Video Transcript
In this video, we will learn how to
draw and use tree diagrams to represent the sample space of an experiment.
Before we look at tree diagrams,
let’s think about the sample space. This is the set of all possible
outcomes or results of an experiment. So, what would that look like? If we spun this spinner, then the
sample space would be one, two, three, and four, since that’s all the possible
results that we could get. On a different spinner with values
of one, two, three repeating, then the sample space here would be one, two, and
three.
Let’s take a look at another
example where we find the sample space.
A bag contains seven balls
which are numbered one to seven. Determine the sample space of
choosing a ball at random.
In order to find the sample
space, we need to list all the outcomes. If we choose a ball at random,
we could choose the ball numbered one, or the ball numbered two, or, in fact,
any of the balls numbered three, four, five, six, and seven. We could, therefore, write the
sample space as the set of the values one, two, three, four, five, six, and
seven.
Let’s have a look at tree diagrams
now and see how they can represent the sample space. Tree diagrams are particularly
useful to find the probabilities of multiple events. Let’s take the example where we
want to find the outcomes that are possible when we flip a coin two times. We begin drawing a tree diagram to
represent the first flip of the coin. We know that there are two possible
outcomes on the coin flip: heads or tails. The probability of getting heads
would be one over two — that’s a half — as there’s one heads out of two
outcomes. It’s the same probability to get
tails. For the second event of flipping
the coin for the second time, once we’ve got heads, there’re still two options,
heads or tails, on the second flip. The probabilities here will still
be the same, since what happened on the first flip doesn’t affect the probability of
what happens on the second flip.
The lower branches represents what
happens on the second flip if we got tails on the first flip. We still have two possible
outcomes. And both of these probabilities are
a half. So, following along, the top branch
gives us the outcome of getting heads and heads. The second outcome is heads
followed by tails. The third outcome is tails followed
by heads. And the fourth outcome is tails,
tails. We can see that there were four
possible outcomes for flipping a coin two times.
Let’s take a look now at an example
of creating a tree diagram for the sample space of a more complex problem.
Suppose 60 percent of single-scoop
ice cream from the local ice cream van is sold in a carton and the other 40 percent
is sold in a cone. Suppose also that the ice cream van
sells three different flavors of ice cream — chocolate, vanilla, and strawberry —
and that 25 percent of single-scoop sales are chocolate, 45 percent are vanilla, and
30 percent are strawberry. Draw a tree diagram to represent
single-scoop ice cream sales from the ice cream van.
Here, we need to draw a tree
diagram to represent the possible outcomes when buying ice cream from the ice cream
van. The first possibilities we have
here are buying ice cream in a carton or ice cream in a cone. We then have different options in
our three different flavors of ice cream. So, looking at the first item of
choosing a carton or a cone, we’re told that 60 percent is sold as a carton. So, we write this along the branch
for the carton. We can write it as a percentage,
but often it’s easier to write it as a decimal or a fraction. And here, 60 percent is equivalent
to 0.6. We’re told that 40 percent is sold
in a cone. 40 percent is equivalent to
0.4. So, we can write that along the
branch for the cone.
We can now look at the different
options for the ice cream flavors. It doesn’t matter whether the
customer chooses a carton or a cone. They’re still going to then choose
between three different flavors of ice cream. They could choose a carton and then
choose chocolate, vanilla, or strawberry ice cream. Or they could choose a cone with
chocolate, vanilla, or strawberry ice cream. We’re told that 25 percent of the
sales are chocolate, 45 percent are vanilla, and 30 percent are strawberry. As an aside, we can see that these
three probabilities would add up to 100 percent as that’s the only options for the
flavors that we have. As a decimal, 25 percent is
0.25. And this will be the same for the
chocolate on the upper branches and the chocolate on the lower branches. The probability of a customer
selecting vanilla would be 0.45 as a decimal and for strawberry will be 0.3.
And we have completed a tree
diagram to show the ice cream sales. We can see here that there are six
possible outcomes: choosing a carton with chocolate ice cream, a carton with vanilla
ice cream, a carton with strawberry ice cream or a cone with chocolate, a cone with
vanilla, or a cone with strawberry ice cream. For a complete tree diagram,
particularly in an exam question, all we need to do is show the different outcomes
along with their probabilities.
We’ll now look at how we use tree
diagrams to actually calculate some probabilities. Let’s return to our example of
flipping the coin twice.
In the tree diagram that we drew
for this earlier in the video, we saw that there were four outcomes. We’ll think about how we calculate
the probabilities of these four outcomes. Starting with the probability of
getting heads and then heads, as we move along the branches, we multiply the
probabilities. So, we have a half multiplied by a
half. And when multiplying fractions, we
multiply the numerators and multiply the denominators. So, we’ve get one-quarter. To find the probability of getting
heads and then tails, we multiply a half by a half which is one-quarter. The probability of getting tails
then heads is also a half times a half which is one-quarter. And the probability of getting two
tails is one-quarter.
One of the key things to highlight
about this is that the sum of all the probabilities of these final outcomes is
one. Let’s imagine we were asked what
the probability is of getting two coins the same. That means we need to think about
the probability of getting heads and heads or the probability of getting tails and
tails. To calculate this probability, we
need to add the probability of getting heads, heads and the probability of getting
tails, tails, which would mean adding one-quarter and one-quarter, giving us an
answer of one-half.
We’ll now look at an example
question, where we need to add the final probabilities at the end of the
branches.
If two spinners are spun, where
the first is numbered from one to two and the second from one to nine, determine
the probability of both spinners stopping at even numbers using a tree
diagram.
There are a number of different
ways we could represent the sample space or different outcomes of these two
spinners. But here, we’re asked to use a
tree diagram. In our tree diagram, we need to
represent the outcomes of the first spinner and then the outcomes of the second
spinner. On our first spinner, there are
just two possible outcomes, the value one or the value two. We can assume for the
probability that these spinners are fair. And therefore, the probability
of getting one will be a half as that’s one outcome out of our two possible
outcomes. And the probability of spinning
a number two is also a half.
Let’s now look at the possible
outcomes on the second spinner. When we’re drawing tree
diagrams, it’s always worth thinking ahead to what the outcomes will look
like. For example, if we drawn our
first spinner values very small and close together, then we would have struggled
to get nine different branches after the one and the two. So, always make sure you have
plenty of space for a tree diagram. As the second spinner is
numbered from one to nine, we have our nine different branches. And we have these nine outcomes
whether or not we get a one or a two on the first spinner. Assuming again that this is a
fair spinner, then the probability of each of the numbers one through nine will
be one-ninth.
We’re asked to calculate the
probability of both spinners stopping at even numbers. We can see the outcomes on the
top set of branches. One, one, for example,
indicates getting a one on the first spinner and a one on the second
spinner. One, two would indicate a one
on the first spinner and a two on the second spinner. As we’re considering the
probability of getting two even numbers, then we can in fact discount all of the
upper branches because all of these will have a number one and that’s an odd
number.
So, to find the probability of
two even numbers, that means we’re interested in the probability of spinning a
two and then a two or the probability of spinning a two on the first spinner and
a four on the second spinner or the probability of getting a two and then a six
or the probability of getting a two and an eight. To work out these probabilities
then, we multiply along the branches. So, the probability of getting
a two and a two will be a half times a ninth. And to multiply fractions, we
multiply the numerators and the denominators. So, we’ll have one
eighteenth. The values are the same for
working out the probability of getting a two and a four. We have a half multiplied by a
ninth, giving us one eighteenth. And the same is true for the
remaining two probabilities. And therefore, to work out the
probability of getting two even numbers, we add each of our four probabilities
together, giving us a final answer of four out of 18 or four eighteenths.
When Daniel gets up on any given
morning, there is a 30 percent chance he will drink tea, a 50 percent chance he will
choose coffee, and a 20 percent chance he will opt for juice. For breakfast, Daniel has toast 40
percent of the time, pancakes 15 percent of the time, cereal 30 percent of the time,
and nothing if he gets up too late and has to rush. Draw a tree diagram to represent
Daniel’s breakfast food and drink options. Find the probability that Daniel
has coffee with toast in the morning. Find the probability that Daniel
has a hot drink with toast. Work out the number of breakfast
options Daniel has each morning.
To start drawing the tree diagram,
we need to consider the different options. First, we have all the options of
drinks for Daniel and, secondly, we have all the options of the things that he
eats. Let’s start our branches
representing the drinks that he has. These will be tea, coffee, or
juice. So, that means that we’ll need
three branches. So, we draw the three branches with
the written options at the end.
We’re told that there’s a 30
percent chance that Daniel will drink tea. This will be equivalent to the
decimal value 0.3, which we write along the branch. There’s a 50 percent chance that
Daniel chooses coffee. And as a decimal, that is 0.5. And we’re told that there is a 20
percent chance that Daniel drinks juice. And as a decimal, that’s 0.2.
We’ll now consider the breakfast
food options that Daniel has. We’re told he could have toast,
pancakes, cereal. But we also need to include the
fact that he could often have nothing to eat. So, if Daniel has tea to drink, he
has four options for his food: toast, pancake, cereal, or nothing to eat. We’re told he has toast 40 percent
of the time. So, this will be 0.4 as a
decimal. He chooses pancakes 15 percent of
the time. That’s 0.15. Cereal is chosen 30 percent of the
time. That’s 0.3. And we’re not given a percentage
for having nothing if Daniel gets up too late. We can, however, work it out by
recalling that each of these four branches will sum to one. Therefore, if we add 0.4, 0.15, and
0.3 and subtract it from one, we’ll get 0.15.
Next, when Daniel has coffee to
drink, he still has the same four options for what he eats. The probabilities of these will all
be the same as in the branches above because what he has to drink doesn’t affect
what he has to eat. The same is true when Daniel has a
juice to drink. He has four different options for
food. We have, therefore, drawn a tree
diagram for the first question. We can use our tree diagram then to
find the probability that Daniel has coffee with toast.
We can find this probability by
going along the branch for coffee and then along the branch for toast. And as we travel along the
branches, we multiply the probabilities. So, we’ll have 0.5 multiplied by
0.4, which is equal to 0.2. And so, our answer to the second
question, the probability that Daniel has coffee with toast, is 0.2.
We’ll now look at the third
question to find the probability that Daniel has a hot drink with toast. As juice isn’t usually a hot drink,
that means we’re looking at the probability of coffee and toast and the probability
of tea and toast. We worked out in the previous
question that the probability of getting coffee and toast is equal to 0.2. And to find the probability of tea
and toast, we multiply along the branches again, giving us 0.3 multiplied by 0.4,
which is 0.12. To find the probability of Daniel
drinking a hot drink and toast, we add together the probability of tea and toast and
the probability of coffee and toast. This is equal to 0.12 plus 0.2. So, our answer to the third
question is 0.32.
Let’s now look at the final part of
this question to work out the number of breakfast options that Daniel has. We have seen in the last two parts
of this question that Daniel has the option of tea and toast or the option of coffee
and toast. In fact, he has all of these other
options listed as well. If we added all of these options
together, we would see that there are 12, which will answer our final question that
he has 12 different breakfast options. We could check this by realizing
that he has three different drink options and four different food options. Multiplying three by four would
also give us 12.
We can now summarize what we’ve
learned in this video. We learned that a tree diagram is a
helpful way to illustrate the sample space of an experiment, where the sample space
is the set of all possible outcomes. We learned that each stage of an
experiment is represented by a set or sets of branches on a tree diagram. Each outcome is represented by a
branch with the probability attached. Each subsequent stage in the tree
diagram is represented by a branch from every outcome of the previous stage. The probability of a final outcome
is found by multiplying along the branches leading to that outcome. The probability of combined final
outcomes is found by adding the probability of individual final outcomes. And a final point about tree
diagrams, it can be useful to plan these out to make sure that we have enough space
on our page, particularly when the tree diagram is very large.
