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Lesson Overview

In this lesson, we derive several results that follow from the definition of probability.

Claim: For any measurable set \(A,\) \(P(A^C) = 1-P(A).\)

By the definition of complement, \(A \cap A^C = \emptyset.\) So, by definition of a probability, \[P(A \cup A^C) = P(A) + P(A^C)\] Let \(\Omega\) be the universal set. By definition of a probability again, \(P(\Omega) = 1.\) So, \begin{align} P(A) + P(A^C) & = P(A \cup A^C) \\ & = P(\Omega) \\ & = 1 \end{align} Subtracting \(P(A)\) from the first line and the last line gives the result: \[P(A^C) = 1 - P(A)\]

Corollary: \(P(\emptyset) = 0.\)

This follows because \(\Omega^C = \emptyset\) and \(P(\Omega) = 1.\) \begin{align} P(\emptyset) & = P(\Omega^C) \\ & = 1 - P(\Omega) \\ & = 1 - 1 \\ & = 0 \end{align}


Inclusion-exclusion: For any measurable sets \(A\) and \(B,\) \(P(A \cup B) = P(A) + P(B) - P(A \cap B).\)

First, we break up the set \(A \cup B\) into \(A\) and \(B \cap A^C\) so that we can use the properties of a probability. The idea to have in mind is that every point in \(A \cup B\) is We can formally show \(A \cup B = A \cup (B \cap A^C)\) by using distribution. Starting with the left hand side: \begin{align} A \cup (B \cap A^C) & = (A \cup B) \cap (A \cup A^C) \\ & = (A \cup B) \cap \Omega \\ & = A \cup B \end{align} Also, \(A \cap (B \cap A^C) = \emptyset,\) because if \(x \in B \cap A^C\) then \(x \in A^C.\) So, by the definition of a probability, \begin{align} P(A \cup B) & = P(A \cup (B \cap A^C)) \\ & = P(A) + P(B \cap A^C) \end{align} Next, we will add \(P(A \cap B) - P(A \cap B),\) which is really another way to add 0. \[P(A) + P(B \cap A^C) = P(A) + P(B \cap A^C) + P(A \cap B) - P(A \cap B)\] The sets in the middle are disjoint. \((B \cap A^C) \cap (A \cap B) = \emptyset\) since no point can be in \(A\) and in \(A^C.\) So, \begin{align} & P(A) + P(B \cap A^C) + P(A \cap B) - P(A \cap B) = \\ & P(A) + P((B \cap A^C) \cup (A \cap B)) - P(A \cap B) \end{align} One can apply distribution twice to see that \[(B \cap A^C) \cup (A \cap B) = B,\] which means \[P(A) + P((B \cap A^C) \cup (A \cap B)) - P(A \cap B) = P(A) + P(B) - P(A \cap B)\] Since we started with \(P(A \cup B),\) we have shown the inclusion-exclusion rule for two sets: \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

Claim: \(P(A \cap B) = P(A) + P(B) - P(A \cup B).\)

Starting from inclusion-exclusion, add \(P(A \cap B)\) to both sides and subtract \(P(A \cup B)\) from both sides.

Claim: If \(A \subset B,\) then \(P(A) \leq P(B).\)

By applying distribution twice, one can show for any sets \(A\) and \(B\) that \[B = (A \cap B) \cup (A^C \cap B)\] Since \(A \subset B,\) \(A \cap B = A.\) So, \(B = A \cup (A^C \cap B).\) Since \(A\) and \(A^C \cap B\) are disjoint, the definition of probability states \[P(B) = P(A) + P(A^C \cap B)\] By definition of a probability, \(P(A^C \cap B) \geq 0.\) Therefore, \[P(B) = P(A) + P(A^C \cap B) \geq P(A)\]

Claim: If \(B \subset A\) then \(P(A-B) = P(A)-P(B).\)

The event \(A\) can be written \(A = (A \cap B) \cup (A \cap B^C).\)

Since \(B \subset A,\) \(A \cap B = B.\) Using the fact that \(A \cap B\) and \(A \cap B^C\) are disjoint events, \begin{align} P(A) & = P(A \cap B) + P(A \cap B^C) \\ & = P(B) + P(A \cap B^C) \end{align} By definition of set subtraction, \(A \cap B^C = A - B.\) So, subtracting \(P(B)\) from both sides, we get \[P(A)-P(B)=P(A-B)\]


You are working for a company that sells auto insurance, life insurance, and home owners insurance. It sells a few less popular products as well. You know the following:

▼ Click for hint 1:
Translate the information into events.
Let \(L\) be the event a customer has life insurance, let \(H\) be the event a customer has home owners insurance, and let \(A\) be the event that a customer has auto insurance. Then you know the following:
▼ Click for hint 2:
After using hint 1 to translate the events, the questions can be translated as follows:
  1. What is \(P(A^C)?\)
  2. What is \(P(A \cup H)?\)
  3. What is \(P((A \cup H \cup L)^C)?\)
  4. Given an event \(E \subset L\), what is a possible value for \(P(E)?\)

1. What is the probability a randomly chosen customer does not have auto insurance?


2. What is the probability a randomly chosen customer has home owners or auto insurance?


3. What is the probability that a randomly chosen customer does not have any of the three types of insurance?


4. Another associate at your company knows the percent of customers that have both life insurance and another one of the less popular products at your company. Which percent is possible?