# 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 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 ExplanationsThe best way to explain the addition rule is to solve the following example using two different methods.
Example 1
## Examples on the Use of the Addition RuleWe now present more examples and questions on how the addition rule is used to solve probability questions.NOTE: Several of the questions below may be solved by other methods which may be faster, but here we use the addition rule when solving these examples in order to learn how to use this rule.
Example 2
Example 3
Example 4
Example 5
Example 6
Solution to Example 6a) Let event \( A \): a black car is selected and event \( B \): a white car is selected. \( \quad \quad \quad P(A \cup B) = P(A) + P(B) - P(A \cap B) \) A car can have one color only and therefore events \( A \) and \( B \) are mutually exclusive; hence \( \quad \quad \quad P(A \cap B) = 0 \) There are 60 Suv's, 40 Sport cars, 50 vans and 50 coupe which gives a total of 200 cars. There is a total of 85 black cars and a total of 50 white cars; hence \( \quad \quad \quad P(A) = 85 / 200 = 17/40 \) \( \quad \quad \quad P(A) = 50 / 200 = 1/4 \) \( \quad \quad \quad P(A \cup B) = P(A) + P(B) - P(A \cap B) = 17/40 + 1/4 - 0 = 27/40\) b) Let event \( C \): a blue car is selected and event \( D \): a coupe is selected. We need to find the probability \( \quad \quad \quad P(C \cup D) = P(C) + P(D) - P(C \cap D) \) Some blue cars are coupe and therefore the events \( C \) and \( D \) are NOT mutually exclusive. There are 30 coupe that are blue therefore \( \quad \quad \quad P(C \cap D) = 30/200 = 3/20\) There is a total of 65 blue cars; hence \( \quad \quad \quad P(C) = 65/200 = 13/40 \) There is a total of 50 coupe cars; hence \( \quad \quad \quad P(D) = 50/200 = 1/4 \) \( \quad \quad \quad P(C \cup D) = P(C) + P(D) - P(C \cap D) = 13/40 + 1/4 - 3/20 = 17/40\) C) Let event \( E \): a black car is selected and event \( F \): an SUV is selected. We need to find the probability \( \quad \quad \quad P(E \cup F) = P(E) + P(F) - P(E \cap F) \) Some black cars are SUV cars and therefore the events \( E \) and \( F \) are NOT mutually exclusive. There are 35 SUV cars that are black therefore \( \quad \quad \quad P(E \cap F) = 35/200 = 7/40\) There is a total of 85 black cars; hence \( \quad \quad \quad P(E) = 85/200 = 17/40 \) There is a total of 60 SUV cars; hence \( \quad \quad \quad P(F) = 60/200 = 3/10 \) \( \quad \quad \quad P(E \cup F) = P(E) + P(F) - P(E \cap F) = 17/40 + 3/10 - 7/40 = 11/20\)
Example 7
Solution to Example 7Let event \( A \): student spends at most 4 hours and event \( B \): student spends at least 11 hours. We need to find the probability \( \quad \quad \quad P(A \cup B) = P(A) + P(B) - P(A \cap B) \) There is a total of 200 students. The first two rows correspond to students who spend 4 hours or less (at most 4 hours) and their total is: 25; hence \( \quad \quad \quad P (A) = 25/200 = 5/20 \) The last two rows correspond to students who spend 11 hours or more (at least 11 hours) and their total is: 90; hence \( \quad \quad \quad P(B) = 90/200 = 9 / 20 \) The two events are mutually exclusive as shown in the table. Hence \( \quad \quad \quad P(A \cap B ) = 0 \) and \( \quad \quad \quad P(A \cup B) = P(A) + P(B) = 5/20 + 9 /20 = 7/10 \)
Example 8
## More References and linksMultiplication Rule for Probabilities of Independent Eventsprobability questionsBinomial Probabilities Examples and Questions classical formula for probability mutually exclusive events Introduction to Probabilities sample space event elementary statistics and probabilities. Home Page |