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Operations with Sets

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In this Concept, you will learn how to combine two or more events by finding the union of the two events or the intersection of the two events. You will also learn how to calculate probabilities related to unions and intersections.

Watch This

For a description of how to find an event given a sample space (1.0) , see patrickJMT, Probability Events (5:40).

Guidance

Union and Intersection

Sometimes we need to combine two or more events into one compound event . This compound event can be formed in two ways.

The union of events A and B occurs if either event A , event B , or both occur in a single performance of an experiment. We denote the union of the two events by the symbol A \cup B . You read this as either “ A union B ” or “ A or B .” A \cup B means everything that is in set A or in set B or in both sets.

The intersection of events A and B occurs if both event A and event B occur in a single performance of an experiment. It is where the two events overlap. We denote the intersection of two events by the symbol A \cap B . You read this as either “ A intersection B ” or “ A and B .” A \cap B means everything that is in set A and in set B . That is, when looking at the intersection of two sets, we are looking for where the sets overlap.

Example A

Suppose you have a standard deck of 52 cards. Let:

& A: {\text{draw a 5}}

& B: {\text{draw a Jack}}

a. Describe A \cup B for this experiment, and find the probability of A \cup B .

b. Describe A \cap B for this experiment, and find the probability of A \cap B .

Solutions:

a. The union of A and B, A \cup B , contains the four 5's and the four Jacks, for 8 total events. Since there are 52 cards, and 8 of them are in A \cup B , then:

P(A \cup B)=\frac{8}{52}=\frac{2}{13}\approx 0.15385

b. There is no card that is a 5 and that is also a Jack. This means that A \cap B is empty, and that P(A \cap B)=0 , since there are no events in the set, the probability of selecting an object from the set A \cap B is 0.

Example B

Consider the throw of a die experiment. Assume we define the following events:

& A: {\text{observe an even number}}

& B: {\text{observe a number less than or equal to } 3}

  1. Describe A \cup B for this experiment.
  2. Describe A \cap B for this experiment.
  3. Calculate P(A \cup B) and P(A \cap B) , assuming the die is fair.

The sample space of a fair die is S = \left \{1,2,3,4,5,6\right \} , and the sample spaces of the events A and B above are A =\left \{2,4,6\right \} and B=\left \{1,2,3\right \} .

1. An observation on a single toss of the die is an element of the union of A and B if it is either an even number, a number that is less than or equal to 3, or a number that is both even and less than or equal to 3. In other words, the simple events of A \cup B are those for which A occurs, B occurs, or both occur:

A \cup B = \left \{2,4,6\right \} \cup \left \{1,2,3\right \} = \left \{1,2,3,4,6\right \}

2. An observation on a single toss of the die is an element of the intersection of A and B if it is a number that is both even and less than 3. In other words, the simple events of A \cap B are those for which both A and B occur:

A \cap B = \left \{2,4,6\right \} \cap \left \{1,2,3\right \} = \left \{2\right \}

3. Remember, the probability of an event is the sum of the probabilities of its simple events. This is shown for A \cup B as follows:

P(A \cup B) & = P(1)+P(2)+P(3)+P(4)+P(6)\\& =\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\\& =\frac{5}{6}

Similarly, this can also be shown for A \cap B :

P(A \cap B) = P(2) = \frac{1}{6}

Intersections and unions can also be defined for more than two events. For example, A \cup B \cup C represents the union of three events.

Example C

Refer to the above example and answer the following questions based on the definitions of the new events C and D .

& C: {\text{observe a number that is greater than } 5}

& D: {\text{observe a number that is exactly } 5}

  1. Find the simple events in A \cup B \cup C .
  2. Find the simple events in A \cap D .
  3. Find the simple events in A \cap B \cap C .

1. Since C = \left \{ 6 \right \}, A \cup B \cup C = \left \{ 2,4,6 \right \} \cup \left \{ 1,2,3 \right \} \cup \left \{ 6 \right \} = \left \{ 1,2,3,4,6 \right \}.

2. Since D = \left \{ 5 \right \}, A \cap D = \left \{ 2,3,6 \right \} \cap \left \{ 5 \right \} = \varnothing , where \varnothing is the empty set. This means that there are no elements in the set A \cap D .

3. Here, we need to be a little careful. We need to find the intersection of the three sets. To do so, it is a good idea to use the associative property by first finding the intersection of sets A and B and then intersecting the resulting set with C .

(A \cap B)\cap C = (\left \{2,4,6\right \} \cap \left \{1,2,3\right \})\cap \left \{6\right \} = \left \{2\right \} \cap \left \{6\right \} = \varnothing

Again, we get the empty set.

Vocabulary

The union of the two events A and B , written A \cup B , occurs if either event A , event B , or both occur on a single performance of an experiment. A union is an 'or' relationship.

The intersection of the two events A and B , written A \cap B , occurs only if both event A and event B occur on a single performance of an experiment. An intersection is an 'and' relationship.

Intersections and unions can be used to combine more than two events.

Guided Practice

Suppose you have a standard deck of 52 cards. Let:

& A: {\text{draw a 7}}

& B: {\text{draw a Diamond}}

a. Describe A \cup B for this experiment, and find the probability of A \cup B .

b. Describe A \cap B for this experiment, and find the probability of A \cap B .

Solutions:

a. The intersection of A and B, A \cup B , contains the four 7's and the thirteen Diamonds, however, one 7 is a diamond, so we don't want to double count it. This means that there are 3+13=16 or 4+12=16 7's and Diamonds. Since there are 52 cards, and 16 of them are in A \cup B , then:

P(A \cup B)=\frac{16}{52}=\frac{4}{13}\approx 0.30769

b. As we discussed above, there is one card that is a both a 7 and a Diamond, so there is 1 event in A \cap B . This means that

P(A \cap B)=\frac{1}{52}\approx 0.01923

Practice

For 1-3, you are given a “fair die” that has six sides with 1 to 6 dots on them. When the die is tossed or rolled, each of the sides is equally likely to come up. Determine the probability of the following outcomes for the number of dots showing on top after a single roll of the die.

  1. 5 dots
  2. two or three dots
  3. all odd dots

Fro 4-8, suppose you draw one card at random from an ordinary card deck. There are 52 possible cards that you can select.

  1. Suppose that all possible card selections are equally likely, what probability should be assigned to each selection?
  2. Find the probability that you choose a face card (J, Q K or A).
  3. Find the probability that you choose a black card (spade or club).
  4. Consider the events ``draw a face card" and ``draw a 5 or smaller". Are these events overlapping or disjoint? Explain.
  5. Find the probability of ``draw a red card" or ``draw a spade".

For 9-12, one of the most popular casino games is roulette. In this game, there is a wheel with 38 numbered metal pockets (1 to 36 plus 0 and 00). The wheel is spun moving a metal ball and the ball comes to rest on one of the 38 pockets. The wheel is balanced so that the ball is equally likely to fall on any one of the 38 possible numbers. You play this game by betting on various outcomes-you win if the particular outcome is spun. What is the probability of winning if you bet on the following?

  1. Number 32
  2. Even numbers (2, 4, etc.)
  3. First 12 numbers (1 to 12)
  4. Four consecutive numbers, such as 20, 21, 23, 24.

Keywords

Event

Experiment

Intersection of events

Mutually exclusive

Sample space

Union of events

Venn diagram

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