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# Operations with Sets

## Relating the union and intersection of events to probability

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Operations with Sets

### Calculating Probabilities Related to Unions and Intersections

#### Union andIntersection

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

The intersection of events A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} occurs if both event A\begin{align*}A\end{align*} and event B\begin{align*}B\end{align*} 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 AB\begin{align*}A \cap B\end{align*}. You read this as either “A\begin{align*}A\end{align*} intersection B\begin{align*}B\end{align*}” or “A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}.” AB\begin{align*}A \cap B\end{align*} means everything that is in set A\begin{align*}A\end{align*} and in set B\begin{align*}B\end{align*}. That is, when looking at the intersection of two sets, we are looking for where the sets overlap.

#### Describing and Finding the Probabilities of Unions and Intersections

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

A:draw a 5\begin{align*}& A: {\text{draw a 5}}\end{align*}

B:draw a Jack\begin{align*}& B: {\text{draw a Jack}}\end{align*}

1. Describe AB\begin{align*}A \cup B\end{align*} for this experiment, and find the probability, P,\begin{align*}P,\end{align*} of AB.\begin{align*}A \cup B.\end{align*}

The union of A and B, AB\begin{align*}A \cup B\end{align*}, contains the four 5's and the four Jacks, for 8 total events. Since there are 52 cards, and 8 of them are in AB\begin{align*}A \cup B\end{align*}, then:

P(AB)=852=2130.15385\begin{align*}P(A \cup B)=\frac{8}{52}=\frac{2}{13}\approx 0.15385 \end{align*}

2. Describe AB\begin{align*}A \cap B\end{align*} for this experiment, and find the probability of AB\begin{align*}A \cap B\end{align*}.

There is no card that is a 5 and that is also a Jack. This means that AB\begin{align*}A \cap B \end{align*} is empty, and that P(AB)=0\begin{align*}P(A \cap B)=0\end{align*} (read "The probability of A intersection B equals zero"), since there are no events in the set, the probability of selecting an object from the set AB\begin{align*}A \cap B\end{align*} is 0.

#### More Describing and Finding

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

A:observe an even number\begin{align*}& A: {\text{observe an even number}}\end{align*}

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

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

1. Describe AB\begin{align*}A \cup B\end{align*} for this experiment.

An observation on a single toss of the die is an element of the union of A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} 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 AB\begin{align*}A \cup B\end{align*} are those for which A\begin{align*}A\end{align*}occurs, B\begin{align*}B\end{align*} occurs, or both occur:

AB={2,4,6}{1,2,3}={1,2,3,4,6}\begin{align*}A \cup B = \left \{2,4,6\right \} \cup \left \{1,2,3\right \} = \left \{1,2,3,4,6\right \}\end{align*}

2. Describe AB\begin{align*}A \cap B\end{align*} for this experiment.

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

AB={2,4,6}{1,2,3}={2}\begin{align*}A \cap B = \left \{2,4,6\right \} \cap \left \{1,2,3\right \} = \left \{2\right \}\end{align*}

3. Calculate P(AB)\begin{align*}P(A \cup B)\end{align*} and P(AB)\begin{align*}P(A \cap B)\end{align*}, assuming the die is fair.

Remember, the probability of an event is the sum of the probabilities of its simple events. This is shown for AB\begin{align*}A \cup B\end{align*} as follows:

P(AB)=P(1)+P(2)+P(3)+P(4)+P(6)=16+16+16+16+16=56\begin{align*}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}\end{align*}

Similarly, this can also be shown for AB\begin{align*}A \cap B\end{align*}:

P(AB)=P(2)=16\begin{align*}P(A \cap B) = P(2) = \frac{1}{6}\end{align*}

Intersections and unions can also be defined for more than two events. For example, ABC\begin{align*}A \cup B \cup C\end{align*} represents the union of three events.

#### Finding Simple Events

Refer to the above example and answer the following questions based on the definitions of the new events C\begin{align*}C\end{align*} and D\begin{align*}D\end{align*}.

C:observe a number that is greater than 5\begin{align*}& C: {\text{observe a number that is greater than } 5}\end{align*}

D:observe a number that is exactly 5\begin{align*}& D: {\text{observe a number that is exactly } 5}\end{align*}

1. Find the simple events in ABC\begin{align*}A \cup B \cup C\end{align*}.

Since C={6},ABC={2,4,6}{1,2,3}{6}={1,2,3,4,6}.\begin{align*}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 \}.\end{align*}

2. Find the simple events in AD\begin{align*}A \cap D\end{align*}.

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

3. Find the simple events in \begin{align*}A \cap B \cap C\end{align*}.

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 \begin{align*}A\end{align*} and \begin{align*}B\end{align*} and then intersecting the resulting set with \begin{align*}C\end{align*}.

\begin{align*}(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\end{align*}

Again, we get the empty set.

### Examples

For the following examples, suppose you have a standard deck of 52 cards. Let:

\begin{align*}& A: {\text{draw a 7}}\end{align*}

\begin{align*}& B: {\text{draw a Diamond}}\end{align*}

#### Example 1

Describe \begin{align*}A \cup B\end{align*} for this experiment, and find the probability of \begin{align*}A \cup B\end{align*}.

The intersection of A and B, \begin{align*}A \cup B\end{align*}, 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 \begin{align*}3+13=16\end{align*} or \begin{align*}4+12=16\end{align*} 7's and Diamonds. Since there are 52 cards, and 16 of them are in \begin{align*}A \cup B\end{align*}, then:

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

#### Example 2

Describe \begin{align*}A \cap B\end{align*} for this experiment, and find the probability of \begin{align*}A \cap B\end{align*}.

As we discussed above, there is one card that is a both a 7 and a Diamond, so there is 1 event in \begin{align*}A \cap B\end{align*}. This means that

\begin{align*}P(A \cap B)=\frac{1}{52}\approx 0.01923\end{align*}

### Review

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.

To view the Review answers, open this PDF file and look for section 3.2.

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Color Highlighted Text Notes

### Vocabulary Language: English

Intersection

Intersection is the probability of both or all of the events you are calculating happening at the same time (less likely).

union

$\cup$ is a symbol that stands for union and is used to connect two groups together. It is associated with the logical term OR.