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Relative Frequency and Expected Outcomes

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How to estimate a probability from the results of an experiment, why more trials give a better estimate, how to work out the number of times you would expect something to happen, and how to decide whether a dice or spinner is fair.

Probability from an experiment

Sometimes you can't work out a probability by counting equally likely outcomes. A drawing pin can land point up or point down, but the two are not equally likely. Instead, you do an experiment: drop the pin lots of times and count what happens. Each drop is called a trial.

A dice or spinner is fair if every outcome is equally likely. If some outcomes are more likely than others, it is biased. For a biased dice, the only way to estimate the probabilities is to roll it lots of times and look at the results.

In the exam you are usually asked to "Work out an estimate for the probability" from someone's results, or to "Work out an estimate for the number of times" something will happen, for 1 to 3 marks. Some questions ask whose results give the better estimate, or whether a dice or spinner is fair, and want a reason.

Relative frequency

The relative frequency of an outcome is the fraction of the trials in which it happened:

\text{relative frequency}=\frac{\text{number of times it happened}}{\text{total number of trials}}

It is your estimate for the probability that it happens next time.

If a drawing pin is dropped 80 times and lands point up 52 times, the relative frequency of point up is \frac{52}{80}=0.65

Give the answer as a fraction or a decimal, such as \frac{13}{20} or 0.65, not as a ratio or as "52 out of 80". It is always between 0 and 1, because the top number can never be bigger than the number of trials.

The pin landed point down 80-52=28 times, so the estimate for point down is \frac{28}{80}=0.35. The two estimates add up to 1, as probabilities do.

Worked example 1: Finding a relative frequency

Tom flips a plastic bottle 50 times. It lands upright 12 times. Work out an estimate for the probability that the bottle will land upright the next time Tom flips it.

  1. There were 50 trials, and the bottle landed upright in 12 of them.
  2. Relative frequency of upright =\frac{12}{50}

  3. Your turn. \frac{12}{50}=\frac{\square}{100}. What number goes in the box?

    Multiply the top and the bottom by 2: \frac{12}{50}=\frac{24}{100}

  4. \frac{24}{100}=0.24
  5. Answer: 0.24

More trials give a better estimate

A relative frequency is only an estimate. With only a few trials it can be a long way from the true probability, just by chance. The more trials there are, the better the estimate is likely to be.

Tess flipped a fair coin 100 times and worked out the relative frequency of heads after every 10 flips. After 10 flips she had 7 heads, a relative frequency of 0.7. After 100 flips she had 49 heads, a relative frequency of 0.49, very close to the true probability of 0.5

So when you compare two people's results, the one with more trials gives the better estimate. When a question says "use all the results", add up all the times it happened and add up all the trials, then divide. Don't find two relative frequencies and take the number halfway between them.

Worked example 2: Using all the results

Ana spins a spinner 30 times. It lands on 5 a total of 9 times. Leo spins the same spinner 50 times. It lands on 5 a total of 19 times. Leo says, "My results give the better estimate, because I got more 5s." (a) Is Leo's reason right? Explain your answer. (b) Use all their results to work out an estimate for the probability that the spinner lands on 5

  1. (a) The number of 5s is not what matters. Leo's results are better because he did more spins: 50, not 30. More trials give a better estimate.
  2. (b) Add up all the trials: 30+50=80 spins

  3. Your turn. How many times did the spinner land on 5 altogether?

    9+19=28 times

  4. Relative frequency =\frac{28}{80}=0.35
  5. Taking the number halfway between \frac{9}{30}=0.3 and \frac{19}{50}=0.38 would give 0.34, which is wrong: Leo's 50 spins should count for more than Ana's 30
  6. Answer: (a) No. His results are better because he did more spins, not because he got more 5s (b) 0.35

Expected outcomes

If you know a probability, you can estimate how many times something will happen. Multiply the probability by the number of trials:

\text{expected number}=\text{probability}\times\text{number of trials}

A fair coin is thrown 60 times. P(\text{heads})=\frac{1}{2}, so you expect \frac{1}{2}\times60=30 heads. A biased dice lands on 6 with probability 0.35, so in 200 rolls you expect 0.35\times200=70 sixes. If 4\% of eggs are cracked, then in 600 eggs you expect 0.04\times600=24 cracked eggs.

The answer is a number of times, not a probability, so write 30, not \frac{30}{60}. It is only an estimate: the coin might land on heads 27 times, or 34 times.

A relative frequency can be used as the probability in the same way. If a pin landed point up 0.65 of the time, then in 400 more drops you expect about 0.65\times400=260 to land point up. You can also work backwards: if the probability is 0.2 and you expect 30, there must be 30\div0.2=150 trials.

Worked example 3: Expected outcomes with a fair dice

Jo is going to roll a fair 6-sided dice 240 times. Work out an estimate for the number of times the dice will land on a number greater than 4

  1. The numbers greater than 4 are 5 and 6, so P(\text{greater than }4)=\frac{2}{6}=\frac{1}{3}
  2. Expected number =\frac{1}{3}\times240

  3. Your turn. Work out \frac{1}{3} of 240

    240\div3=80

  4. Answer: 80

Worked example 4: A biased spinner

A biased spinner can land on red, blue, green or white. The probability that it lands on red is 0.32, on blue is 0.4 and on green is 0.13. Mia is going to spin the spinner 120 times. Work out an estimate for the number of times the spinner will land on white.

  1. Put the probabilities in a table. All four add up to 1. Red, blue and green add up to 0.32+0.4+0.13=0.85

  2. Your turn. Work out the probability that the spinner lands on white.

    P(\text{white})=1-0.85=0.15

  3. Expected number of whites =0.15\times120. Multiply by the number of spins; don't divide.
  4. 0.1\times120=12 and 0.05\times120=6, so 0.15\times120=12+6=18
  5. Answer: 18

Fair or biased

For a fair dice, each number has probability \frac{1}{6}. So in 120 rolls you expect each number about \frac{1}{6}\times120=20 times. You will hardly ever get exactly 20 of each, because small differences happen by chance.

To decide whether a dice or spinner is fair, compare the results with what you would expect if it were fair. Work out the expected number, or compare the relative frequency with the probability for a fair one.

  • A small difference, such as 23 sixes when you expect 20, does not show it is biased.
  • A big difference after lots of trials, such as 45 sixes when you expect 20, suggests it is biased.
  • With only a few trials, such as 12 rolls, you can't tell. It needs to be rolled many more times.

Always give a reason with numbers in it, for example: "Biased, because 45 is a lot more than the 20 you would expect."

Worked example 5: Deciding whether a dice is fair

Sanjay rolls an ordinary dice 300 times. It lands on 6 a total of 81 times. (a) How many times would you expect a fair dice to land on 6 in 300 rolls? (b) Work out the relative frequency of a 6 in Sanjay's rolls. (c) Do you think Sanjay's dice is fair? Give a reason for your answer.

  1. For a fair dice, P(6)=\frac{1}{6}

  2. Your turn. (a) How many times would you expect a fair dice to land on 6 in 300 rolls?

    \frac{1}{6}\times300=300\div6=50 times

  3. (b) Relative frequency =\frac{81}{300}=0.27
  4. (c) 81 is a lot more than 50, and 0.27 is a lot more than \frac{1}{6}, which is about 0.17. 300 rolls is a lot of trials, so this is very unlikely to be just chance.
  5. Answer: (a) 50 (b) 0.27 (c) Probably not fair, because it landed on 6 a lot more often than expected over many rolls.

Worked example 6: Estimating a profit

Ella has a "ball in the bucket" game at a charity fun day. Each go costs £1. A player whose ball lands in the bucket wins a prize of £4. Before the fun day, Ella's friends try the game 50 times, and the ball lands in the bucket 8 times. Ella expects 200 goes on the day. Use the results to work out an estimate for the profit Ella will make.

  1. Relative frequency of a win =\frac{8}{50}=0.16

  2. Your turn. How many wins should Ella expect in 200 goes?

    0.16\times200=32 wins

  3. Money taken: 200\times\pounds 1=\pounds 200
  4. Prizes paid: 32\times\pounds 4=\pounds 128
  5. Profit =\pounds 200-\pounds 128=\pounds 72
  6. Answer: £72

Common mistakes

  • Giving a probability when the question asks for a number of times. An expected number is a count, such as 25 heads, not \frac{25}{50}.
  • Dividing by the probability instead of multiplying. Multiply the probability by the number of trials: 0.4\times200=80, not 200\div0.4=500
  • Choosing the results with more successes, or fewer trials. The results with more trials give the better estimate.
  • Taking the number halfway between two relative frequencies. To use all the results, add up all the successes and all the trials, then divide.
  • Forgetting the missing probability. The probabilities of all the outcomes add up to 1, so work out the missing one before you multiply.
  • Expecting an estimate to be exact. A fair coin thrown 100 times will not always give 50 heads, so a small difference doesn't mean it is biased.

Practice questions

More on this topic: Relative Frequency and Expected Outcomes worksheet with full solutions