How do I construct a tree diagram?

In this explainer, we will learn how to draw tree diagrams representing two or more successive experiments and use them to calculate probabilities.

Recall that an experiment is a repeatable process with a known set of possible outcomes. An experiment can comprise just a single event, such as rolling a die once, but the term is more often used to describe a combination of two or more events. The repeatable and predictable nature of these processes means that they are best analyzed using probability.

When calculating probabilities for experiments involving simultaneous events, Venn diagrams can be a helpful way to show the possible outcomes. For example, we could use a Venn diagram to work out the probability that a randomly chosen student in a given school year studies both mathematics and history.

Sometimes, however, we need to represent events that happen one after another, called successive events. The first event in a sequence of successive events may or may not affect the probabilities of the later ones, but either way, we need a simple means to represent all the possible outcomes. This will then enable us to calculate probabilities. It turns out that the most convenient method for doing this is drawing a tree diagram.

Properties: Tree Diagram

A tree diagram is a means of representing two or more successive events.

Each branch of a tree diagram has its outcome written at its right-hand end and is labeled with the probability of that outcome.

For example, the tree diagram below represents the experiment of flipping a fair coin twice.

We start on the left with the first flip of the coin. This is represented by a pair of branches coming from a single point that correspond to the outcomes “heads” and “tails.” Since the coin is fair, or unbiased, each outcome has a probability of 0.5, which is shown as the label on each branch. Note that these are the only two possible outcomes of flipping a fair coin, so their probabilities sum to 1.

For the second flip of the coin, we move right and draw similar pairs of branches starting from each outcome of the first flip. In this way, we can represent all four possible outcomes of flipping a fair coin twice, namely “heads, heads,” “heads, tails,” “tails, heads,” and “tails, tails.” To calculate the probability of any of these four outcomes, we multiply the probabilities along the branches leading to that outcome. So, for example, to calculate the probability of “heads, heads,” we multiply the probabilities along the branches highlighted below to get 0.5×0.5=0.25.

In this particular case, all four possible outcomes have the same probability of 0.25 because they give rise to four identical calculations. This will not happen in general. However, it is significant that 0.25+0.25+0.25+0.25=1. In other words, the probabilities of all the possible outcomes of the experiment sum to 1, which is true for all tree diagrams.

Note that, in this simple example, the two successive events are independent, since the outcome of the first flip of the coin has no effect upon the probabilities of the second flip. Often, however, tree diagrams are used to represent successive events where the outcome of the first event has an impact on the probabilities of the later event (or events); these are called dependent events. With questions on this topic, it is always important to read them carefully to work out whether the given events are independent or dependent, as we need to make sure that we assign the correct probabilities to the branches of the tree diagram.

Let us start with an example to check our understanding of how to construct a tree diagram.

Example 1: Drawing a Tree Diagram

Fady chooses a card from a standard deck of 52 playing cards. He records whether it is one of the 12 face cards, that is, king, queen, or jack, or just an ordinary number card. Then, without putting the card back in the deck, he chooses another one. Which tree diagram correctly represents these two successive events?

Answer

Recall that a tree diagram is a means of representing two (or more) successive events. Each branch has its outcome written at its right-hand end and is labeled with the probability of that outcome. From the question wording, we must work out what the two successive events are, together with the probabilities of their associated outcomes. We can then pick the correct tree diagram from the four answer options.

Here, the first event is choosing a card at random from a standard deck of 52 cards, with the two possible outcomes being either a face card or a number card. We are told that this card is not put back in the deck afterward, so the second event is choosing a card at random from the remaining 51 cards, again with the two possible outcomes of it being either a face card or a number card.

For the first event, the probability of choosing a face card is given by numberoffacecardstotalnumberofcards=1252=313.

Similarly, the probability of choosing a number card is given by numberofnumbercardstotalnumberofcards=4052=1013.

Reviewing the answer options, we can immediately rule out D because it has these two probabilities the wrong way around.

Options A, B, and C all have the correct probabilities for the outcomes in the first event, so now we consider the second event. As the first card is not put back in the deck, this means that the outcome of the first event affects the probability of the second event, so these events are dependent. Since there are now only 51 cards to choose from, our denominator must therefore be 51. Consequently, we can also rule out option A because it shows the same pairs of probabilities for both events. This tree diagram would be correct for the equivalent experiment where the first card was put back in the deck before the second one was chosen.

We now only have options B and C to consider. If the first card was a face card, then we would be left with 11 face cards and 40 number cards out of a total of 51 cards. Thus, for the second event, the probability of choosing a face card would be numberoffacecardstotalnumberofcards=1151 and the probability of choosing a number card would be numberofnumbercardstotalnumberofcards=4051.

On the other hand, if the first card was a number card, then we would be left with 12 face cards and 39 number cards out of a total of 51 cards. In this case, for the second event, the probability of choosing a face card would be numberoffacecardstotalnumberofcards=1251=417 and the probability of choosing a number card would be numberofnumbercardstotalnumberofcards=3951=1317.

Comparing these probabilities with those in the remaining answer options, we see that B shows pairs of probabilities for the second event that contradict the outcomes in the first event. For example, the probability 417 (=1251) on the first branch of the second event indicates that there are 12 face cards still remaining after one has already been selected in the first event, which is not the case. Meanwhile, C shows exactly the probabilities obtained above, all in the correct places.

We conclude that the correct tree diagram to represent these two successive events is C.

In the next example, we will practice analyzing a tree diagram in order to identify a specific outcome and calculate its probability. In general, we will use the notation 𝑃(𝑋) to refer to the probability of the outcome 𝑋.

Example 2: Using a Tree Diagram to Compute Probabilities

The probability that a randomly selected student reviews for a mathematics examination is 0.8. If a student does review, the probability that they pass the examination is 0.9. If they do not review, the probability that they pass the examination is 0.25.

Using the tree diagram above, calculate the probability that a student chosen at random does not review and still passes.

Answer

Recall that a tree diagram is a means of representing two (or more) successive events. To calculate the probability of an individual outcome, we multiply the probabilities along the branches leading to that outcome.

Notice that this tree diagram shows how the probability of a student passing their mathematics examination is affected by whether or not they review, meaning that these two events are dependent. Of course this is not really surprising, as we would expect there to be a link between reviewing and passing exams.

If a student does not review for their mathematics examination and still passes it, then the outcome of the first event is that they do not review and the outcome of the second event is that they pass. Therefore, to calculate the required probability, we must multiply the probabilities along the branches shown below.

Hence, we have 𝑃()=0.2×0.25=0.05,“doesnotreview”and“pass” which is our answer.

The probability that the student does not review and still passes the mathematics examination is 0.05.

Tree diagrams also enable us to answer more complex questions that involve calculating the probability of a set of outcomes instead of just a single one. For instance, returning to our simple coin-flipping experiment, we might be asked to calculate the probability of getting the same result from both coin flips, which corresponds to the outcomes “heads, heads” or “tails, tails.” In situations like these, we multiply along the branches to get the probabilities of the individual outcomes and then do the extra step of adding those probabilities together. This process is illustrated below.

In the following example, we need to read each part of the question carefully to identify the required outcome, or set of outcomes, before calculating the associated probabilities.

Example 3: Using a Tree Diagram to Compute Probabilities

The given probability tree shows two events: 𝐴 and 𝐵. 𝐴 is the event of it raining and 𝐵 is the event of a group of friends playing football.

1. Determine the probability of the friends playing football and it raining.
2. Determine the probability of the group of friends playing football irrespective of whether it rains.

Answer

Recall that a tree diagram is a way of representing two (or more) successive events. To calculate the probability of an individual outcome, we multiply the probabilities along the branches leading to that outcome. To calculate the probability of a set of outcomes, we calculate the probabilities of the individual outcomes within the set and then add them all together.

In the above tree diagram, notice first the notation 𝐴 and 𝐵, which means “not 𝐴” and “not 𝐵.” This is called set notation and is often used in tree diagrams because many events have only two possible outcomes: either they happen or they do not. Here, since 𝐴 is the event of it raining and 𝐵 is the event of a group of friends playing football, then 𝐴 means that it does not rain and 𝐵 means that the group of friends do not play football.

Part 1

For part 1, we must determine the probability of the friends playing football and it raining. This corresponds to the single outcome “𝐴 and 𝐵” shown below.

To calculate the required probability, we multiply the probabilities along the branches to get 𝑃(𝐴𝐵)=0.3×0.2=0.06.and

Part 2

For part 2, we need to calculate the probability of the group of friends playing football irrespective of whether it rains. This is equivalent to calculating the probability of the two outcomes below where the friends play football, with or without it raining.

To calculate the probability of each individual outcome, we multiply the probabilities along the branches leading to that outcome. Then, we add the two results together. Therefore, the required probability is 𝑃(𝐴𝐵)+𝑃(𝐴𝐵)=0.3×0.2+0.7×0.7=0.06+0.49=0.55.andand

Notice that we could have shortened this calculation slightly by substituting 𝑃(𝐴𝐵)=0.06and directly from part 1, which would give the same answer.

Our next example is a word problem from which we must construct a tree diagram and then use it to compute some probabilities.

Example 4: Drawing a Tree Diagram and Using It to Compute Probabilities

Shady has entered a tennis tournament and wants to practice as often as possible to prepare for it. However, whether his coach, Maged, plays tennis with him on any given day depends on the weather. Based on past data, the probability that it will rain on any given day is 0.4. If it rains, the probability that Maged will play tennis with Shady is 0.05. If it does not rain, however, the probability that Maged will play tennis with Shady is 0.9.

1. Draw a tree diagram to represent Shady’s tennis preparation possibilities.
2. What is the chance that on any given day it rains and Shady and Maged play tennis?
3. What is the probability that Shady and Maged play tennis on any given day?

Answer

Part 1

Recall that a tree diagram is a way of representing two (or more) successive events. Reading through the question, we see that this scenario comprises two successive events. The first event concerns the weather conditions, namely whether it rains or not on any given day. The second event concerns whether or not Shady and Maged play tennis on that day.

For the first event, we are told that the probability that it rains is 0.4. Therefore, since the probabilities of the outcomes of any event must sum to 1, the probability that it does not rain must be 1−0.4=0.6.

Note also that the two events are dependent because the outcome of the first event affects the probabilities of the outcomes of the second event. If it rains, then the probability of Shady and Maged playing is only 0.05. Hence, the probability of them not playing is 1−0.05=0.95. If it does not rain, then the probability of Shady and Maged playing is 0.9. Hence, in this case, the probability of them not playing is 1−0.9=0.1. Writing “rain” and “no rain” for the weather outcomes and “yes” and “no” for the tennis-playing outcomes, we can construct the following tree diagram.

Part 2

To work out the chance that on any given day it rains and Shady and Maged play tennis, we must first identify the correct outcome (or outcomes) from the tree diagram. As shown below, this corresponds to the single outcome of “rain” followed by “yes.”

Recall that to calculate the probability of an individual outcome, we multiply the probabilities along the branches leading to that outcome. Therefore, the required probability is 𝑃()=0.4×0.05=0.02.“rain”and“yes”

Part 3

Finally, to work out the probability that Shady and Maged play tennis on any given day, we must again identify the correct outcome (or outcomes). Shady and Maged playing tennis regardless of the weather conditions corresponds to the two outcomes of “rain” followed by “yes” and “no rain” followed by “yes” as shown below.

Recall that to calculate the probability of a set of outcomes, we calculate the probabilities of the individual outcomes within the set and then add them all together. Hence, in this case we have 𝑃()+𝑃()=0.4×0.05+0.6×0.9=0.02+0.54=0.56.“rain”and“yes”“norain”and“yes”

Notice that to shorten this calculation, we could have substituted 𝑃()=0.02“rain”and“yes” directly from part 2, which would give the same result.

In our final example, we analyze a tree diagram that represents three successive events. As these scenarios give rise to a large number of possible outcomes, we can sometimes find useful shortcuts to reduce the amount of calculation needed when working out probabilities.

Example 5: Drawing a Tree Diagram and Using It to Compute Probabilities

A bag contains 20 gold coins and 11 silver coins. Three coins are chosen at random from the bag, without replacement. All of the possible outcomes are illustrated in the tree diagram shown below, where G represents a selected gold coin and S represents a selected silver coin.

1. Use the tree diagram to find the probability that all 3 of the chosen coins are gold. Give your answer as a fraction in its simplest form.
2. Find the probability that at least 1 of the chosen coins is silver. Give your answer as a fraction in its simplest form.

Answer

Recall that a tree diagram is a way of representing two or more successive events. To calculate the probability of an individual outcome, we multiply the probabilities along the branches leading to that outcome. To calculate the probability of a set of outcomes, we calculate the probabilities of the individual outcomes within the set and then add them all together.

The experiment shown in this tree diagram comprises three successive events of choosing a gold or a silver coin at random from a bag. In addition, note that the coins are chosen without replacement, so the events are dependent, meaning that the outcome of each event affects the probabilities of later ones.

Part 1

To work out the probability that all 3 of the chosen coins are gold, we first identify the correct outcome from the tree diagram, as shown below.

To calculate the probability, we multiply the probabilities along the relevant branches and then simplify the resulting fraction: 𝑃()=2031×1930×1829=684026970=228899.GGG

Part 2

To work out the probability that at least 1 of the chosen coins is silver, we again need to identify the correct outcome (or outcomes). The phrase “at least” is important here because it tells us that we need the set of outcomes that feature 1, 2, or 3 silver coins. However, since seven of the eight possible outcomes have this property, instead of multiplying along the branches and adding the seven results together for the required probability, we can apply the following simple trick.

As the probabilities of all the possible outcomes sum to 1, if we calculate the probability of the single outcome that does not have the property we want, we can then subtract it from 1 to get our answer. This single outcome is the one where 3 gold coins (and hence no silver coins) are chosen, and we already know that 𝑃()=228899GGG from part 1. Therefore, the probability that at least 1 of the chosen coins is silver is given by 1−𝑃()=1−228899=899899−228899=671899,GGG which is a fraction in its simplest form.

Note that, as expected, we obtain the same result if we sum the probabilities of the seven outcomes where at least 1 silver coin is selected, as shown below.

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