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# 2.1: Tree Diagrams

Difficulty Level: Basic Created by: CK-12
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Suppose you flip a coin three times in succession. How can you illustrate all your possible outcomes? What is the probability that your coin will come up heads three times in a row?

### Watch This

First watch this video to learn about tree diagrams.

Then watch this video to see some examples.

Watch this video for more help.

### Guidance

In previous Concepts, we studied independent and dependent events, as well as mutually inclusive and mutually exclusive events. We used the Addition Rule for dependent events, as well as mutually inclusive and mutually exclusive events. The Addition Rule, or Addition Principle, is used to find P(A or B)\begin{align*}P(A \ or \ B)\end{align*}, while the Multiplication Rule is used for independent events.

Addition Rule – For 2 events, A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, the probability of selecting one event or another is given by: P(A or B)=P(A)+P(B)P(A and B)\begin{align*}P(A \ \text{or} \ B) = P(A) + P(B) - P(A \ \text{and} \ B)\end{align*}.

Multiplication Rule – For 2 independent events, A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, where the outcome of A\begin{align*}A\end{align*} does not change the probability of B\begin{align*}B\end{align*}, the probability of A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} is given by: P(A and B)=P(A)×P(B)\begin{align*}P(A \ \text{and} \ B) = P(A) \times P(B)\end{align*}.

Tree diagrams are another way to show the outcomes of simple probability events. In a tree diagram, each outcome is represented as a branch on a tree.

#### Example A

If you toss a coin 2 times, what is the probability of getting 2 heads? Use a tree diagram to find your answer.

This is an example of independent events, because the outcome of one event does not affect the outcome of the second event. What does this mean? Well, when you flip the coin once, you have an equal chance of getting a head (H) or a tail (T). On the second flip, you also have an equal chance of getting a a head or a tail. In other words, whether the first flip was heads or tails, the second flip could just as likely be heads as tails. You can represent the outcomes of these events on a tree diagram.

From the tree diagram, you can see that the probability of getting a head on the first flip is 12\begin{align*}\frac{1}{2}\end{align*}. Starting with heads, the probability of getting a second head will again be 12\begin{align*}\frac{1}{2}\end{align*}. But how do we calculate the probability of getting 2 heads? These are independent events, since the outcome of tossing the first coin in no way affects the outcome of tossing the second coin. Therefore, we can calculate the probability as follows:

P(A and B)P(A and B)=12×12=14\begin{align*}P(A \ \text{and} \ B) &= \frac{1}{2} \times \frac{1}{2}\\ P(A \ \text{and} \ B) &= \frac{1}{4}\end{align*}

Therefore, we can conclude that the probability of getting 2 heads when tossing a coin twice is 14\begin{align*}\frac{1}{4}\end{align*}, or 25%. Let’s try an example that is a little more challenging.

#### Example B

Irvin opens up his sock drawer to get a pair socks to wear to school. He looks in the sock drawer and sees 4 red socks, 8 white socks, and 6 brown socks. Irvin reaches in the drawer and pulls out a red sock. He is wearing blue shorts, so he replaces it. He then draws out a white sock. What is the probability that Irvin pulls out a red sock, replaces it, and then pulls out a white sock?

First let’s draw a tree diagram.

There are 18 socks in Irvin’s sock drawer. The probability of getting a red sock when he pulls out the first sock is:

P(red)P(red)=418=29\begin{align*}P(\text{red}) &= \frac{4}{18}\\ P(\text{red}) &= \frac{2}{9}\end{align*}

Irvin puts the sock back in the drawer and pulls out the second sock. The probability of getting a white sock on the second draw is:

P(white)P(white)=818=49\begin{align*}P(\text{white}) &= \frac{8}{18}\\ P(\text{white}) &= \frac{4}{9}\end{align*}

Therefore, the probability of getting a red sock and then a white sock when the first sock is replaced is:

P(red and white)P(red and white)=29×49=881\begin{align*}P(\text{red and white}) &= \frac{2}{9} \times \frac{4}{9}\\ P(\text{red and white}) &= \frac{8}{81}\end{align*}

One important part of these types of problems is that order is not important.

Let’s say Irvin picked out a white sock, replaced it, and then picked out a red sock. Calculate this probability.

P(white and red)P(white and red)=49×29=881\begin{align*}P(\text{white and red}) &= \frac{4}{9} \times \frac{2}{9}\\ P(\text{white and red}) &= \frac{8}{81}\end{align*}

So regardless of the order in which he takes the socks out, the probability is the same. In other words, P(red and white)=P(white and red)\begin{align*}P(\text{red and white}) = P(\text{white and red})\end{align*}.

#### Example C

In Example B, what happens if the first sock is not replaced?

The probability that the first sock is red is:

P(red)P(red)=418=29\begin{align*}P(\text{red}) &= \frac{4}{18}\\ P(\text{red}) &= \frac{2}{9}\end{align*}

The probability of picking a white sock on the second pick is now:

So now, the probability of selecting a red sock and then a white sock, without replacement, is:

P(red and white)P(red and white)P(red and white)=29×617=12153=451\begin{align*}P(\text{red and white}) &= \frac{2}{9} \times \frac{6}{17}\\ P(\text{red and white}) &= \frac{12}{153}\\ P(\text{red and white}) &= \frac{4}{51}\end{align*}

If the first sock is white, will P(red and white)=P(white and red)\begin{align*}P(\text{red and white}) = P(\text{white and red})\end{align*} as we found in Example 1? Let's find out.

P(white)P(white)=618=13\begin{align*}P(\text{white}) &= \frac{6}{18}\\ P(\text{white}) &= \frac{1}{3}\end{align*}

The probability of picking a red sock on the second pick is now:

As with the last example, P(red and white)=P(white and red)\begin{align*}P(\text{red and white}) = P(\text{white and red})\end{align*}. So when does order really matter? We'll find out in the next concept.

Points to Consider

• How are tree diagrams helpful for determining probabilities?

### Vocabulary

Tree diagrams are a way to show the outcomes of simple probability events, where each outcome is represented as a branch on a tree.

### Guided Practice

In a survey, baseball fans were asked who they would like to win the National League playoffs. 54% responded that they would like the Phillies to win, and 46% responded that they would like the Giants to win. The fans were then asked who they would like to win the American League playoffs if the Phillies win the National League playoffs, and who they would like to win the American League playoffs if the Giants win the National League playoffs. If the Phillies win the National League playoffs, 42% of the fans responded that they want the Rangers to win the American League playoffs, and 58% said that they want the Yankees to win. If the Giants win the National League playoffs, 48% of the fans responded that they want the Rangers to to win the American League Playoffs, and 52% said that they want the Yankees to win. The results of the survey are shown in the following tree diagram:

According to the survey, what percentage of the fans want each of the possible World Series match-ups? Do all of the probabilities add up to 100%?

The percentage of the fans wanting each of the possible World Series match-ups can be calculated as follows:

P(Phillies and Rangers)P(Phillies and Rangers)P(Phillies and Rangers)=0.54×0.42=0.2268=22.68%\begin{align*}P(\text{Phillies and Rangers}) &= 0.54 \times 0.42\\ P(\text{Phillies and Rangers}) &= 0.2268\\ P(\text{Phillies and Rangers}) &= 22.68\%\end{align*}

P(Phillies and Yankees)P(Phillies and Yankees)P(Phillies and Yankees)=0.54×0.58=0.3132=31.32%\begin{align*}P(\text{Phillies and Yankees}) &= 0.54 \times 0.58\\ P(\text{Phillies and Yankees}) &= 0.3132\\ P(\text{Phillies and Yankees}) &= 31.32\%\end{align*}

P(Giants and Rangers)P(Giants and Rangers)P(Giants and Rangers)=0.46×0.48=0.2208=22.08%\begin{align*}P(\text{Giants and Rangers}) &= 0.46 \times 0.48\\ P(\text{Giants and Rangers}) &= 0.2208\\ P(\text{Giants and Rangers}) &= 22.08\%\end{align*}

P(Giants and Yankees)P(Giants and Yankees)P(Giants and Yankees)=0.46×0.52=0.2392=23.92%\begin{align*}P(\text{Giants and Yankees}) &= 0.46 \times 0.52\\ P(\text{Giants and Yankees}) &= 0.2392\\ P(\text{Giants and Yankees}) &= 23.92\%\end{align*}

Now let's add up all the probabilities.

22.68%+31.32%+22.08%+23.92%=100%\begin{align*}22.68\%+31.32\%+22.08\%+23.92\% = 100\%\end{align*}

All of the probabilities do, in fact, add up to 100%.

### Practice

1. A bag contains 3 red balls and 4 blue balls. Thomas reaches in the bag and picks a ball out at random from the bag. He places it back into the bag. Thomas then reaches in the bag and picks another ball at random.
1. Draw a tree diagram to represent this problem.
2. What is the probability that Thomas picks:
1. 2 red balls
2. a red ball in his second draw
2. A teacher has a prize box on her front desk for when students do exceptional work in math class. Inside the box there are 20 math pencils and 10 very cool erasers. Janet completed a challenge problem for Ms. Cameron, and Ms. Cameron rewarded Janet’s innovative problem-solving approach with a trip to the prize box. Janet reaches into the box and picks out a prize and then drops it back in. Then she reaches in again and picks out a prize a second time.
1. Draw a tree diagram to represent this problem.
2. What is the probability that Janet reaches into the box and picks out an eraser on the second pick?

3. Students in BDF High School were asked about their preference regarding the new school colors. They were given a choice between green and blue as the primary color and red and yellow as the secondary color. The results of the survey are shown in the tree diagram below. You can see that 75% of the students choose green as the primary color. Of this 75%, 45% chose yellow as the secondary color. What is the probability that a student in BDF High School selected red as the secondary color if he or she chose blue as the primary color?
4. In question 3, what color combination do the students in BDF High School want the most?
5. According to the following tree diagram, what is the probability of getting 1 head and 1 tail when flipping a coin 2 times? Note that 1 head and 1 tail can mean a head and then a tail or a tail and then a head.
6. If a coin is flipped 4 times, how many branches will the tree diagram have that represents this situation?
7. If a die is rolled 2 times, how many branches will the tree diagram have that represents this situation?
8. Suppose a tree diagram has 4 branches, and the probabilities of the outcomes representing 3 of the branches are 0.12, 0.53, and 0.28, respectively. What is the probability of the outcome representing the remaining branch?
9. The following tree diagram represents the responses to 2 yes/no questions:
10. If the probability of answering yes to both questions is 0.1947, what is the probability of answering yes to the second question if the answer to the first question was yes?
11. For question 9, if the probability of answering yes to both questions is 0.1947, what is the probability of answering no to the second question if the answer to the first question was yes?

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English Spanish

tree diagrams

Tree diagrams are a way to show the outcomes of simple probability events where each outcome is represented as a branch on a tree.

If events A and B are mutually inclusive, then P(A or B) = P(A) + P(B) – P(A and B)

Multiplication Rule

States that for 2 events (A and B), the probability of A and B is given by: P(A and B) = P(A) x P(B).

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