Repeated chance experiments & independent events
Theory
Repeated chance experiments are a Year 4 chance skill: running the same experiment many times, working out the expected results and comparing them with what really happens. This page also explains independent events, where one result does not change the next (coins and dice have no memory, and counters are put back), and dependent events, where it does (counters are not put back). Curriculum: VC2M4P02, VC2M4P01 / AC9M4P02, AC9M4P01.
A repeated chance experiment is one you do many times, such as spinning a spinner \(30\) times. Each go is a trial. The expected result is what should happen on average: with \(3\) equal colours and \(30\) spins, about \(10\) of each. The observed results are what really happened, and they usually vary a little from the expected.
Two events are independent when the first does not change the chance of the second. Tossing a coin and then rolling a die are independent. A coin that has landed on tails six times in a row is still just as likely to land on heads or tails next, because it has no memory.
Two events are dependent when the first changes the chance of the second. If you take a counter from a bag and do not put it back, the bag has changed, so the chances for the next draw have changed too. If you put it back, the chances stay the same.
These facts help you plan a chance experiment and think about what to expect.
How to plan and run a repeated chance experiment
- Plan it. Choose the object (coin, die, spinner or bag), how many trials to do, and whether counters are put back.
- Predict. Work out the expected count for each outcome: trials \(\div\) number of equal parts (or half, a quarter, and so on).
- Record the results. Tally every trial in a table, then count each outcome and check the total.
- Compare and explain. Compare observed with expected. Small differences are normal variation; a very big difference may mean the object is not fair. More trials give a better picture.
No. The die and the spinner do not affect each other. Events like this are independent: the first result does not change the chance of the second.
No. The die does not remember earlier rolls. Each roll is independent, so every number, including \(3\), still has the same chance.
| Red | Blue | |
|---|---|---|
| Heads (H) | H, red | H, blue |
| Tails (T) | T, red | T, blue |
Pair each coin result with each spinner result. A table makes sure none is missed.
There are \(4\) outcomes: H and red, H and blue, T and red, T and blue. They are all equally likely, because the coin and the spinner are both fair.
(i) If it is put back, which colour is more likely next?
(ii) If it is not put back, which colour is more likely next?
(i) Green: the bag still has \(3\) green and \(2\) yellow, so the chances have not changed (independent).
(ii) Neither: now there are \(2\) green and \(2\) yellow, so they are equally likely. Not putting it back changed the chances, so the second draw depends on the first.
Picking C next is impossible, because the C card is no longer there. A, B and D are left, so A is now one of \(3\) equally likely cards.
| \(36 \div 3\) | \(=\) | \(12\) |
Expect each colour about \(12\) times. The real results will usually be a little different.
| \(\text{half of }40\) | \(=\) | \(20\text{ yellow}\) |
| \(\text{a quarter of }40\) | \(=\) | \(10\text{ red}\) |
| \(\text{a quarter of }40\) | \(=\) | \(10\text{ green}\) |
Expect about \(20\) yellow, \(10\) red and \(10\) green. Check: \(20 + 10 + 10 = 40\).
(i) Which numbers came up more than expected?
(ii) How many fewer \(5\)s came up than expected?
| \(30 \div 6\) | \(=\) | \(5\) |
| \(\text{more than 5}\) | \(2\ (6\text{ times}),\ 4\ (7\text{ times})\) | |
| \(5 - 3\) | \(=\) | \(2\) |
(i) \(2\) and \(4\).
(ii) \(2\) fewer. The results are close to expected but not exact — that is normal variation.
| \(\text{Sam expects}\) | \(5,\ \text{got } 7\) | |
| \(\text{Ana expects}\) | \(50,\ \text{got } 52\) |
Ana's result is closer to half. With more trials, results usually get closer to what is expected; a small number of trials can vary a lot.
| \(60 \div 3\) | \(=\) | \(20\) |
| \(41 - 20\) | \(=\) | \(21\) |
Each colour was expected about \(20\) times. Red came up \(21\) more times than expected, which is far too many to be normal variation, so the spinner probably is not fair.
Common pitfalls
Frequently asked questions
What are independent events?
Independent events do not affect each other. Rolling a die and then spinning a spinner are independent, because the die result cannot change the spinner.
What are dependent events?
Dependent events affect each other. Taking a counter out of a bag and not putting it back changes what is left, so the chance of the next draw changes.
Does a coin remember what it landed on before?
No. A fair coin has no memory, so after any number of tails in a row, heads and tails are still equally likely on the next toss.
How do you work out the expected results of an experiment?
Divide the number of trials by the number of equal parts. A spinner with \(5\) equal colours spun \(50\) times should give about \(50 \div 5 = 10\) of each colour.
Why are the observed results different from the expected results?
Each trial is up to chance, so real results vary. They are usually close to the expected results, and they tend to get closer when you do more trials.
How can results show that a spinner might not be fair?
If one colour comes up far more often than expected over many spins, much more than normal variation, the spinner is probably not fair.