Chapter 4: Probability
Section 2: The Addition Rules of Probability
Concepts of Mutually Exclusive Events
Definition of Mutually Exclusive Events (Disjoint Events):
Events that CANNOT occur at the same time. This means that they have no common outcomes.
If one event occurs, the other cannot occur simultaneously.
The intersection of two mutually exclusive events is empty, meaning the probability of both occurring together is zero: P(A \text{ and } B) = 0.
Definition of Non-Mutually Exclusive Events (Overlapping Events):
Events that CAN occur at the same time. They share one or more common outcomes.
If one event occurs, the other might also occur.
The intersection of two non-mutually exclusive events is not empty, meaning the probability of both occurring together is greater than zero: P(A \text{ and } B) \neq 0.
Examples of Mutually Exclusive Events
Example 4-15: Determining Mutually Exclusive Events
a. Randomly selecting a female student and randomly selecting a student who is a junior
Solution: Not mutually exclusive. A student can indeed be both female and a junior, sharing common characteristics. For example, a female student in her junior year satisfies both conditions.
b. Randomly selecting a person with type A blood and randomly selecting a person with type O blood
Solution: Mutually exclusive. A person's blood type is unique; they cannot simultaneously have type A blood and type O blood. These events cannot co-occur.
c. Rolling a die and getting an odd number and rolling a die and getting a number less than 3
Solution: Not mutually exclusive. The number 1 is an outcome that is both an odd number (1, 3, 5) and less than 3 (1, 2). Since a single roll can result in 1, which satisfies both conditions, they are not mutually exclusive.
d. Randomly selecting a person who is under 21 years of age and randomly selecting a person who is over 30 years of age
Solution: Mutually exclusive. A person's age cannot simultaneously be under 21 years and over 30 years. These age ranges are distinct and do not overlap.
Example 4-16: Drawing a Card
a. Getting a king and getting a diamond
Solution: Not mutually exclusive. The King of Diamonds is a single card that is both a king and a diamond. This overlap means they can occur at the same time.
b. Getting a 4 and getting a king
Solution: Mutually exclusive. When drawing a single card, it is impossible for that card to be both a 4 and a king. There is no card that satisfies both conditions simultaneously.
c. Getting a face card and getting a club
Solution: Not mutually exclusive. There are three face cards (Jack, Queen, King) that are also clubs (Jack of Clubs, Queen of Clubs, King of Clubs). These cards represent outcomes where both events occur.
d. Getting a face card and getting a 10
Solution: Mutually exclusive. A face card (Jack, Queen, King) by definition is not a 10. Therefore, a single card cannot be both a face card and a 10.
Addition Rule for Mutually Exclusive Events
Addition Rule 1 (For "OR" events that are mutually exclusive):
When two events A and B are mutually exclusive, the probability that A OR B will occur (meaning at least one of them happens) is found by simply summing their individual probabilities. This is because there is no overlap to consider.
The formula is expressed as:
P(A \text{ or } B) = P(A) + P(B)This rule applies specifically when P(A \text{ and } B) = 0.
Venn Diagram for Rule 1 for Mutually Exclusive Events:
A Venn diagram for mutually exclusive events shows two distinct circles that do not intersect each other. This visual separation clearly illustrates that there are no common outcomes between event A and event B.
The total probability of P(A \text{ or } B) is the sum of the areas of the two separate circles, representing P(A) + P(B).
The total probability of the sample space is P(S) = 1.
Example 4-17: Coffee Shop Selection
Scenario: A city has 9 coffee shops. These include 3 Starbucks, 2 Caribou Coffees, and 4 Crazy Mocho Coffees.
Question: Find the probability of randomly selecting either a Starbucks or a Crazy Mocho Coffee.
Solution:
First, identify the total possible outcomes: Total coffee shops = 9.
Calculate the individual probabilities:
P(\text{Starbucks}) = \frac{3}{9} = \frac{1}{3}
P(\text{Crazy Mocho}) = \frac{4}{9}
Determine if the events are mutually exclusive: Selecting a Starbucks and selecting a Crazy Mocho are mutually exclusive, as a single coffee shop cannot be both at the same time.
Apply Addition Rule 1:
P(\text{Starbucks or Crazy Mocho}) = P(\text{Starbucks}) + P(\text{Crazy Mocho})
P(\text{Starbucks or Crazy Mocho}) = \frac{1}{3} + \frac{4}{9} = \frac{3}{9} + \frac{4}{9} = \frac{7}{9} \text{ or approximately } 0.778
Example 4-19: Favorite Ice Cream
Survey Results: A survey indicates that 8% of respondents favor cookies and cream ice cream, and 6% favor mint chocolate chip ice cream.
Question: Find the probability that a randomly selected person likes either cookies and cream or mint chocolate chip.
Solution:
Identify individual probabilities as decimals:
P(\text{cookies and cream}) = 0.08
P(\text{mint chocolate chip}) = 0.06
Determine if the events are mutually exclusive: Assuming that a person can only state one favorite flavor, these events are mutually exclusive. A person cannot simultaneously favor both as their single favorite.
Apply Addition Rule 1:
P(\text{cookies and cream or mint chocolate chip}) = P(\text{cookies and cream}) + P(\text{mint chocolate chip})
P(\text{cookies and cream or mint chocolate chip}) = 0.08 + 0.06 = 0.14 \text{ or } 14\%.
Addition Rule for Three Mutually Exclusive Events
For three mutually exclusive events A, B, and C: The principle extends directly. If no two events can occur simultaneously, their probabilities are simply added.
P(A \text{ or } B \text{ or } C) = P(A) + P(B) + P(C)
Addition Rule for Non-Mutually Exclusive Events
Addition Rule 2 (General Addition Rule for "OR" events):
When two events A and B are not mutually exclusive (meaning they have common outcomes), the probability that A OR B will occur is given by the sum of their individual probabilities minus the probability of both occurring simultaneously. The subtraction term corrects for the 'double counting' of the overlap when P(A) and P(B) are added.
The formula is expressed as:
P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
Note: Addition Rule 2 is a more general rule and can also be applied to mutually exclusive events. If A and B are mutually exclusive, then P(A \text{ and } B) = 0, which simplifies Addition Rule 2 back to Addition Rule 1: P(A \text{ or } B) = P(A) + P(B) - 0 = P(A) + P(B).
Venn Diagram for Rule 2 for Non-Mutually Exclusive Events:
A Venn diagram for non-mutually exclusive events shows two circles that overlap. The overlapping region represents the outcomes where both event A and event B occur, which is P(A \text{ and } B).
When you add P(A) and P(B), the overlapping region is counted twice. Therefore, to get the probability of P(A \text{ or } B) (the entire area covered by both circles), you must subtract P(A \text{ and } B) once to correct for this double-counting.
The total probability of the sample space is P(S) = 1.
Example 4-20: Drawing a Card
Scenario: A single card is drawn at random from a standard deck of 52 cards.
Question: Find the probability that the card drawn is either an ace or a black card.
Solution:
Identify the total possible outcomes: 52 cards in a standard deck.
Calculate individual probabilities:
Total aces = 4 (Ace of Spades, Ace of Clubs, Ace of Hearts, Ace of Diamonds)
P(\text{ace}) = \frac{4}{52}
Total black cards = 26 (13 spades + 13 clubs)
P(\text{black}) = \frac{26}{52}
Identify the overlap: Are these events non-mutually exclusive? Yes, there are cards that are both an ace and a black card.
Black aces = 2 (Ace of Spades + Ace of Clubs)
P(\text{black aces}) = P(\text{ace and black card}) = \frac{2}{52}
Apply Addition Rule 2:
P(\text{ace or black card}) = P(\text{ace}) + P(\text{black}) - P(\text{ace and black card})
P(\text{ace or black card}) = \frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13} \text{ or approximately } 0.538
Example 4-21: Selecting a Medical Staff Person
Scenario: In a hospital, there are 8 nurses and 5 physicians, making a total of 13 staff members. Among them, 7 nurses are female, and 3 physicians are female.
Question: Find the probability that a randomly selected person from the staff is either a nurse or a male.
Solution:
First, determine the total composition of the staff:
Total Staff: 8 Nurses + 5 Physicians = 13 Staff
Females: 7 Nurses (F) + 3 Physicians (F) = 10 Females
Males: (8 Total Nurses - 7 Female Nurses) = 1 Male Nurse; (5 Total Physicians - 3 Female Physicians) = 2 Male Physicians; Total Males = 1 Male Nurse + 2 Male Physicians = 3 Males
Calculate individual probabilities:
P(\text{nurse}) = \frac{8}{13}
P(\text{male}) = \frac{3}{13}
Identify the overlap: Is there a person who is both a nurse and a male? Yes, there is 1 male nurse.
P(\text{male and nurse}) = \frac{1}{13}
Apply Addition Rule 2:
P(\text{nurse or male}) = P(\text{nurse}) + P(\text{male}) - P(\text{male and nurse})
P(\text{nurse or male}) = \frac{8}{13} + \frac{3}{13} - \frac{1}{13} = \frac{10}{13} \text{ or approximately } 0.769
Example 4-22: Driving While Intoxicated
Scenario: On New Year's Eve, historical data provides the following probabilities:
The probability of a person driving while intoxicated is P(\text{intoxicated}) = 0.32.
The probability of a person having a driving accident is P(\text{accident}) = 0.09.
The probability of a person driving while intoxicated and having a driving accident is P(\text{intoxicated and accident}) = 0.06.
Question: What is the probability of a person driving while intoxicated or having a driving accident?
Solution:
The events are clearly non-mutually exclusive because there's a non-zero probability of both happening (P(\text{intoxicated and accident}) = 0.06).
Apply Addition Rule 2:
P(\text{intoxicated or accident}) = P(\text{intoxicated}) + P(\text{accident}) - P(\text{intoxicated and accident})
P(\text{intoxicated or accident}) = 0.32 + 0.09 - 0.06 = 0.35
Addition Rule for Three Non-Mutually Exclusive Events
For three events A, B, and C that are not mutually exclusive: This rule, also known as the Principle of Inclusion-Exclusion for three sets, accounts for all individual probabilities, subtracts the probabilities of all pairwise intersections (double-counted regions), and then adds back the probability of the triple intersection (which was subtracted too many times).
The formula is:
P(A \text{ or } B \text{ or } C) = P(A) + P(B) + P(C) - P(A \text{ and } B) - P(A \text{ and } C) - P(B \text{ and } C) + P(A \text{ and } B \text{ and } C)