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 to understand the examples, is the concept of sample space of an experiment and the event of interest. Also 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 to Example 1
Two methods are suggested.

Method 1: Use the sample space
The sample space S, which is the set of all possible outcomes, of the experiment in rolling a die is given by
\( \quad \quad \quad S = \{1,2,3,4,5,6\} \)
The number of elements \( n(S) \) in the set \( S \) is given by \( n(S) = 6 \)
Let E be the event "getting and odd number or a number less than or equal to \( 3 \)". Check each element of the sample space\( S \) to see if it is odd or less than or equal to \( 3 \) to end up with all outcomes belonging to the set E given by
\( \quad \quad \quad E = \{1,2,3,5\} \)
The number of elements \( n(E) \) in the set \( E \) is given by
\( \quad \quad \quad n(E) = 4 \)
Let \( P(E) \) be the probability that event E, defined above, occurs. We now use the formula of the classical probability to find \( P(E) \) as:
\( \quad \quad \quad P(E) = \dfrac{n(E)}{n(S)} = \dfrac{4}{6} = \dfrac{2}{3} \)


Method 2: Use addition formula
E is the event "getting an odd number or a number less than or equal to \( 3 \)" is in fact the union of two events: event \( A \) corresponding to "getting an odd number" and event \( B \) corresponding to "getting a number less than or equal to \( 3 \)".
\( \quad \quad \quad A = \{1,3,5\} \)
\( \quad \quad \quad B = \{1,2,3\} \)
such that
\( \quad \quad \quad E = A \cup B = \{1,3,5\} \cup \{1,2,3\} = \{1,2,3,5\} \)
The Venn diagrams below show set \( A \) and set \( B \) and their union \( A \cup B \). Note also the intersection of \( A \) and \( B \) has two elements: \( A \cap B = \{1,3\} \)



Let \( n(E), n(A) \), \( n(B) \) and \( n(A \cap B) \) be the numbers of elements in the sets \( E \), \( A \), \( B \) and \( (A \cap B) \) respectively.
We know from above that \( \quad \quad \quad n(E) = 4 \) , \( n(A) = 3 \) , \( n(B) = 3 \) and \( n(A \cap B) = 2 \)
We can write: \( n(E) = n(A) + n(B) - n(A \cap B) = 3 + 3 - 2 = 4 \)
The reason we subtracted \( n(A \cap B) \) in the expression of \(n(E) \) above, is because \( n(A \cap B) \) is counted twice: once in \( n(A) \) and once in \( n(B) \)
The probability \( P(E) \) of event \( E = A \cup B \) is given by
\( \quad \quad \quad P(E) = \dfrac{n(E)}{n(S)} = \dfrac{n(A)+ n(B) - n(A \cap B) }{n(S)} = \dfrac{n(A)}{ n(S)} + \dfrac{n(B)}{n(S)} - \dfrac{n(A \cap B)}{n(S)} = P(A) + P(B) - P(A \cap B) \)
Hence the general rule of addition in probabilities is given by
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] or \[ P(A \; \text{or} \; B) = P(A) + P(B) - P(A \; \text{and} \; B) \]
where \( p(A) \) is the probability of \( A \) happening, \( P(B) \) is the probability of \( B \) happening, and \( P(A \cap B) \) is the probability of both \( A \) and \( B \) happening at the same time.
\( \quad \quad \quad p(A) = \dfrac{n(A)}{n(S)} = \dfrac{3}{6} = \dfrac{1}{2}\)
\( \quad \quad \quad P(B) = \dfrac{n(B)}{n(S)} = \dfrac{3}{6} = \dfrac{1}{2}\)
\( \quad \quad \quad P(A \cap B) = \dfrac{n(A \cap B)}{n(S)} = \dfrac{2}{6} = \dfrac{1}{3} \)
Use the addition rule above
\( \quad \quad \quad P(A \cup B) = P(A) + P(B) - P(A \cap B) = 1/2 + 1/2 - 1/3 = 2/3\)
NOTE: If events \( A \) and \( B \) are mutually exclusive, meaning that they cannot happen at the same time, the intersection \( A \cap B = \phi \), empty set, and therefore \( P(A \cap B) = 0 \) which simplifies the addition rule 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

Example 2
A fair die is rolled one time, find the probability of getting a "\( 1 \)" or a "\( 5 \) ".

Solution to Example 2
Let event \( A \): getting a "\( 1 \)" and event \( B \): getting a "\( 5 \) ". We are then asked to find \( P(A \cup B) \) which given by
\( \quad \quad \quad P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
These two events are mutually exclusive because you cannot get "\( 1 \)" and a "\( 5 \) " at the same time. Hence
\( \quad \quad \quad P(A \cap B) = 0 \)
and
\( \quad \quad \quad P(A \cup B) = P(A) + P(B) \)
The probability of getting a \( 1 \) (event A) when rolling a die is
\( \quad \quad \quad P(A) = \dfrac {1}{6} \)
The probability of getting a \( 5 \) (event B) when rolling a die is
\( \quad \quad \quad P(B) = \dfrac {1}{6} \)
\( \quad \quad \quad P(A \cup B) = P(A) + P(B) = \dfrac{1}{6}+\dfrac{1}{6} = \dfrac{1}{3} \)

Example 3
A box contains 3 red balls, 2 green balls and 5 blue balls. A ball is drawn at random from the box. Find the probability that the ball is either green or blue.

Solution to Example 3
Let event \( A \): the ball is green and event \( B \): the ball is blue. We are then asked to find \( P(A \cup B) \) which is given by ".
\( \quad \quad \quad P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
The probability of getting a green (event A) is calculated as follows
There is a total of 10 balls and 2 are green; hence
\( \quad \quad \quad P(A) = \dfrac {2}{10} = 1/5 \)
The probability of getting a blue (event B) is calculated as follows
There is a total of 10 balls and 5 are blue; hence
\( \quad \quad \quad P(B) = \dfrac {5}{10} = 1/2 \)
These two events are mutually exclusive: we cannot get a green and a blue ball at the same time. Hence
\( \quad \quad \quad P(A \cap B) = 0 \)
and
\( \quad \quad \quad P(A \cup B) = P(A) + P(B) = 1/5 + 1/2 = 7/10 \)

Example 4
In a school of 100 students, 50 play football, 20 play basketball and 10 play both football and basketball. If a student is selected at random, what is the probability that he/she plays at least one of the two sports?

Solution to Example 4
Let event \( A \): the student plays football , Let event \( B \): the student plays basketball
We are looking for the probability that the selected student plays football, basketball or both written as
\( \quad \quad \quad P( A \cup B) = P(A) + P(B) - P(A \cap B) \)
If 50 out of 100 students play football, then
\( \quad \quad \quad P(A) = \dfrac{50}{100} = 1/2 \)
If \( 20 \) out of \( 100 \) play basketball, then
\( \quad \quad \quad P(B) = \dfrac{20}{100} = 1/5 \)
If 10 play both, then
\( \quad \quad \quad P(A \cap B) = \dfrac{10}{100} = 1/10 \)
and the probability we are calculating is given by
\( \quad \quad \quad P( A \cup B) = P(A) + P(B) - P(A \cap B) = 1/2 + 1/5 - 1/10 = 3/5 \)

Example 5
A single card is drawn from a deck. Find the probability of selecting the following.
a) a "2" or a "5"
b) An "8" or a "heart"
c) A "Queen" or a "red card"

Solution to Example 5

a)
Let event \( A \): selecting a "2" and event \( B \): selecting a "5".
In a deck of 52 cards, there 4 "2"s and 4 "5"s.
Events \( A \) and \( B \) are mutually exclusive you cannot get a "2" and "5" at the same time if you draw one card. Hence
\( \quad \quad \quad P(A \cap B) = 0 \)
\( \quad \quad \quad P(A) = 4/52 = 1/3 \)
\( \quad \quad \quad P(B) = 4/52 = 1/3 \)
\( \quad \quad \quad P( A \cup B) = P(A) + P(B) = 1/13 + 1/13 = 2/13 \)
b)
Let event \( C \): selecting an "8" and event \( D \): selecting a "heart".
In a deck of 52 cards, there 4 "8"s ; hence
\( \quad \quad \quad P(C) = 4/52 = 1/13 \)
and 13 "hearts"; hence
\( \quad \quad \quad P(D) = 13/52 = 1/4 \)
Events \( C \) and \( D \) are NOT mutually exclusive; you can get a "8" and a "heart" at the same time if you draw an "8" of "hearts". Hence
\( \quad \quad \quad P(C \cap D) = 1/52 \)
\( \quad \quad \quad P( C \cup D) = P(C) + P(D) - P( C \cup D) = 1/13 + 1/4 - 1/52 = 4/13 \)
c)
Let event \( E \): selecting a "Queen" and event \( F \): selecting a "red card".
In a deck of 52 cards, there 4 "Queens"s ; hence
\( \quad \quad \quad P(E) = 4/52 = 1/13 \)
and 26 "red cards"; hence
\( \quad \quad \quad P(F) = 26/52 = 1/2 \)
Events \( C \) and \( D \) are NOT mutually exclusive; you can get a "Queen" of "Hearts" which a "red card " and you can also get "Queen" of "Diamonds" which a "red card " . Hence
\( \quad \quad \quad P(E \cap F) = 2/52 = 1/26 \)
\( \quad \quad \quad P( E \cup F) = P(E) + P(F) - P( E \cup F) = 1/13 + 1/2 - 2/52 = 7/13 \)

Example 6
A car dealer has the cars listed in the table below classified by type and color. If a car is selected at random, what is the probability that it is
a) a black or a white car?
b) blue car or a coupe?
c) a black car or an SUV?

Example 7
The table below shows the number of hours students spend on their homework every week.

Example 8
The customers of an insurance company have at least one of two insurances: home, auto or both. 80% of the customers have home insurance and 60% have auto insurance. If we select one person at random, what is the probability that this person has both insurances with the company?


Solution to Example 8
Let event \( A \): have home insurance and event \( B \) : have auto insurance.
\( \quad \quad \quad P( A \cup B ) = P(A) + P(B) - P(A \cap B) \)
We need to find \( P(A \cap B) \)
100% of the customers have either one or both insurances; hence
\( \quad \quad \quad P( A \cup B ) = 100\% \)
80% of customers have home insurance, hence
\( \quad \quad \quad P(A) = 80\% \)
60% of customers have auto insurance, hence
\( \quad \quad \quad P(B) = 60\% \)
\( \quad \quad \quad 100\% = 80\% + 60\% - P(A \cap B) \)
\( \quad \quad \quad P(A \cap B) = 140\% - 100\% = 40\% = 0.4 \)

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