Inclusion-Exclusion Principle: Actuary Exam P

ByCL

The Inclusion-Exclusion Principle is a counting technique in Probability and Combinatorics. The idea is simple, yet it always feels more difficult when solving problems. However, It does not have to be difficult.

We use sample questions for Exam P from the Society of Acturaries to illustrate how we can break things down and how it helps with problem-solving.

The first step is always to read the question and figure out “what is given” and “what is the goal”. Once we have a clear starting point and goal, the problem becomes easier. When doing so, we might feel it takes so long to solve a problem. Over time, you will master the process and become faster and faster. Remember,

Be Clear, Not Clever.

Question 1

A survey of a group’s viewing habits over the last year revealed the following information:

  • 28% watched gymnastics
  • 29% watched baseball
  • 19% watched soccer
  • 14% watched gymnastics and baseball
  • 12% watched baseball and soccer
  • 10% watched gymnastics and soccer
  • 8% watched all three sports

Calculate the percentage of the group that watched none of the three sports during the last year.

  1. 24%
  2. 36%
  3. 41%
  4. 52%
  5. 60%

Key notes

  • Reading through the question, we understand that the question is about the relations among three sets: viewers who watched gymnastics, baseball, and soccer, respectively.
  • Use the inclusion-exclusion Principle of three sets (or events, which is the terminology in Probability)

Solution

Given:

Let G = viewer watched gymnastics, B = viewer watched baseball, S = viewer watched soccer. We know that

  • P[G] = 0.28
  • P[B] = 0.29
  • P[S] = 0.19
  • P[G\cap B] = 0.14
  • P[B \cap S] = 0.12
  • P[G \cap S] = 0.10
  • P[G \cap B \cap S] = 0.08
Goal:

Find the percentage of the group that watched none of the three sports, that is, P[ (G \cup B \cup S)’ ]

Let us try to achieve the goal:


Notice that P[ (G \cup B \cup S)’ ] = 1 – P[ (G \cup B \cup S) ] . By the Inclusion-Exclusion formula, we have
\begin{aligned} P[ G \cup B \cup S ] &= P[G] + P[B] + P[S] – P[G \cap B] – P[B \cap S] – P[G \cap S] + P[G \cap B \cap S] \\ &= 0.28 + 0.29 + 0.19 – 0.14 – 0.12 – 0.10+ 0.08 \\ & = 0.48 \end{aligned}
Thus, P[ (G \cup B \cup S)’ ] = 1 – P[ (G \cup B \cup S) ] = 1 – 0.48 = 0.52 . The percentage of the group that watched none of the three sports is 52%.

Question 2

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.

  1. 0.05
  2. 0.12
  3. 0.18
  4. 0.25
  5. 0.35

Key notes

  • There are two possible events: lab work and referral to a specialist
  • Use the inclusion-exclusion Principle of two events

Solution

Given:

Let L = lab work and R = referral to a specialist. We are given that

  • The probability that a visit results in neither lab work nor referral to a specialist is P[(L \cup R)’] = 35\%
  • 30\% are referred to specialists, P[R] = 30\%
  • 40\% require lab work, P[L] = 40\%
Goal:

Find the probability that a visit results in both lab work and referral to a specialist, P[L \cap R]

Let us try to achieve the goal:

\begin{aligned} P[L \cap R] &= P[L] + P[R] – P[ L \cup R] \\ &= P[L] + P[R] – (1 – P[ (L \cup R)’ ] ) \mbox{ because we only know } P[ (L \cup R)’ = 0.35 \\ &= P[L] + P[R] – 1 + P[ (L \cup R)’ ] \\ &= 0.30 + 0.40 -1 + 0.35 = 0.05 \end{aligned}

The probability that a visit results in both lab work and referral to a specialist is 5%.

Question 3

You are given P[ A \cup B ] = 0.7 and P[ A \cup B’ ] = 0.9 .

Calculate P[A] .

  1. 0.2
  2. 0.3
  3. 0.4
  4. 0.6
  5. 0.8

Key notes

  • This question has a simple structure. We can go ahead and use the given information to achieve the goal P[A] .
  • Use the inclusion-exclusion Principle of two sets

Solution

Let us try to achieve the goal:

Note that

  • P[ A \cup B] = P[A] + P[B] – P[ A \cap B]
  • P[A \cup B’] = P[A] + P[B’] – P[ A \cap B’]

Adding the two equations, we get
\begin{aligned} P[ A \cup B] + P[A \cup B’] &= 0.7 + 0.9 \\ &= P[A] + P[B] – P[ A \cap B] + P[A] + P[B’] – P[ A \cap B’] \\ &= 2P[A] + (P[B] + P[B’]) – (P[ A \cap B] + P[ A \cap B’]) \\ &= 2P[A] + 1 – P[A] \end{aligned}

Therefore, 1.6 = P[A] + 1 and P[A] = 0.6

Similar Posts