Addition Rule for Probabilities
The addition rule of probabilities is used to solve probability questions and problems. Several examples are presented along with their detailed solutions.
The minimum background needed is the concept of sample space and event. Reviewing basic probability questions could be helpful.
In what follows, \( n(S) \) is the number of elements in the sample space \( S \) and \( n(E) \) is the number of elements in the event \( E \).
Addition Rule Explanations
The best way to explain the addition rule is to solve the following example using two different methods.
Example 1
A fair die is rolled one time. Find the probability of getting an odd number or a number less than or equal to \( 3 \).
Solution
Two methods are suggested.
Method 1: Use the sample space
The sample space \( S \) is:
\[ S = \{1,2,3,4,5,6\} \]
The number of elements \( n(S) = 6 \).
Let \( E \) be the event "getting an odd number or a number less than or equal to \( 3 \)". The outcomes in \( E \) are:
\[ E = \{1,2,3,5\} \]
So, \( n(E) = 4 \).
The probability is:
\[ P(E) = \frac{n(E)}{n(S)} = \frac{4}{6} = \frac{2}{3} \]
Method 2: Use addition formula
Let \( A \): getting an odd number, \( B \): getting a number ≤ 3.
\[ A = \{1,3,5\}, \quad B = \{1,2,3\} \]
Then \( E = A \cup B = \{1,2,3,5\} \).
We have:
\[ n(A)=3, \; n(B)=3, \; n(A \cap B)=2 \]
\[ n(E) = n(A) + n(B) - n(A \cap B) = 3+3-2=4 \]
The probability:
\[ P(E) = \frac{n(E)}{n(S)} = \frac{n(A)}{n(S)} + \frac{n(B)}{n(S)} - \frac{n(A \cap B)}{n(S)} = P(A) + P(B) - P(A \cap B) \]
Hence the general addition rule:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
or
\[ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \]
Here:
\[ P(A) = \frac{3}{6} = \frac{1}{2}, \quad P(B) = \frac{3}{6} = \frac{1}{2}, \quad P(A \cap B) = \frac{2}{6} = \frac{1}{3} \]
\[ P(A \cup B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{3} = \frac{2}{3} \]
NOTE: If events \( A \) and \( B \) are mutually exclusive (cannot happen together), then \( P(A \cap B) = 0 \) and the rule simplifies to:
\[ P(A \cup B) = P(A) + P(B) \]
or
\[ P(A \text{ or } B) = P(A) + P(B) \]
Examples on the Use of the Addition Rule
More examples and questions on using the addition rule.
Example 2
A fair die is rolled once. Find the probability of getting a "\( 1 \)" or a "\( 5 \)".
Solution:
Let \( A \): getting a "1", \( B \): getting a "5". These are mutually exclusive.
\[ P(A) = \frac{1}{6}, \quad P(B) = \frac{1}{6}, \quad P(A \cap B) = 0 \]
\[ P(A \cup B) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \]
Example 3
A box has 3 red, 2 green, and 5 blue balls. One ball is drawn randomly. Find the probability it is green or blue.
Solution:
Let \( A \): ball is green, \( B \): ball is blue. Mutually exclusive.
\[ P(A) = \frac{2}{10} = \frac{1}{5}, \quad P(B) = \frac{5}{10} = \frac{1}{2} \]
\[ P(A \cup B) = \frac{1}{5} + \frac{1}{2} = \frac{7}{10} \]
Example 4
In a school of 100 students, 50 play football, 20 play basketball, and 10 play both. A student is selected at random. What is the probability that he/she plays at least one sport?
Solution:
Let \( A \): plays football, \( B \): plays basketball.
\[ P(A) = \frac{50}{100} = \frac{1}{2}, \quad P(B) = \frac{20}{100} = \frac{1}{5}, \quad P(A \cap B) = \frac{10}{100} = \frac{1}{10} \]
\[ P(A \cup B) = \frac{1}{2} + \frac{1}{5} - \frac{1}{10} = \frac{3}{5} \]
Example 5
A single card is drawn from a deck. Find the probability of:
- a "2" or a "5"
- An "8" or a "heart"
- A "Queen" or a "red card"
Solution:
- Let \( A \): "2", \( B \): "5". Mutually exclusive.
\[ P(A) = \frac{4}{52} = \frac{1}{13}, \quad P(B) = \frac{4}{52} = \frac{1}{13} \]
\[ P(A \cup B) = \frac{1}{13} + \frac{1}{13} = \frac{2}{13} \]
- Let \( C \): "8", \( D \): "heart". Not mutually exclusive.
\[ P(C) = \frac{4}{52} = \frac{1}{13}, \quad P(D) = \frac{13}{52} = \frac{1}{4}, \quad P(C \cap D) = \frac{1}{52} \]
\[ P(C \cup D) = \frac{1}{13} + \frac{1}{4} - \frac{1}{52} = \frac{4}{13} \]
- Let \( E \): "Queen", \( F \): "red card". Not mutually exclusive.
\[ P(E) = \frac{4}{52} = \frac{1}{13}, \quad P(F) = \frac{26}{52} = \frac{1}{2}, \quad P(E \cap F) = \frac{2}{52} = \frac{1}{26} \]
\[ P(E \cup F) = \frac{1}{13} + \frac{1}{2} - \frac{1}{26} = \frac{7}{13} \]
Example 6
A car dealer has cars as in the table. A car is selected at random. Find the probability that it is:
- a black or a white car?
- blue car or a coupe?
- a black car or an SUV?
| SUV | Sport Car | Van | Coupe | Total |
|---|
| Black | 35 | 10 | 25 | 15 | 85 |
| White | 10 | 15 | 20 | 5 | 50 |
| Blue | 15 | 15 | 5 | 30 | 65 |
| Total | 60 | 40 | 50 | 50 | 200 |
Solution:
- Let \( A \): black, \( B \): white. Mutually exclusive.
\[ P(A) = \frac{85}{200} = \frac{17}{40}, \quad P(B) = \frac{50}{200} = \frac{1}{4} \]
\[ P(A \cup B) = \frac{17}{40} + \frac{1}{4} = \frac{27}{40} \]
- Let \( C \): blue, \( D \): coupe. Not mutually exclusive.
\[ P(C) = \frac{65}{200} = \frac{13}{40}, \quad P(D) = \frac{50}{200} = \frac{1}{4}, \quad P(C \cap D) = \frac{30}{200} = \frac{3}{20} \]
\[ P(C \cup D) = \frac{13}{40} + \frac{1}{4} - \frac{3}{20} = \frac{17}{40} \]
- Let \( E \): black, \( F \): SUV. Not mutually exclusive.
\[ P(E) = \frac{85}{200} = \frac{17}{40}, \quad P(F) = \frac{60}{200} = \frac{3}{10}, \quad P(E \cap F) = \frac{35}{200} = \frac{7}{40} \]
\[ P(E \cup F) = \frac{17}{40} + \frac{3}{10} - \frac{7}{40} = \frac{11}{20} \]
Example 7
The table shows hours students spend on homework weekly. A student is selected at random. Find the probability that the student spends at most 4 hours or at least 11 hours.
| Time (hours) | Number of students |
|---|
| 0-2 | 5 |
| 3-4 | 20 |
| 5-7 | 35 |
| 8-10 | 50 |
| 11-12 | 60 |
| 13+ | 30 |
Solution:
Let \( A \): at most 4 hours, \( B \): at least 11 hours. Mutually exclusive.
\[ P(A) = \frac{5+20}{200} = \frac{25}{200} = \frac{1}{8}, \quad P(B) = \frac{60+30}{200} = \frac{90}{200} = \frac{9}{20} \]
\[ P(A \cup B) = \frac{1}{8} + \frac{9}{20} = \frac{5}{40} + \frac{18}{40} = \frac{23}{40} \]
Example 8
An insurance company's customers have home insurance (80%), auto insurance (60%), or both. One person is selected at random. What is the probability they have both insurances?
Solution:
Let \( A \): home insurance, \( B \): auto insurance.
\[ P(A \cup B) = 1 \quad (\text{all customers have at least one}) \]
\[ P(A) = 0.8, \quad P(B) = 0.6 \]
\[ 1 = 0.8 + 0.6 - P(A \cap B) \]
\[ P(A \cap B) = 1.4 - 1 = 0.4 \]
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