Probabilistic Principle of Inclusion and Exclusion

The probabilistic principle of inclusion and exclusion (PPIE for short) is a method used to calculate the probability of unions of events. For two events, the PPIE is equivalent to the probability rule of sum:

Let $A$ and $B$ be events. The probability of the union of these events is:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

The PPIE is closely related to the principle of inclusion and exclusion in set theory. The formulas for probabilities of unions of events are very similar to the formulas for the size of unions of sets.

The PPIE for two events is equivalent to the probability rule of sum.

A card is drawn from a standard deck of cards. What is the probability that the card drawn is a queen or a heart?

---

Let $A$ be the event that the card is a queen, and let $B$ be the event that the card is a heart. Then $P(A \cup B) = P(A) + P(B) - P(A \cap B).$ Since there are 13 different ranks of cards in the deck, $P(A) = \frac{1}{13} ,$ and since there are 4 suits in the deck, $P(B) = \frac{1}{4}.$ There is only one card that is both a queen and a heart, so $P(A \cap B) = \frac{1}{52}.$ Therefore,

$$P(A \cup B) = \frac{1}{4} + \frac{1}{13} - \frac{1}{52} = \frac{16}{52} = \boxed{\dfrac{4}{13}}.$$

When events are independent, the rule of product can be used to find the probability of an intersection of events. Then, the rule of sum can be used to find the probability of the union of those events.

A fair 6-sided die and a fair 8-sided die are rolled. What is the probability that one of the dice rolls is a 6?

---

Let $A$ be the event that the 6-sided die is 6. $P(A)=\dfrac{1}{6}$.

Let $B$ be the event that the 8-sided die is 6. $P(B)=\dfrac{1}{8}$.

The events are independent, so by the rule of product, $P(A\cap B)=\dfrac{1}{6}\times\dfrac{1}{8}=\dfrac{1}{48}$.

By the rule of sum, $P(A\cup B)=\dfrac{1}{6}+\dfrac{1}{8}-\dfrac{1}{48}=\dfrac{13}{48}$

The probability that either dice roll is 6 is $\boxed{\dfrac{13}{48}}$.

When the three events are independent, the probability of the union of those events can be found using complement probabilities and the rule of product:

Given three independent events $A$, $B$, and $C$, the probability of the union of these events is:

$P(A\cup B\cup C)=1-P(A^c)P(B^c)P(C^c)$

A fair 6-sided die is rolled three times. Let $A$ be the event that the first roll is 1, Let $B$ be the event that the second roll is 3, and let $C$ be the event that the third roll is 3. What is $P(A\cup B\cup C)$?

---

P(A)=$\dfrac{1}{6}$, and so $P(A^c)=\dfrac{5}{6}$.

P(B)=$\dfrac{1}{6}$, and so $P(B^c)=\dfrac{5}{6}$.

P(C)=$\dfrac{1}{6}$, and so $P(C^c)=\dfrac{5}{6}$.

These events are independent, so the above formula can be used:

$P(A\cup B\cup C)=1-\left(\dfrac{5}{6}\right)^3=\boxed{\dfrac{91}{216}}$

When events are dependent, then the probability of each intersection of events must be either known or calculated. Then, the probability of the union of events can be calculated using the following formula:

Given three dependent events $A$, $B$, and $C$, the probability of the union of these events is:

$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B \cap C)+P(A\cap B\cap C)$

When events are independent, the probability of a union of those events can be found using complement probabilities and the rule of product:

If $\{A_1, \dots ,A_n\}$ is a set of mutually independent events, then the probability of the union of those events is:

$$\large P\left(\bigcup\limits_{i=1}^{n}{A_i}\right)=1-\prod\limits_{i=1}^{n}{P(A_i^c)}$$

This identity is a direct result of De Morgan's Laws.

When events are dependent, the general form of the PPIE for any number of events involves adding or subtracting intersections of events:

If $\{A_1, \dots A_n\}$ is a set of dependent events, then the probability of the union of those events is:

$$\large P\left(\bigcup\limits_{i=1}^{n}{A_i}\right)=\sum\limits_{k=1}^{n}(-1)^{k+1}\left(\sum\limits_{1\le i_1< \dots < i_k\le n}P\left(A_{i_1}\cap \dots \cap A_{i_k}\right) \right)$$

Let $A$, $B$, $C$, and $D$ be pairwise dependent events. What is the formula for the probability of the union of these events?

---

By the general formula above, this is:

$\begin{array}{ll} P(A\cup B\cup C\cup D)= & P(A)+P(B)+P(C)+P(D) \\ & -\ P(A\cap B)-P(A\cap C)-P(A\cap D)-P(B\cap C)-P(B\cap D)-P(C\cap D) \\ & +\ P(A\cap B\cap C)+P(A\cap B\cap D)+P(A\cap C\cap D)+P(B\cap C\cap D) \\ & -\ P(A\cap B\cap C\cap D) \end{array}$