Probability
UK Secondary (7-11)
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Frequency Trees
Lesson

Frequency trees provide a way of calculating probabilities when a chain of outcomes leads to a compound event.

In statistics, the word frequency means the number of occurrences of something.

Example 1

Look at the following diagram. The labels on the tree branches indicate the rules used to split a main group into smaller frequency sub-groups. The numbers are the frequencies for each outcome.

 

At each stage, there is a column of boxes forming the tree. The numbers in the boxes for each column have to sum to the same total- the number in the far left-hand box. This is because at each stage we are just splitting the numbers into smaller frequency groups.

In the fictitious experiment illustrated by the diagram, a sample of $56$56 experimental subjects (people) was partitioned into categories in two steps. People in the experiment were considered to be either left- or right-handed and people were considered to be tall or short depending on whether their height was above or below some chosen medium height.

We can draw some conclusions about the proportions of subjects in each category. Here are some examples: 

  • The proportion of left-handed subjects is $\frac{17}{56}$1756.
  • The proportion of tall subjects among the right-handed group is $\frac{25}{39}$2539.
  • The proportion of tall subjects among the whole group is $\frac{25+11}{56}=\frac{36}{56}=\frac{9}{14}$25+1156=3656=914.
  • The proportion of short left-handed people in the whole group is $\frac{6}{56}$656.

 

In a properly constructed statistical experiment, the proportions deduced from the frequency tree would be taken to be estimates of the probabilities associated with each compound outcome. Note that this technique can help to answer questions in probability where outcomes are connected by the words AND and OR.

Example 2

With reference to the diagram above, we might ask: What is the probability that a person is tall AND left-handed? Or, we might ask: What is the probability that a person is tall OR left-handed.

We can see immediately that there are $11$11 out of the $56$56 people who are both tall and left-handed. So, the probability estimate for being tall AND left-handed is $\frac{11}{56}$1156.

There are $25+17=42$25+17=42 people who are either tall or left-handed or both. So, the probability estimate for being tall OR left-handed is $\frac{42}{56}=\frac{3}{4}$4256=34.

 

Worked Examples

Question 1

There are $20$20 teachers at a high school, and $13$13 of them like football.

  1. Complete this frequency tree:

    like football $\editable{}$
    $20$20
    don't like football $\editable{}$

Question 2

There are $44$44 chimpanzees in an enclosure at the zoo.

There are $23$23 males in the enclosure.

Of the males, $19$19 like ice cream.

Of the females, $5$5 do not like ice cream.

  1. Complete this frequency tree:

    like ice cream $19$19
    males $23$23
    don't like ice cream $\editable{}$
    $44$44
    like ice cream $\editable{}$
    females $\editable{}$
    don't like ice cream $5$5
  2. Find the probability that a chimpanzee chosen at random is a female who likes ice cream.

Question 3

The local cinema kept track of every customer who came to see a new film, recording their gender, and whether they bought no popcorn, a small popcorn, or a large popcorn. Both male and female customers were split in a ratio $4$4:$5$5:$2$2 between buying no popcorn, a small popcorn, and a large popcorn. The two-way table below shows some information about the results.

  1. Complete the two-way table:

      No popcorn Small popcorn Large popcorn Total
    Men $\editable{}$ $\editable{}$ $28$28 $154$154
    Women $\editable{}$ $115$115 $\editable{}$ $253$253
    Total $148$148 $185$185 $\editable{}$ $\editable{}$
  2. Complete this frequency tree:

    large popcorn $\editable{}$
    women $\editable{}$ small popcorn $\editable{}$
    no popcorn $\editable{}$
    $\editable{}$
    large popcorn $\editable{}$
    men $\editable{}$ small popcorn $\editable{}$
    no popcorn $\editable{}$
  3. Find the probability that a cinema customer chosen at random will be a man who bought a small popcorn.

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