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Year 5 Maths Chance

Repeated chance experiments: record & compare frequencies

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Theory

When a chance experiment is repeated many times, the results are recorded in a frequency table. This Year 5 maths page shows how to record and read the results, compare them with what you expect, and why the results vary but get closer to what is expected with more trials.

Each go is a trial. The frequency of an outcome is how many times it happened. The frequencies add up to the number of trials.

A result can be written as a fraction of the trials: \(18\) reds in \(50\) spins is \(\dfrac{18}{50}\).

We expect equally likely outcomes to happen about equally often: a fair coin about half heads, a fair die about \(\dfrac{1}{6}\) of the time for each number.

Results vary: two groups doing the same experiment usually get different counts. With more trials the results usually get closer to what is expected.

If results are very different from what is expected after many trials, the spinner, coin or die may not be fair.

Column graph of 50 spins A column graph of the results of 50 spins: red 18, blue 13, green 11, yellow 8. Scale 0 to 20 in steps of 2. 50 spins 02468101214161820 Frequency Red Blue Green Yellow Colour
Red came up most: \(18\) of \(50\) spins.
Column graph of 36 die rolls A column graph of 36 rolls of a die: 1 came up 5 times, 2 came up 7, 3 came up 6, 4 came up 4, 5 came up 8 and 6 came up 6. Scale 0 to 10 in steps of 2. 36 rolls of a die 0246810 Frequency 1 2 3 4 5 6 Number rolled
Expected about \(36 \div 6 = 6\) of each. The results vary around \(6\).
Spinner with a blue half A spinner with four equal sectors: two blue, one red, one yellow. Blue covers half the spinner. Blue Blue Red Yellow
Blue covers half, so expect blue about half the time.

Comparing results with expectations:

\[ \text{fraction} = \dfrac{\text{frequency}}{\text{number of trials}} \]
fraction = frequency ÷ trials
\[ \text{expected} \approx \text{trials} \times \text{chance} \]
expected ≈ trials × chance
\[ 36 \times \dfrac{1}{6} = 6 \]
36×1/6=6
Results vary. Getting 9 heads in 20 tosses is not unusual. Exact halves are rare.
More trials, better picture. Combine results or do more trials to get closer to what is expected.

Running and comparing an experiment

  1. Record each trial with a tally.
  2. Count the frequencies and check they add to the number of trials.
  3. Write results as a fraction of the trials, or graph them.
  4. Compare with what you expected, and decide whether the difference is small (normal variation) or large (perhaps unfair).
Example 1 — Total trials
Use the first diagram. How many spins were there, and how many more reds than yellows?
Solution
\(18 + 13 + 11 + 8\)\(=\)\(50\)
\(18 - 8\)\(=\)\(10\)
18+13+11+8=50,18−8=10

\(50\) spins; \(10\) more reds.

Example 2 — As a fraction
Use the first diagram. What fraction of the spins were blue?
Solution

\(\dfrac{13}{50}\).

Example 3 — Expected count
A fair die is rolled \(36\) times. About how many times do you expect each number?
Solution
\(36 \div 6\)\(=\)\(6\)
36÷6=6

About \(6\) each. The second diagram shows real results: between \(4\) and \(8\).

Example 4 — Most likely colour
Use the third diagram. Which colour is most likely, and about how many times in \(40\) spins?
Solution
\(40 \div 2\)\(=\)\(20\)
40÷2=20

Blue, about \(20\) times, because it covers half the spinner.

Example 5 — Combine results
Four groups each toss a coin \(20\) times and get \(9\), \(12\), \(10\) and \(11\) heads.
(i) How many heads altogether?
(ii) Out of how many tosses?
Solution
\(9 + 12 + 10 + 11\)\(=\)\(42\)
\(4 \times 20\)\(=\)\(80\)
42 heads in 80 tosses

(i) \(42\)
(ii) \(80\). Half of \(80\) is \(40\), so \(42\) is close to half.

Example 6 — Estimate a bag
A bag has \(6\) counters, blue or white. Draw one, record, put it back, \(30\) times: blue \(20\) times. About how many counters are blue?
Solution

Blue came out \(\dfrac{20}{30}\), about two-thirds of the time. Two-thirds of \(6\) is \(4\), so about \(4\) are blue.

Example 7 — Fair or not?
A spinner is half green and half orange. In \(60\) spins, green comes up \(47\) times. Does it seem fair?
Solution
\(60 - 47\)\(=\)\(13\)
60−47=13

No. We expect about \(30\) each, but green came up \(47\) times and orange only \(13\).

Example 8 — Find the mistake
Nia tossed a coin \(10\) times and got \(7\) heads. She says the coin must be unfair. Is she right?
Solution

Not necessarily. \(10\) tosses is very few, and results vary. She should do many more tosses.

Common pitfalls

Expecting exact results. Chance results vary around what is expected.
Judging from a few trials. Small numbers of trials can be misleading.
Frequencies that do not add up. Check the total equals the number of trials.
Ignoring sector sizes. A bigger sector should come up more often.

Frequently asked questions

What is a frequency?

How many times an outcome happened.

How do I write a result as a fraction?

Frequency over the number of trials: \(\dfrac{13}{50}\).

Why do two groups get different results?

Chance results vary every time.

Why are more trials better?

With more trials, unusual results have less effect, so the results get closer to what is expected.

How can I tell if a spinner is unfair?

After many spins, the results are very far from what its sectors would lead you to expect.

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