math]L[/math] for lab work) is crucial for translating word problems into mathematical expressions.
  • Understanding how to interpret "neither A nor B" is key. It translates to [math]P(A^c \cap B^c)[/math], which, by De Morgan's Law, is equivalent to [math]P((A \cup B)^c)[/math].
  • The complement rule, [math]P(E^c) = 1 - P(E)[/math], is fundamental for finding the probability of an event when the probability of its complement is known.
  • The Principle of Inclusion-Exclusion for two events, [math]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/math], is a versatile tool for calculating probabilities of unions and intersections.
  • Probability formulas can be rearranged to solve for unknown components, such as [math]P(A \cap B)[/math] when [math]P(A)[/math], [math]P(B)[/math], and [math]P(A \cup B)[/math] are known.
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Apr 28'23

Solution: A

Let

[math]R[/math] = event of referral to a specialist

[math]L[/math] = event of lab work

We want to find

[[math]] \begin{align*} \operatorname{P}[R∩L] &= \operatorname{P}[R] + \operatorname{P}[L] – \operatorname{P}[R∪L] \\ &= \operatorname{P}[R] + \operatorname{P}[L] – 1 + \operatorname{P}[~(R∪L)] \\ &= \operatorname{P}[R] + \operatorname{P}[L] – 1 + \operatorname{P}[~R∩~L] \\ &= 0.30 + 0.40 – 1 + 0.35 \\ &= 0.05 . \end{align*} [[/math]]

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

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