This year, a medical insurance policyholder has probability 0.70 of having no emergency room visits, 0.85 of having no hospital stays, and 0.61 of having neither emergency room visits nor hospital stays

Calculate the probability that the policyholder has at least one emergency room visit and at least one hospital stay this year.

• 0.045
• 0.060
• 0.390
• 0.667
• 0.840

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

Step 1: Define Events and Given Probabilities

We define the following events:

• E: The policyholder has at least one emergency room visit.
• H: The policyholder has at least one hospital stay.

From the problem statement, we are given probabilities related to the complements of these events:

Given Probabilities

Event	Description	Probability
[math]P(E')[/math]	No emergency room visits	0.70
[math]P(H')[/math]	No hospital stays	0.85
[math]P(E' \cap H')[/math]	Neither emergency room visits nor hospital stays	0.61

Our goal is to calculate [math]P(E \cap H)[/math], the probability that the policyholder has at least one emergency room visit and at least one hospital stay.

Step 2: Calculate Probabilities of At Least One Event

We can find the probabilities of event [math]E[/math] and event [math]H[/math] occurring by using the complement rule: [math]P(A) = 1 - P(A')[/math].

• Probability of at least one emergency room visit:

[[math]]P(E) = 1 - P(E') = 1 - 0.70 = 0.30[[/math]]

• Probability of at least one hospital stay:

[[math]]P(H) = 1 - P(H') = 1 - 0.85 = 0.15[[/math]]

Step 3: Calculate Probability of At Least One of Either Event Occurring

We are given the probability that neither event occurs, which is [math]P(E' \cap H') = 0.61[/math]. Using De Morgan's Laws, we know that the event "neither E nor H" is the complement of the event "E or H". Therefore, [math](E' \cap H') = (E \cup H)'[/math]. We can find the probability of having at least one emergency room visit OR at least one hospital stay using the complement rule:

[[math]]P(E \cup H) = 1 - P((E \cup H)') = 1 - P(E' \cap H')[[/math]]

[[math]]P(E \cup H) = 1 - 0.61 = 0.39[[/math]]

Step 4: Calculate the Probability of Both Events Occurring

We can use the Principle of Inclusion-Exclusion to relate the probabilities we have calculated:

[[math]]P(E \cup H) = P(E) + P(H) - P(E \cap H)[[/math]]

We want to find [math]P(E \cap H)[/math], so we rearrange the formula:

[[math]]P(E \cap H) = P(E) + P(H) - P(E \cup H)[[/math]]

Substitute the values calculated in Step 2 and Step 3:

• [math]P(E) = 0.30[/math]
• [math]P(H) = 0.15[/math]
• [math]P(E \cup H) = 0.39[/math]

[[math]]P(E \cap H) = 0.30 + 0.15 - 0.39[[/math]]

[[math]]P(E \cap H) = 0.45 - 0.39[[/math]]

[[math]]P(E \cap H) = 0.06[[/math]]

Thus, the probability that the policyholder has at least one emergency room visit and at least one hospital stay this year is 0.06.

Key Insights

• The complement rule ([math]P(A) = 1 - P(A')[/math]) is fundamental for converting probabilities of "no events" to "at least one event".
• De Morgan's Laws are crucial for relating the complement of a union of events to the intersection of their complements: [math](A \cup B)' = A' \cap B'[/math]. This allows us to interpret "neither A nor B" as the complement of "A or B".
• The Principle of Inclusion-Exclusion ([math]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/math]) is essential for solving problems involving the union and intersection of events. It allows us to calculate any one of the four probabilities if the other three are known or can be derived.
• Careful definition of events and their complements is critical to correctly apply probability rules.

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Solution: B

P(at least one emergency room visit or at least one hospital stay) = 1 – 0.61 = 0.39 = P(at least one emergency room visit) + P(at least one hospital stay) – P(at least one emergency room visit and at least one hospital stay).

P(at least one emergency room visit and at least one hospital stay) = 1 – 0.70 + 1 – 0.85 – 0.39 = 0.060.

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