In this lesson, we derive several results that follow from the definition of probability.
Claim: For any measurable set \(A,\) \(P(A^C) = 1-P(A).\)
▼ Proof:
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:
\[P(A^C) = 1 - P(A)\]
Inclusion-exclusion: For any measurable sets \(A\) and \(B,\) \(P(A \cup B) = P(A) + P(B) - P(A \cap B).\)
▼ Proof:
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
an element of \(A,\) or
not an element of \(A.\) If it is not an element of \(A\) then it must be an element of \(B,\) so it is an element of \(B \cap A^C.\)
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:
\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]
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 \(P(A) \leq P(B).\)
▼ Proof:
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 \(P(A-B) = P(A)-P(B).\)
▼ Proof:
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
\[P(A)-P(B)=P(A-B)\]
Quiz:
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:
40% of the customers have life insurance
55% of the customer have home owners insurance
55% of the customers have auto insurance
20% of the customers have both auto and home owners insurance
15% of the customer have both life and home owners insurance
30% of the customers have both auto and life insurance
10% of the customers have all three types of insurance
▼ 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:
\(P(L) = 0.4\)
\(P(H) = 0.55\)
\(P(A) = 0.55\)
\(P(H \cap A) = 0.2\)
\(P(L \cap H) = 0.15\)
\(P(A \cap L) = 0.3\)
\(P(L \cap H \cap A) = 0.1\)
▼ Click for hint 2:
After using hint 1 to translate the events, the questions can be translated as follows:
What is \(P(A^C)?\)
What is \(P(A \cup H)?\)
What is \(P((A \cup H \cup L)^C)?\)
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?
Unanswered
Solution: Let \(A\) be the set of customers that have auto insurance. Then \(P(A) = 0.55.\) The set of customers that do not have auto insurance is \(A^C.\) So, the probability that a randomly chosen customer does not have auto insurance is
\[P(A^C) = 1 - P(A) = 1 - 0.55 = 0.45\]
2. What is the probability a randomly chosen customer has home owners or auto insurance?
Unanswered
Solution: Let \(H\) be the set of customers that have home owners insurance and \(A\) be the set of customers that have auto insurance. The set of customers that have home owners or auto insurance is \(H \cup A.\) By inclusion-exclusion,
\begin{align}
P(H \cup A) & = P(H) + P(A) - P(H \cap A) \\
& = 0.55 + 0.55 - 0.2 \\
& = 0.9
\end{align}
3. What is the probability that a randomly chosen customer does not have any of the three types of insurance?
Unanswered
Solution: Let \(L\) be the set of customers that have life insurance, \(H\) be the set of customers that have home owners insurance and \(A\) be the set of customers that have auto insurance. Using inclusion-exclusion for \(3\) sets, we have
\begin{align}
P(L \cup H \cup A) & = P(L) + P(H) + P(A) - P(L \cap H) - P(L \cap A) - P(H \cap A) + P(L \cap H \cap A) \\
& = 0.4 + 0.55 + 0.55 - 0.15 - 0.3 - 0.2 + 0.1 \\
& = 0.95
\end{align}
The probability that a customer has none of life insurance, health insurance, or auto insurance, is
\[P((L \cup H \cup A)^C) = 1 - P(L \cup H \cup A) = 1 - 0.95 = 0.05\]
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?
Unanswered
Solution: Let \(B\) be the set of customers that have both life insurance and a less popular product, and let \(L\) be the set of customer that have life insurance. Then \(B \subset L,\) so \(P(B) \leq P(L).\) Since \(P(L) = 0.4,\) the only possible answer among the choices given is \(20\%.\)