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Year 6 Maths Chance

Conduct chance experiments; observed vs expected

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Theory

In a chance experiment each go is a trial. The expected number of times an outcome happens is its probability times the number of trials; the observed number is what actually happens. This Year 6 maths page shows how to work out expected results, compare them with observed results, and why more trials give a clearer picture.

A trial is one go of the experiment: one coin flip, one roll, one spin.

Expected number \(=\) probability \(\times\) number of trials. A die rolled \(120\) times: each number is expected \(\dfrac{1}{6} \times 120 = 20\) times.

Observed results are what really happens. They are usually close to, but not exactly, the expected results.

The more trials you do, the closer the observed fractions usually get to the expected probabilities.

Each trial is independent: a coin does not remember. After four tails in a row, the chance of a head is still \(\dfrac{1}{2}\).

Expected and observed results of 120 die rolls A side-by-side column graph for a die rolled 120 times. Expected: 20 for every number. Observed: 1 came up 18 times, 2 came up 23, 3 came up 19, 4 came up 17, 5 came up 24 and 6 came up 19. Expected Observed 051015202530 Number of rolls 20 18 1 20 23 2 20 19 3 20 17 4 20 24 5 20 19 6 Number rolled
The observed columns are close to the expected 20, but not exactly 20.
Spinner with two yellow and three blue sectors A spinner with five equal sectors: two yellow and three blue. yellow blue yellow blue blue
\(P(\text{yellow}) = \dfrac{2}{5}\). In \(50\) spins, expect \(20\) yellows.
Coin results from few and many flips A table: Lily flipped a coin 10 times and got 7 heads, a fraction of 0.7. Noah flipped 500 times and got 243 heads, a fraction of 0.486. Student Flips Heads Fraction Lily 10 7 0.7 Noah 500 243 0.486
Noah’s many flips give a fraction much closer to \(0.5\).

Expected results:

\[ \text{expected} = \text{probability} \times \text{number of trials} \]
expected = probability × number of trials
\[ \text{difference} = \text{observed} - \text{expected} \]
difference = observed − expected
ExperimentTrialsExpected
coin: heads\(50\)\(25\)
die: a four\(120\)\(20\)
3 equal sectors: red\(45\)\(15\)
Close, not exact. Observed results that are a little above or below the expected number are normal.
Far off is a warning. If one outcome is far from expected over many trials, the spinner or die may not be fair.

How to find an expected number

  1. Find the probability of the outcome.
  2. Multiply by the number of trials.

How to compare with observed results

  1. Record the results in a table (tally, then count).
  2. Subtract to see how far each count is from the expected number.
  3. Decide: small differences are normal; large differences over many trials suggest it is not fair.
Example 1 — Coin
A coin is flipped \(50\) times. How many heads would you expect?
Solution
\(\dfrac{1}{2} \times 50\)\(=\)\(25\)
1/2×50=25

\(25\) heads.

Example 2 — Die
A die is rolled \(120\) times. How many fours would you expect?
Solution
\(\dfrac{1}{6} \times 120\)\(=\)\(20\)
1/6×120=20

\(20\) fours.

Example 3 — Three-colour spinner
A spinner with \(3\) equal sectors is spun \(45\) times. How many times would you expect each colour?
Solution
\(45 \div 3\)\(=\)\(15\)
45÷3=15

\(15\) times each.

Example 4 — Unequal sectors
Use the spinner with \(2\) yellow and \(3\) blue sectors. It is spun \(50\) times. How many yellows would you expect?
Solution
\(\dfrac{2}{5} \times 50\)\(=\)\(20\)
2/5×50=20

\(20\) yellows (and \(30\) blues).

Example 5 — Compare with expected
Use the graph of \(120\) die rolls.
(i) How many more fives than expected came up?
(ii) How many fewer fours than expected?
Solution
\(24 - 20\)\(=\)\(4\)
\(20 - 17\)\(=\)\(3\)
24−20=420−17=3

(i) \(4\) more
(ii) \(3\) fewer

Example 6 — Is the die fair?
Do the \(120\) rolls suggest the die is fair?
Solution

Yes. Every count is within \(4\) of the expected \(20\); small differences like these are normal.

Example 7 — A missing count
A spinner with \(4\) colours was spun \(60\) times: red \(17\), blue \(14\), green \(12\). How many yellows?
Solution
\(60 - 17 - 14 - 12\)\(=\)\(17\)
60−17−14−12=17

\(17\) yellows.

Example 8 — Not fair
A spinner with \(5\) equal colours is spun \(100\) times and lands on red \(47\) times. What does this suggest?
Solution
\(100 \div 5\)\(=\)\(20\)
\(47 - 20\)\(=\)\(27\)
100÷5=20,47−20=27

Red came up \(27\) more times than expected in many trials, so the spinner is probably not fair.

Example 9 — More trials
Use the coin table. Whose results are closer to the expected fraction of heads, and why?
Solution

Noah’s: \(0.486\) is much closer to \(0.5\) than \(0.7\). More trials usually bring results closer to what is expected.

Example 10 — Find the mistake
A coin has landed tails \(4\) times in a row. Ben says the next flip is very likely to be heads. What is right?
Solution

Each flip is independent. The chance of heads is still \(\dfrac{1}{2}\).

Common pitfalls

Expecting exact results. \(23\) fives instead of \(20\) does not mean the die is unfair.
Judging from a few trials. \(7\) heads in \(10\) flips can easily happen with a fair coin.
Thinking a coin remembers. Past results do not change the chance of the next trial.
Forgetting unequal sectors. With \(2\) of \(5\) sectors yellow, expect \(\dfrac{2}{5}\) of the spins, not half.

Frequently asked questions

What is an expected result?

The number of times an outcome should happen: probability times the number of trials.

What is an observed result?

What actually happens when you do the experiment.

Why are observed results different from expected results?

Chance varies from trial to trial, so the counts are rarely exact.

Why do more trials help?

With more trials, the observed fractions usually get closer to the true probabilities.

After 4 tails in a row, is a head more likely?

No. Each flip is independent; the chance is still \(\dfrac{1}{2}\).

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