Probability: Addition Rules and Complements

Probability: Addition Rules and Complements

Disjoint (Mutually Exclusive) Events

In probability theory, understanding the relationship between events is crucial for calculating probabilities accurately. Two events are disjoint (or mutually exclusive) if they have no outcomes in common. This means that the occurrence of one event excludes the possibility of the other occurring at the same time.

• Definition: Events E and F are disjoint if $E \cap F = \varnothing$.

• Venn Diagrams: Disjoint events are represented as non-overlapping circles within a rectangle (the sample space).

Example: Disjoint Events

Suppose we randomly select a chip from a bag labeled 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

• Let E = "choose a number less than or equal to 2" (i.e., 0, 1, 2).

• Let F = "choose a number greater than or equal to 8" (i.e., 8, 9).

These events are disjoint because they have no numbers in common.

• $P(E) = \frac{3}{10} = 0.3$

• $P(F) = \frac{2}{10} = 0.2$

• $P(E \text{ or } F) = P(E) + P(F) = 0.3 + 0.2 = 0.5$

Addition Rule for Disjoint Events

When two events are disjoint, the probability that either event occurs is the sum of their individual probabilities.

• Formula: $P(E \text{ or } F) = P(E) + P(F)$

• For multiple pairwise disjoint events: $P(E_1 \text{ or } E_2 \text{ or } \ldots) = P(E_1) + P(E_2) + \ldots$

Benford’s Law

Benford’s Law describes the frequency distribution of leading digits in many real-life sets of numerical data. The law states that lower digits (especially 1) occur as the first digit more frequently than higher digits.

• Discovered by Simon Newcomb (1881) and Frank Benford (1938).

• Probabilities for the first digit (from Benford’s Law):

Digit	1	2	3	4	5	6	7	8	9
Probability	0.30	0.18	0.12	0.10	0.08	0.07	0.06	0.05	0.04

• Example Questions:

  • Probability that X is 1 or 2: $0.30 + 0.18 = 0.48$

  • Probability that X is even: $0.18 + 0.10 + 0.07 + 0.05 = 0.40$ (for digits 2, 4, 6, 8)

  • Probability that X is either 1 or even: $0.30 + 0.18 + 0.10 + 0.07 + 0.05 = 0.70$

Probabilities for Non-Disjoint Events

When two events are not disjoint (i.e., they can occur together), simply adding their probabilities will double-count the outcomes they share. The general addition rule corrects for this by subtracting the probability of their intersection.

• Formula: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

Example: Drawing a card from a standard deck:

• Event A: The card is a Diamond ($P(A) = \frac{13}{52}$)

• Event B: The card is a Face Card ($P(B) = \frac{12}{52}$)

• Event A and B: The card is a Diamond and a Face Card ($P(A \text{ and } B) = \frac{3}{52}$)

• $P(A \text{ or } B) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}$

Compound Events: Unions and Intersections

Compound events involve combining two or more events using unions or intersections.

• Union (A ∪ B): The event that at least one of A or B occurs. $A \cup B = \{x : x \in A \text{ or } x \in B\}$

• Intersection (A ∩ B): The event that both A and B occur. $A \cap B = \{x : x \in A \text{ and } x \in B\}$

Mutually Exclusive Events: If $A \cap B = \varnothing$, then A and B are mutually exclusive (disjoint).

General Addition Rule: Proof

The general addition rule for any two events A and B is:

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

Short Proof:

• Let $n(S)$ be the number of outcomes in the sample space S.

• $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{n(A) + n(B) - n(A \cap B)}{n(S)} = P(A) + P(B) - P(A \cap B)$

Example: Cards

Suppose a single card is selected from a standard 52-card deck. Compute the probability of the event E = "drawing a king" or F = "drawing a diamond."

• $P(E) = \frac{4}{52}$ (4 kings)

• $P(F) = \frac{13}{52}$ (13 diamonds)

• $P(E \cap F) = \frac{1}{52}$ (king of diamonds)

• $P(E \cup F) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$

Example: Dice

Suppose a pair of dice are thrown. Let E = "the first die is a two" and F = "the sum of the dice is less than or equal to 5." Find $P(E \cup F)$.

• $P(E) = \frac{6}{36}$ (first die is 2, six possible outcomes)

• $P(F) = \frac{10}{36}$ (sum ≤ 5, ten possible outcomes)

• $P(E \cap F) = \frac{3}{36}$ (first die is 2 and sum ≤ 5, three possible outcomes)

• $P(E \cup F) = \frac{6}{36} + \frac{10}{36} - \frac{3}{36} = \frac{13}{36}$

Complement of an Event

The complement of an event E, denoted $E^c$, consists of all outcomes in the sample space S that are not in E.

• Formula: $P(E^c) = 1 - P(E)$

Complement Rule: Example

According to the National Gambling Impact Study Commission, 52% of Americans have played state lotteries. What is the probability that a randomly selected American has not played a state lottery?

• $P(\text{not played}) = 1 - P(\text{played}) = 1 - 0.52 = 0.48$

Therefore, the probability is 0.48.