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4.3: Addition Rules for Probabilities

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    58264

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    Learning Objectives
    • Understand the two addition rules for probability to calculate the chance of either one event or another occurring.
    • Apply the addition rule for mutually exclusive events by summing their probabilities.
    • Apply the addition rule for events that can occur together by adding their probabilities and subtracting the overlap to avoid double-counting.

    When computing the probability that at least one of several events will occur, addition rules can be applied to facilitate the computation of these probabilities. For example, if a card is selected from a standard deck, what is the chance that the card is a jack or a queen? This probability can be computed rapidly by computing the possibility of getting a jack and a queen separately and then adding the two probabilities together. This process can be done because the events are mutually exclusive. In other words, the events have no outcomes in common.

    Definition: \(\PageIndex{1}\)

    Mutually exclusive events (disjoint) are events with no outcomes in common.

    Two mutually exclusive events for jacks and kings.
    Figure \(\PageIndex{1}\): Two Mutually Exclusive Events
    Example \(\PageIndex{1}\)
    1. Roll a single die. The events are to get a 2 or a 5.
    2. Select a single card. The events are to get a jack or a queen.
    3. A room has 5 math majors, 12 chemistry majors, 6 biology majors, and 4 physics majors. A single person is selected, and no person is double-majoring. The person is a math major or a biology major.
    Solution

    None of the events has outcomes in common.

    Definition: \(\PageIndex{2}\)
    Addition Rule I

    For two events A and B that are mutually exclusive, the probability that A or B will occur is P(A or B) = P(A) + P(B).

    Example \(\PageIndex{2}\)

    Use the addition rule I to compute the following probabilities. Write the final answers as fractions or decimals rounded to three places.

    1. Roll a single die. What is the chance of getting a 2 or a 5?
    2. Select a single card. What is the chance the card is a jack or a queen?
    3. A room has 5 math majors, 12 chemistry majors, 6 biology majors, and 4 physics majors. A single person is selected, and no person is double-majoring. What is the chance the person is a math or biology major?
    Solutions
    1. P( 2 or 5) = \(\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3} = 0.333\)
    2. P(ace or king) = \(\dfrac{4}{52} + \dfrac{4}{52} = \dfrac{8}{52} = \dfrac{2}{13} = 0.154\)
    3. P(math major or biology major) = \(\dfrac{5}{27} + \dfrac{6}{27} = \dfrac{11}{27} = 0.407\)
    Definition: \(\PageIndex{3}\)

    Two events are not mutually exclusive if they can happen at the same time, meaning they have some outcomes in common.

    Two events that are not mutually exclusive.
    Figure \(\PageIndex{2}\): Two Not Mutually Exclusive Events

    When two or more events have outcomes in common, they are labeled non-mutually exclusive events. These types of events have one or more outcomes in common. Care must be taken not to over-count these outcomes. For example, suppose a card is selected from a deck. What is the chance that the card is a jack or a black card? If addition rule I is applied, this would result in the wrong answer. The reason is that two blackjacks are being over-counted. To correctly compute the probability, these extra two blackjacks must be subtracted from the total number of outcomes. This process is outlined in Addition Rule II.

    Definition: \(\PageIndex{4}\)
    Addition Rule II

    For two events A and B that are not mutually exclusive, the probability that A or B will occur is P(A or B) = P(A) + P(B) – P(A and B)

    Example \(\PageIndex{3}\)

    Use the addition rule II to compute the following probabilities. Write the final answers as fractions or decimals rounded to three place places.

    1. Select a single card. What is the chance the card is a jack or a black card?
    2. Select a single card. What is the chance the card is a king or a diamond?
    3. A room has 5 math majors, 12 chemistry majors, 6 biology majors, and 4 physics majors. In this group, 2 people are double-majoring in math and physics. A single person. What is the chance the person is a math or physics major?
    Solutions
    1. P( Jack or black) = \(\dfrac{4}{52} + \dfrac{26}{52} - \dfrac{2}{52} = \dfrac{28}{52} = \dfrac{7}{13} = 0.538\)
    2. P(king or diamond) = \(\dfrac{4}{52} + \dfrac{13}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13} = 0.308\)
    3. P(math major or physics major) = \(\dfrac{5}{27} + \dfrac{4}{27} - \dfrac{2}{27} = \dfrac{7}{27} = 0.259\)

    Sometimes data is presented in a tabular form. This type of table is called a contingency table, and the table is used to show a relationship between two data types. The data are listed in rows and columns. We can treat the contingency table as two sets where the column data types represent one set and the row data types represent another data type. To compute probabilities using a contingency table, the first step is to find the sum of each row and column. For example, students are categorized into those who don’t work, work part-time, and work full-time, along with the modality of the class they are taking for Math 165: regular, with support, hybrid, or online. The contingency table is displayed below for 120 randomly selected students.

    Contingency Table Involving Modality of Classes and Work Status
    Regular With Support Hybrid Online Total
    Does not work 15 14 5 3 37
    Works part-time 9 16 10 5 40
    Works full-time 7 8 13 15 43
    Total 31 38 28 23 120

    Table \(\PageIndex{1}\): Contingency Table Involving Modality of Classes and Work Status

    Example \(\PageIndex{4}\)

    Using the contingency table above, compute the probability for one randomly selected student. Write the final answers as fractions or decimals rounded to three places.

    1. The student works full-time or takes a class online.
    2. The student is part-time and takes a hybrid class.
    3. The student does not work or take a class with support.
    Solutions
    1. P( works full-time or takes an online class) = \(\dfrac{43}{120} + \dfrac{23}{120} - \dfrac{15}{120} = \dfrac{51}{120} = 0.425\)
    2. P( part-time and takes a hybrid class) = \(\dfrac{10}{120} = 0.083\)
    3. P( Does not work or takes a class with support) =\(\dfrac{37}{120} + \dfrac{38}{120} - \dfrac{14}{120} = \dfrac{61}{120} = 0.508\)

    Exercises

    1. A pet shop has the following puppy types for sale: 6 French Bulldogs, 4 Golden Retrievers, 3 Labrador Retrievers, 8 Poodles, and 2 Beagles. If a customer purchases one puppy randomly, compute the following probabilities. Write the final answers as fractions or decimals rounded to three place values.
      1. The puppy is a Golden Retriever.
      2. The puppy is a French Bulldog or a Labrador Retriever.
      3. The puppy is a Poodle or a Beagle, or a Golden Retriever.

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    1. Two dice are tossed one time. Compute the following probabilities. Write the final answers as fractions or decimals rounded to three place values.
      1. Get a sum of 5
      2. Get a sum of 4 or 8.
      3. Get a sum greater than 6 or even.

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    1. An award ceremony has 200 student-athletes in attendance. A prestigious award is to be given to one of these students. The students consist of 55 football players, 28 baseball players, 25 basketball players, 19 soccer players, and other athletes. Moreover, 8 athletes play both football and baseball. If one student is selected to win the award, compute the following probabilities. Write the final answers as fractions or decimals rounded to three place values.
      1. The student is a basketball player or soccer player.
      2. The student is a football player or baseball player.
      3. The student is neither a basketball player nor a football player.

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    1. A card is selected from a standard deck of 52 cards. Compute the following probabilities. Write the final answers as fractions or decimals rounded to three place values.
      1. The card is an ace or king.
      2. The card is a queen or a spade.
      3. The card is a diamond and a jack.
      4. The card is a heart or a 5 or a red card.

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    1. A survey was conducted of a group of students to identify their gender and eye color. The data is represented in the contingency table below.
    Contingency Table Involving Eye Color and Gender
    Brown Black Grey Green Blue Total
    Female 22 9 3 5 10 49
    Male 27 11 4 3 6 51
    Total 49 20 7 8 16 100

    Table \(\PageIndex{2}\): Contingency Table Involving Gender and Eye Color

    If a student is selected at random, compute the following probabilities. Write the final answers as fractions or decimals rounded to three place values.

    1. The student has grey eyes or black eyes.
    2. The student is a male and has blue eyes.
    3. The student is a female or has grey eyes.
    4. The student is a male or has green eyes.

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    Answers

    If you are an instructor and want the solutions to all the exercise questions for each section, please email Toros Berberyan.

    This page titled 4.3: Addition Rules for Probabilities is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Toros Berberyan, Tracy Nguyen, and Alfie Swan.