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8 Probability
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Australia
Year 9

8.03 Two-step experiments

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Introduction

Two-step experiments are those that incorporate two simple experiments, for example tossing a coin and rolling a die, or tossing a coin twice. Finding probabilities of two-step experiments is easier if we use a list, table, or tree diagram to show all possible outcomes.

Tables

A table is useful for showing all possible outcomes of two events in the rows and columns.

123456
\text{H}\text{H}1\text{H}2\text{H}3\text{H}4\text{H}5\text{H}6
\text{T}\text{T}1\text{T}2\text{T}3\text{T}4\text{T}5\text{T}6

For example, if we tossed 1 coin and 1 die we can show the outcomes for the coin along the first column on the left: \text{H}, \, \text{T}, and the outcomes for the die across the top row: 1, 2, 3, 4, 5, 6.

Each cell in the table is an outcome of rolling a die and a coin. There are 12 possible outcomes in the sample space.

Examples

Example 1

A player is rolling 2 dice and looking at their sum. They draw up a table of all the possible dice rolls for two dice and what they sum to.

123456
1234567
2345678
3456789
45678910
567891011
6789101112

What is the probability the dice will sum to 8?

Worked Solution
Create a strategy

Use the formula \text{Probability}=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}

Apply the idea

From the table we can see that there are 6\times 6=36 possible outcomes in total, 5 of which are an 8.

\displaystyle \text{Probability} \displaystyle =\displaystyle \dfrac{5}{36}Substitute the values
Idea summary

A table is useful for showing all possible outcomes of two events in the rows and columns.

Tree diagrams

A tree diagram is useful in tracking two-step experiments. It is named because the diagram that results looks like a tree.

It is useful if the events have different weightings or are unequal events.

This image shows tree diagram with labels Branch, Probability, and Outcome. Ask your teacher for more information.

The important components of the tree diagram are:

  • Branch

  • Probability (used when the events have unequal probabilities)

  • Outcome

When the outcomes are not equally likely the probability will be written on the branches.

The sum of the probabilities on branches from a single node should sum to 1.

When a single trial is carried out, we have just one column of branches.

Here are some examples. None of these have probabilities written on the branches because the outcomes are equally likely.

This image shows 3 tree diagrams with one column of outcomes. Ask your teacher for more information.

Here are some examples that have probabilities on the branches, because they do not have an equal chance of occurring.

3 tree diagrams with one column of outcomes and probabilities on the branches. Ask your teacher for more information.

When looking at a group of branches that come from a single point, the sum of the group always adds to 1 (or 100\%). This indicates that all the outcomes are listed.

When more than one experiment is carried out, we have two (or more) columns of branches.

A tree diagram for tossing a coin twice. The first column of outcomes is T and H, which both branch out to another T and H.

Here is an example. This one does not have the probabilities written, because the outcomes are equally likely.

This tree diagram shows the outcomes and sample space for tossing a coin twice. To calculate the probability of HH we would get \dfrac{1}{4}, or the probability of just 1 head, would be \dfrac{2}{4} as the outcomes HT and TH both have just one head.

This image shows a tree diagram on the outcomes of playing two games of tennis. Ask your teacher for more information.

Here is an example that have probabilities on the branches. This probability tree diagram shows the outcomes of playing two games of tennis where the probability of winning is 0.3 and the probability of losing is 0.7.

The probabilities of the events are multiplied along each branch, for example the probability of winning both games is 9\% which is found by \\ 0.3 \times 0.3 = 0.09

To find the probability of at least 1 win, we could do either

a) \text{P(win, lose)} + \text{P(lose, win)} + \text{P(win, win)} = 21\% + 21\% + 9\% = 51\% or

b) Use the complementary event of losing both games and calculate:

1 - \text{P (lose, lose)} = 1- 49\% = 51\%

For multistage events where the next stage is affected by the previous stage, we call these dependent events. We need to take care when drawing the tree diagram accordingly.

One type of experiment that is dependent on previous trials is an experiment without replacement. This means that the object selected (e.g. card, marble, person) is not able to be selected in the second selection.

This image shows a tree diagram for drawing two cards from a pack of 52 cards. Ask your teacher for more information.

For example, the probability of drawing a red card from a standard pack of 52 cards is \dfrac{26}{52} = \dfrac{1}{2}. If we do draw a red card and choose to select a second card without replacement there are only 25 red cards left, but there are still 26 black cards. And there are only 51 cards left in the entire deck. The probability of selecting a second red card is \dfrac{25}{51}. This can be seen in the top branches of the tree diagram.

Examples

Example 2

This image shows a tree diagram on the outcomes of two sets of traffic lights shows red, yellow or green light.

On the island of Timbuktoo, the probability that a set of traffic lights shows red, yellow or green is equally likely. Christa is travelling down a road where there are two sets of traffic lights.

What is the probability that both sets of traffic lights will be red?

Worked Solution
Create a strategy

Find the number of favourable outcomes red and red or RR in the probability tree.

Apply the idea

From the tree diagram below we can see that there is only 1 outcome for "red and red" or RR, which is shown in the bottom branch.

We can also see that there are 9 possible outcomes.

P(RR)=\dfrac{1}{9}

Example 3

Han owns four green ties and three blue ties. He selects one of the ties at random for himself and then another tie at random for his friend.

A tree diagram where the first column of outcomes is G and B, which both branch out to another G and B.

The colour combinations of the two ties he selects are shown in this tree diagram.

a

Write the probabilities for the outcomes on the edges of the probability tree diagram.

Worked Solution
Create a strategy

The probability of picking a tie will be given by:\dfrac{\text{Number of coloured ties}}{\text{Total number of ties}}

Apply the idea

Initially Han has a choice of 4 green and 3 blue ties, out of a total of 7 ties.

So on the first set of branches: P(G)=\dfrac{4}{7} and P(B)=\dfrac{3}{7}.

Assuming his first selection was green, Han would then have a choice of 3 green and 3 blue ties, out of a total of 6 ties.

So on the top set of branches after G: P(G)=\dfrac{3}{6} and P(B)=\dfrac{3}{6}.

Assuming his first selection was blue, Han would then have a choice of 4 green and 2 blue ties, out of a total of 6 ties.

So on the bottom set of branches after B: P(G)=\dfrac{4}{6} and P(B)=\dfrac{2}{6}.

So our probability tree would be as follows:

A probability tree diagram with two columns of outcomes G and B. Ask your teacher for more information.
b

What is the probability that Han selects a blue tie for himself?

Worked Solution
Create a strategy

Follow the path to the first B in the probability tree diagram.

Apply the idea

Using the tree diagram from part (a), the probability that Han selects a blue tie for himself is \dfrac{3}{7}.

c

Calculate the probability that Han selects two green ties

Worked Solution
Create a strategy

To find the probabilities follow the top path to GG and multiply the probabilities.

Apply the idea

The probability that the first tie was green is \dfrac{4}{7}, and the probability that the second tie was also green is \dfrac{1}{2}.

\displaystyle P(GG)\displaystyle =\displaystyle \dfrac{4}{7} \times \dfrac{1}{2}Multiply the probabilities
\displaystyle =\displaystyle \dfrac{4}{14} Evaluate
\displaystyle =\displaystyle \dfrac{2}{7} Simplify

The probability that Han selects two green ties is \dfrac{2}{7}.

Idea summary

A tree diagram is useful in tracking two-step experiments. It is named because the diagram that results looks like a tree.

For two-step experiments:

Multiply along the branches to calculate the probability of individual outcomes.

Add down the list of outcomes to calculate the probability of multiple options.

The final percentage should add to 100, or the final fractions should add to 1 - this is useful to see if you have calculated everything correctly.

Outcomes

ACMSP225

List all outcomes for two-step chance experiments, both with and without replacement using tree diagrams or arrays. Assign probabilities to outcomes and determine probabilities for events

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