On any given night, the probability that Nick has a cookie for dessert is 10%. The probability that Nick has ice cream for dessert is 50%. The probability that Nick has a cookie or ice cream is 55%. What is the probability that Nick has a cookie **and** ice cream for dessert?

#### Guidance

Consider a sample space with events \begin{align*}A\end{align*} and \begin{align*}B\end{align*}.

Recall that the **union of events \begin{align*}A\end{align*} and \begin{align*}B\end{align*}** is an event that includes all the outcomes in either event **\begin{align*}A\end{align*}**, event **\begin{align*}B\end{align*}**, or both. The symbol \begin{align*}\cup\end{align*} represents union. Below, \begin{align*}A \cup B\end{align*} is shaded.

How do you find the **number of outcomes** in a union of events? If you find the sum of the number of outcomes in event \begin{align*}A\end{align*} and the number of outcomes in event \begin{align*}B\end{align*}, you will have counted some of the outcomes twice. *In fact, you will have counted the outcomes that are in both event \begin{align*}A\end{align*} and event \begin{align*}B\end{align*} twice.* Therefore, in order to correctly count the number of outcomes in the union of two events, you must count the number of outcomes in each event separately and *subtract* the number of outcomes shared by both events (so these are not counted twice). Generalizing to probability:

**\begin{align*}P(A \cup B)=P(A)+P(B)-P(A \cap B)\end{align*}. This is called the Addition Rule for Probability.**

Note that \begin{align*}(A \cap B)\end{align*} is the intersection of the two events. It contains all the outcomes that are shared by both events and is the intersection of the two circles in the Venn diagram.** **

Suppose that in your class of 30 students, 8 students are in band, 15 students play a sport, and 5 students are both in band and play a sport. Let \begin{align*}A\end{align*} be the event that a student is in band and let \begin{align*}B\end{align*} be the event that a student plays a sport. Create a Venn diagram that models this situation.

*In order to fill in the Venn diagram, remember that the total of the numbers in circle \begin{align*}A\end{align*} must be 8 and the total of the numbers in circle B must be 15. The intersection of the two circles must contain a 5.*

\begin{align*}P(A \cup B)\end{align*} is the probability that a student is in band **or** plays a sport **or** both. With the help of the Venn diagram, this is not too difficult to calculate:

\begin{align*}P(A \cup B)=\frac{3+10+5}{30}=\frac{18}{30}=\frac{3}{5}\end{align*}

You could also compute this probability using the Addition Rule:

\begin{align*}P(A \cup B)&=P(A)+P(B)-P(A \cap B)\\&=\frac{8}{30}+\frac{15}{30}-\frac{5}{30}\\&=\frac{8+15-5}{30}\\&=\frac{18}{30}\\&=\frac{3}{5}\\\end{align*}

Note that by using the Addition Rule, you avoid having to determine that there are 3 people who are in band and don't play a sport and 10 people who play a sport but are not in band. The Addition Rule is easier when you have not created a Venn diagram.

**Example A**

Two events \begin{align*}C\end{align*} and \begin{align*}D\end{align*} are disjoint. Explain why \begin{align*}P(C \cup D)=P(C)+P(D)\end{align*}.

**Solution:** If two events are disjoint (also known as mutually exclusive), then they share no outcomes. Therefore, the probability of both events occurring simultaneously is 0 \begin{align*}(P(C \cap D)=0)\end{align*}. By the Addition Rule:

\begin{align*}P(C \cup D) &= P(C)+P(D)-P(C \cap D)\\ P(C \cup D) &= P(C)+P(D)-0\\ P(C \cup D) &= P(C)+P(D)\end{align*}

**Example B**

Suppose that today there is a 90% chance of snow, a 20% chance of a strong winds, and a 15% chance of both snow and strong winds. What is the probability of snow or strong winds?

**Solution:** Use the Addition Rule:

\begin{align*}P(Snow \cup Strong \ Winds) &= P(Snow)+P(Strong \ Winds)-P(Snow \cap Strong \ Winds)\\ \end{align*}

\begin{align*}P(Snow \cup Strong \ Winds) &= 0.90+0.20-0.15\\ P(Snow \cup Strong \ Winds) &= 0.95\end{align*}

There is a 95% chance of either snow, strong winds, or both.

**Example C**

Now suppose that today there is a 60% chance of snow, an 85% chance of snow or strong winds, and a 25% chance of snow and strong winds. What is the chance of strong winds?

**Solution:** Once again you can use the Addition Rule, because it relates the probabilities in the problem.

\begin{align*}P(Snow \cup Strong \ Winds) &= P(Snow)+P(Strong \ Winds)-P(Snow \cap Strong \ Winds)\\ \end{align*}

\begin{align*}0.85 &= 0.60+P(Strong \ Winds)-0.25\\ 0.50 &= P(Strong \ Winds)\end{align*}

There is a 50% chance of strong winds.

**Concept Problem Revisited**

On any given night, the probability that Nick has a cookie for dessert is 10%. The probability that Nick has ice cream for dessert is 50% . The probability that Nick has a cookie or ice cream is 55%. What is the probability that Nick has a cookie and ice cream for dessert?

Let \begin{align*}C\end{align*} be the event that Nick has a cookie and let I be the event that Nick has ice cream. The given probabilities are:

- \begin{align*}P(C)=10\%\end{align*}
- \begin{align*}P(I)=50\%\end{align*}
- \begin{align*}P(C \cup I)=55\%\end{align*}

You are looking for \begin{align*}P(C \cap I)\end{align*}. By the Addition Rule, you know that \begin{align*}P(C \cup I)=P(C)+P(I)-P(C \cap I)\end{align*}. Substitute in the values you know in order to solve for \begin{align*}P(C \cap I)\end{align*}.

\begin{align*}P(C \cup I) &= P(C)+P(I)-P(C \cap I)\\ 0.55 &= 0.10+0.50-P(C \cap I)\\ 0.55 &= 0.60-P(C \cap I)\\ P(C \cap I) &= 0.05\end{align*}

The probability that Nick has a cookie *and* ice cream is 5%.

#### Vocabulary

The ** union** of two events is the event that includes all outcomes that are in either or both of the original events. The symbol for union is \begin{align*}\cup\end{align*}.

The ** intersection** of two events is the event that includes all outcomes that are in both of the original events. The symbol for intersection is \begin{align*}\cap\end{align*}.

The ** probability** of an event is the chance of the event occurring.

The ** Addition Rule** states that for two events \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, \begin{align*}P(A \cup B)=P(A)+P(B)-P(A \cap B)\end{align*}

A ** Venn diagram** is a way to visualize sample spaces, events, and outcomes.

#### Guided Practice

Consider the experiment of rolling a pair of dice. There are 36 outcomes in the sample space. You are interested in the sum of the numbers. Let \begin{align*}A\end{align*} be the event that the sum is even and let B be the event that the sum is less than 5.

- Create a Venn diagram that models this situation. The Venn diagram should contain 36 numbers.
- Find \begin{align*}P(A \cup B)\end{align*} using the Venn diagram.
- Find \begin{align*}P(A \cup B)\end{align*} using the Addition Rule and explain why the answer makes sense.

**Answers:**

1. Find all 36 outcomes, and then find the sum of each pair of numbers. Sort the numbers into the Venn diagram so that even numbers are in circle \begin{align*}A\end{align*} and numbers less than 5 are in circle \begin{align*}B\end{align*}. Any other numbers should appear outside of the circles.

2. There are 20 numbers within the circles and 36 numbers total. Therefore, \begin{align*}P(A \cup B)=\frac{20}{36}=\frac{5}{9}\end{align*}.

3. To use the Addition Rule, you need to know \begin{align*}P(A), P(B)\end{align*} and \begin{align*}P(A \cap B)\end{align*}.

- \begin{align*}P(A)=\frac{18}{36}=\frac{1}{2}\end{align*}
- \begin{align*}P(B)=\frac{6}{36}=\frac{1}{6}\end{align*}
- \begin{align*}P(A \cap B)=\frac{4}{36}=\frac{1}{9}\end{align*}

Now, use the Addition Rule: \begin{align*}P(A \cup B)=\frac{18}{36}+\frac{6}{36}-\frac{4}{36}=\frac{20}{36}=\frac{5}{9}\end{align*}. This answer is the same as the answer to #2, as it should be. This calculation makes sense because both \begin{align*}P(A)\end{align*} and \begin{align*}P(B)\end{align*} include the 4 numbers in the intersection of the circles. You need to subtract \begin{align*}P(A \cap B)\end{align*}, the probability of those 4 numbers, so that you do not count the probability of those numbers twice.

#### Practice

1. State the Addition Rule for probability and explain when it is used.

2. What happens to the Addition Rule when the two events considered are disjoint?

3. Use a Venn diagram to help explain why there is subtraction in the Addition Rule.

4. Sarah tells her mom that there is a 40% chance she will clean her room, a 70% she will do her homework, and a 24% chance she will clean her room and do her homework. What is the probability of Sarah cleaning her room or doing her homework?

5. You dad only ever makes one meal for dinner. The probability that he makes pizza tonight is 30%. The probability that he makes pasta tonight is 60%. What is the probability that he makes pizza or pasta?

6. After your little sister has gone trick-or-treating for Halloween, your mom says she is allowed to have 2 pieces of candy. The probability of her having a Snickers is 50%. The probability of her having a peanut butter cup is 60%. The probability of her having a Snickers or a peanut butter cup is 100%. What is the probability of her having a Snickers and a peanut butter cup?

7. Deanna sometimes likes honey or lemon in her tea. There is a 50% chance that she will have honey and lemon, a 95% chance that she will have honey or lemon, and a 80% chance that she will have honey. What is the probability that she will have lemon?

Consider the experiment of drawing a card from a standard deck. Let \begin{align*}A\end{align*} be the event that the card is a diamond. Let \begin{align*}B\end{align*} be the event that the card is a Jack. Let \begin{align*}D\end{align*} be the event that the card is a four.

8. Find \begin{align*}P(A), P(D), P(A \cap D)\end{align*}.

9. Find \begin{align*}P(A \cup D)\end{align*}. What does this probability represent compared to \begin{align*}P(A \cap D)\end{align*}?

10. To find \begin{align*}P(B \cup D)\end{align*}, all you need to do is add \begin{align*}P(B)\end{align*} and \begin{align*}P(D)\end{align*}. Why is this and why do you not have to subtract anything?

Consider the experiment of flipping three coins and recording the sequence of heads and tails. Let \begin{align*}B\end{align*} be the event that there is at least one heads. Let \begin{align*}C\end{align*} be the event that the third coin is a tails. Let \begin{align*}D\end{align*} be the event that the first coin is a heads.

11. Find \begin{align*}P(C \cup D)\end{align*}. What does this probability represent?

12. Create a Venn diagram to show events \begin{align*}B\end{align*} and \begin{align*}C\end{align*} for this experiment.

13. Find \begin{align*}P(B \cup C)\end{align*} and \begin{align*}P(B \cap C)\end{align*}. Compare and contrast these two probabilities.

The Addition Rule can be extended for three events. Consider three events that all share outcomes, as shown in the Venn diagram below.

14. Label the shaded part of the diagram in terms of \begin{align*}A, B, C\end{align*}.

15. Find a rule for \begin{align*}P(A \cup B \cup C)\end{align*} in terms of \begin{align*}P(A), P(B), P(C), P(A \cap B), P(B \cap C), P(A \cap C), P(A \cap B \cap C)\end{align*}.