7.3: Probability and Probability Models


This tutorial: Part B: Probability of Unions, Intersections, and Complements

First, a warm-up on unions, intersections, complements, and mutually exclusive events to get started... (To review these topics, go to the tutorial for Section 7.1.)

If E is the event that and F is the event that , then is the event that:

For the next warmup exercise, recall from the tutorial for Section 7.1 that two events E and F are mutually exclusive if their intersection is empty; that is, E \cap F = Ø.

You select an exercise at random from . (The book has some exercises marked as "applications" and others marked as "communication and reasoning exercises".) Which of the following are mutually exclusive pairs of events?


 

We now see how to calculate the probability of a union of two or more events:
 
Addition Principle for Mutually Exclusive Events

If E and F are mutually exclusive events, then

    P(E \cup F) = P(E) + P(F).

In words: The probability that either E or F occurs is the probability that E occurs, plus the probability that F occurs.

Examples

1. Cast two dice and add the numbers facing up.

Since these are mutually exclusive events, the probability that the sum is either 5 or even is

    P(E \cup F)=P(E) + P(F)
    = \frac{1}{9} + \frac{1}{2} = \frac{11}{18}

2. Cast two dice and add the numbers facing up. Let

    E: The sum is or less.
    F: The sum is or more.
    P(E \cup F) =

3. In a survey of voters, of the respondents reported supporting the Republican candidate for governor, and reported supporting the Democrat candidate. (Voting machines made it impossible to "spoil the ballot" by voting for both candidates.)

    Based on this survey, the estimated probability that a voter will vote for a candidate of either major political party is:
                   

 

Q: Pairs of events are not always mutually exclusive as we saw above. How do we calculate the probability of the union of two (not necessarily mutually exclusive) events E and F?
A: The following formula generalizes the one we have seen to cover all pairs of events; mutually exclusive or not:

 
General Addition Principle

If E and F are any two events (mutually exclusive or not) then:

    P(E \cup F) = P(E) + P(F) - P(E \cap F)
(For an explanation of where this formula come from, consult Section 7.3 in or .)

Note When E and F are mutually exclusive, P(E \cap F) = 0 and so the above formula reduces to the formula we saw earlier:

    P(E \cup F) = P(E) + P(F)          Mutually exclusive events

Examples

1. Cast a single die and note the number facing up. Take

    E: The outcome is 1, 2 or 3; P(E) = \frac{1}{2}
    F: The outcome is even. P(F) = \frac{1}{2}

We will also need:

    P(E \cap F): The outcome is 2; P(E \cap F) = \frac{1}{6}         Press here to see how we got that.

Because E and F are not mutually exclusive, we use

    P(E \cup F) = P(E) + P(F) - P(E \cap F)
    = \frac{1}{2} + \frac{1}{2} - \frac{1}{6} = \frac{5}{6}

2. Cast two dice (one green, one red) and add the numbers facing up. Let

    E: The green die shows .
    F: The sum is .
    P(E \cup F) =

The next quiz is similar to Example 1 in Section 7.3 in or


               

What is the probability that but
               

Further Principles of Probability
Here is an expanded table which lists all the principles of probability we are interested in here. You will recognize the first and second from above.
Addition Principle for Mutually Exclusive Events

If E and F are mutually exclusive events, then

    P(E \cup F) = P(E) + P(F).

In words: The probability that either E or F occurs is the probability that E occurs, plus the probability that F occurs.
 

Addition Principle for Any Number of Mutually Exclusive Events

If E_1, E_2, ..., E_n are (pairwise) mutually exclusive events, then

    P(E_1 \cup E_2 \cup ... \cup E_n) = P(E_1) + P(E_2) + ... + P(E_n).
Example

Cast two dice and add the numbers facing up.

    E_1: The sum is 5. Thus, P(E_1) = \frac{1}{9}
    E_2: The sum is 7. Thus, P(E_2) = \frac{1}{6}
    E_3: The sum is even. Thus, P(E_3) = \frac{1}{2}

Since these are mutually exclusive events, the probability that the sum is either 5, 7, or even is

    P(E \cup E_2 \cup E_3)= P(E_1) + P(E_2) + P(E_3)
    = \frac{1}{9} + \frac{1}{6} + \frac{1}{2} = \frac{7}{9}
General Addition Principle

If E and F are not mutually exclusive events, then we must use the following more general formula:

    P(E \cup F) = P(E) + P(F) - P(E \cap F)
Example

Cast a single die and note the number facing up. Take

    E: The outcome is 1, 2 or 3; P(E) = \frac{1}{2}
    F: The outcome is even. P(F) = \frac{1}{2}
    P(E \cap F): The outcome is 2; P(E \cap F) = \frac{1}{6}

    P(E \cup F) = P(E) + P(F) - P(E \cap F)
    = \frac{1}{2} + \frac{1}{2} - \frac{1}{6} = \frac{5}{6}

Probability of S and Ø

    P(S) = 1
    P() = 0

In words: The probability that something happens is 1; the probability that nothing happens is 0.

Example

In the experiment immediately above,

    S = \{\.1, 2, 3, 4, 5, 6\.\};
    P(S) = 1

(The outcome is always a number in the range 1-6.)

Complement

For all events E,

    P(E') = 1 - P(E)
In words: The probability that E does not happen is 1 minus the probability of E.. 		Example

In the experiment immediately above, take

    E: The outcome is 6; P(E) = \frac{1}{6}.
    E': The outcome is not 6; P(E') = 1 - P(E) = 1 - \frac{1}{6} = \frac{5}{6}

We now put all of these principles to use:
 

In a survey of voters who were eligible to vote in both of the last two gubernatorial elections, reported voting in both, reported voting in the first, while did not vote at all.

The next quiz is similar to Example 3 in Section 7.3 in or The following table shows sales of new recreational boats in the U.S. during the period 1999-2001. (Numbers are in thousands.)*

Motor boats    Jet skis      Sailboats   Total
1999
2000
2001
Total

Consider the experiment in which a recreational boat is selected at random from those in the table. Let E be the event that the boat was a let F be the event that the boat was purchased in and let G be the event that the boat was a Calculate the following:

For more practice, try some of the questions in the chapter quiz (Warning: it covers the whole of Chapter 7) by pressing the button on the sidebar. Then try the exercises in Section 7.3 of Finite Mathematics and Finite Mathematics and Applied Calculus

Last Updated: June, 2009
Copyright © 2009 Stefan Waner

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