# Tree Diagrams

## Tree Diagrams Revision

**Tree Diagrams**

**Tree diagrams** are used to display **probabilities**, and describe how** likely** it is that events will happen.

**Introduction: Combined Probability**

Combined **probability** means there are two or more events occurring.

For two events, these can be split into two **SEPARATE** events, and then a rule can be applied to work out the overall **probability**:

- The
**AND**rule - The
**OR**rule

Events may be** independent** – the outcome of one does not affect the outcome of another; or **dependent** – the outcome of one does affect the outcome of another.

**The AND Rule**

Where event one and event two are** independent** events, the** AND** rule means both event one **AND** event two occur. Let’s call event one A and event two B.

The** probability** of Event A and Event B **both** happening is the same as the **probability** of Event A happening multiplied by the **probability** of Event B happening, or:

\text{P}(A \text{ and } B) = \text{P}(A)\times\text{P}(B)

**Example:**

If the **probability** of rain is 0.7 and the** probability** of thunder is 0.2, what is the** probability** of rain** AND** thunder?

Call** probability** of rain, \text{P}(R) = 0.7, and thunder, \text{P}(T)=0.2

Use the **AND** equation:

0.7\times0.2=0.14

So, the **probability** of rain **AND** thunder is 0.14

**The OR Rule**

The** OR** rule means either Event A **OR** Event B occur, where both Events are mutually exclusive (this means, the events cannot happen at the same time)

The** probability** of Event A **OR** Event B happening is the same as the **probability** of Event A happening added to the **probability** of Event B happening, or:

\text{P}(A \text{ and } B) = \text{P}(A)+\text{P}(B)

**Example:**

A biased dice is thrown with the **probabilities** given below,

Calculate the** probability** of a 4 or a 6 being rolled.

Use the **OR** equation:

0.31+0.16=0.47

So, the **probability** of rolling a 4 **OR** a 6 is 0.47

**Tree Diagrams – Independent Events**

Firstly, let’s look at how to construct and understand** tree diagrams** where the events are **independent**.

For **example**, a bag contains 3 **red** counters and 7 **blue** counters.

Lauren picks 2 counters at random. After she picks the first, she puts it back in the bag and then randomly takes the next. This means these events are independent, as the probabilities will be the same the second time she chooses a counter.

To construct a tree diagram, we need to consider **probabilities**. With total 10 counters in the bag,

\text{P}(R)=\dfrac{3}{10}\\

\text{P}(B)=\dfrac{7}{10}\\

These **probabilities** remain the same for the second selection as Lauren puts the ball back in the bag, so the **tree diagram** would look like this:

**Using the AND rule:**

**Example**: Find the** probability** that Lauren chooses a red counter **AND** then a blue counter.

So, on the first selection on the diagram you can see that the **probability** of red is \dfrac{3}{10}

Then, we can follow this red line to the second selection, where we meet the options red and blue again. If Lauren gets a blue counter on her second selection, this **probability** is \dfrac{7}{10}.

Therefore, using the **AND** rule:

**Using the OR rule:**

Example: find the **probability** Lauren chooses two counters of the same colour.

There are two ways Lauren could do this, she could get two red counters, or two blue counters.

To find the probability of two red counters, we need to firstly use the **AND** rule (as she is getting a red counter **AND** a red counter):

\dfrac{3}{10}\times\dfrac{3}{10}=0.09

Same for two blue:

\dfrac{7}{10}\times\dfrac{7}{10}=0.49

Now, she either gets two reds OR two blues, so let’s use the **OR** rule:

\text{P}(A)+\text{P}(B)=0.09+0.49=0.58

**Tree Diagrams – Dependent Events**

To draw **tree diagrams** for dependent events, you need to be more careful when putting the** probabilities** onto the diagram.

Let’s use the same example as earlier, but let’s say that Lauren does not replace the counters after the first selection.

Remember, the bag contains 3 **red** counters and 7 **blue** counters.

On the first selection, with all the counters in the bag, the** probabilities** will remain,

\text{P}(R)=\dfrac{3}{10}\\

\text{P}(B)=\dfrac{7}{10}\\

If Lauren picks a red counter on her first selection, there will only be 2 red counters left in the bag for her second selection, out of 9 possible counters, so the **probabilities** would be:

\text{P}(R)=\dfrac{2}{9}\\

\text{P}(B)=\dfrac{7}{9}\\

If Lauren picks a blue counter on her first selection, there will only be 6 bluecounters left in the bag for her second selection, out of 9 possible counters, so the **probabilities** would be:

\text{P}(R)=\dfrac{3}{9}\\

\text{P}(B)=\dfrac{6}{9}\\

And the** tree diagram** would be:

**The AND rule and the OR rule for dependent events**

Earlier, you learnt that the **AND** rule only work for independent events. However, tree diagrams are slightly different.

When plotting a dependent tree diagram, you take into account dependence when drawing probabilities onto the tree, and this means you can use the **AND** rule for any tree diagram.

Hence, you can use the **AND** rule for both independent and dependent tree diagrams.

For the **OR** rule, dependent events on a tree diagram are always mutually exclusive,

Hence, you can also use the **OR** rule for both independent and dependent tree diagrams.

**Example 1: Using the AND rule**

Lara is randomly choosing two crayons out of a crayon pot. The pot contains 11 green crayons and 9 pink crayons.

Work out the **probability** of neither of the crayons being pink.

**[4 marks]**

These events are not independent, as Lara does not replace the crayons, therefore we need to be careful when thinking of the **probabilities**. First selection** probabilities**:

\text{P}(G)=\dfrac{11}{20}\\

\text{P}(P)=\dfrac{9}{20}\\

If Lara picks a green counter first, there will only be 10 red counters left in the bag for her second selection, out of 19 possible counters, so the **probabilities** would be:

\text{P}(G)=\dfrac{10}{19}\\

\text{P}(P)=\dfrac{9}{19}\\

If Lara picks a pink counter first, there will only be 8 blue counters left in the bag for her second selection, out of 19 possible counters, so the **probabilities** would be:

\text{P}(G)=\dfrac{11}{19}\\

\text{P}(P)=\dfrac{8}{19}\\

And the **tree diagram** would be:

Now, back to the question. The probability that no pink crayons are chosen means Lara must choose two green crayons. Therefore, we can use the** AND** rule (Lara is picking one green** AND** then another green).

**Probability** of green on first selection =\dfrac{11}{20}

**Probability** of green on second selection = \dfrac{10}{19}

\text{P}(G\text{ and }G)=\dfrac{11}{20}\times\dfrac{10}{19}=\dfrac{11}{38}

**Example 2: Using the OR Rule**

Brendon and Kylie are taking maths exams. The **probability** than Brendon passes his maths exam is 0.2. The **probability** that Kyle passes her maths exam is 0.73. Work out the **probability** that exactly one of Brenden and Kylie pass their maths exam.

**[4 marks]**

These events are independent, so we do not need to work out **probabilities** to put onto the tree diagram.

But we need to work out the **probabilities** of Brenden and Kylie failing their exams (remember, **probabilities** add to 1):

\text{P(Brenden Fails)}=1-0.2=0.8

\text{P(Kylie Fails)}=1-0.73=0.27

We can now plot the **tree** with either Brenden first or Kyle first:

Now, the** probability** of exactly one of them passing their exam means either Brenden fails and Kylie passes, or Brenden passes and Kylie fails.

Let’s work out the **probability** of each using the AND rule:

\text{P}(\text{Brenden Fails AND Kylie Passes})=0.8\times0.73=0.584

\text{P}(\text{Kylie Fails AND Brenden Passes})=0.2\times0.27=0.054

So, using the **OR** rule, the **probability** of either of these ocurring is:

0.584+0.054=0.638

## Tree Diagrams Example Questions

**Question 1**: On a beach, the chance of rain is 0.3. If it rains, the chance of there being over 1000 people surfing is 0.28 whereas if there is no rain, the probability of over 1000 people surfing is 0.56. Express this using a probability tree.

**[2 marks]**

Find the missing probabilities,

\text{P(Not raining)}=1-0.3=0.7

\text{P(Not over }1000\text{ surfing if raining)}=1-0.28=0.72

\text{P(Not over }1000\text{ surfing if not raining)}=1-0.56=0.44

**Question 2**: A biased coin is tossed twice. The probability of it landing on heads is 0.6. Use a probability tree to work out the probability of it landing on tails twice.

**[4 marks]**

Probability tree:

Use the AND rule to find the probability of tails twice:

\text{P(Tails and Tails)}=0.4\times0.4=0.16

**Question 3**: There is a lucky dip at a school fair with 100 prizes in. There are 43 rubber balls and 57 lollipops in the lucky dip. Each person gets two goes to pick out a prize at the lucky dip. Find the probability that the first person who chooses will get two of the same prizes.

**[5 marks]**

Let’s express this as a tree diagram, remembering that the prize will not be replaced, so the second probabilities will be different:

Firstly, work out the probability of two balls being picked:

\dfrac{43}{100}\times\dfrac{42}{99}=\dfrac{301}{1650}

Probability of two lollypops being picked:

\dfrac{57}{100}\times\dfrac{56}{99}=\dfrac{266}{825}

Using OR rule, probability of the same prize:

\dfrac{301}{1650}+\dfrac{266}{825}=\dfrac{833}{1650}