Experimental Probability

Estimating a probability from the results of an experiment: relative frequency, what happens as the number of trials grows, fair or biased, and using an estimate.

What You Should Already Know

From lesson 13.2: for equally likely outcomes, $P(E) =$ number of favourable outcomes $\div$ total number of possible outcomes. For a fair coin $P(\text{head}) = \frac{1}{2}$ ; for a fair die $P(6) = \frac{1}{6}$ .
Writing a fraction as a decimal, rounded when needed: $\frac{1}{6} = 0.1666\ldots \approx 0.167$ (to 3 decimal places).

Theoretical and Experimental Probability

The probabilities of lessons 13.2 and 13.3 were worked out from the sample space, before anything was tossed or drawn. Such a probability is called a theoretical probability.

We can also do the experiment many times and count. If a fair coin is tossed 10 times, will it give exactly 5 heads? Sometimes, but often not: 4 or 6 or 7 heads are quite usual.

Relative frequency

Each time an experiment is carried out is a trial.

The relative frequency of an event is

$\text{relative frequency} = \dfrac{\begin{matrix}\text{number of times} \\ \text{the event happens}\end{matrix}}{\text{number of trials}}$

The relative frequency is also called the experimental probability. It is an estimate of the probability of the event.

Example 1: A coin tossed 50 times

Sarah tosses a fair coin 50 times and gets 28 heads. (a) What is the theoretical probability of a head? (b) Find the relative frequency of a head in her experiment. (c) Are the two the same?

Solution

(a) A fair coin has 2 equally likely outcomes, one of them a head: $P(\text{head}) = \frac{1}{2} = 0.5\text{.}$

(b) Relative frequency $= \frac{28}{50} = 0.56$ .

(a) $0.5$ ; (b) $0.56$ ; (c) no, they differ by $0.06$ .

What Happens as the Trials Grow

Example 2: A results table

Nadia tosses a fair coin 1000 times. At several points she writes down how many heads she has had so far. Complete the relative-frequency column, to 3 decimal places where needed, and describe what happens.

Tosses	Heads	Relative frequency
20	14
50	26
100	48
200	105
400	198
700	354
1000	503

Solution

Divide the number of heads by the number of tosses in each row. The column reads, from the top: $\frac{14}{20} = 0.7$ , $\frac{26}{50} = 0.52$ , $\frac{48}{100} = 0.48$ , $\frac{105}{200} = 0.525$ , $\frac{198}{400} = 0.495$ , $\frac{354}{700} = 0.5057\ldots \approx 0.506\text{,}$ $\frac{503}{1000} = 0.503$ .

The graph plots each relative frequency against its number of tosses. The dashed line is the theoretical value, 0.5.

After few tosses the relative frequency is far from 0.5 (0.7 after 20). As the tosses grow it settles close to 0.5.

It does not get closer every time: after 100 tosses it is $0.5 - 0.48 = 0.02$ from 0.5, and after 200 tosses $0.525 - 0.5 = 0.025\text{,}$ a little further.

0.7, 0.52, 0.48, 0.525, 0.495, 0.506, 0.503: the relative frequency varies a lot at first and settles near 0.5 as the number of tosses grows, though not closer at every step.

Many trials

As the number of trials grows, the relative frequency tends to settle close to the probability.

It need not get closer at every step, and it may still differ a little after very many trials.

After only a few trials it may be far off, or it may happen to equal the probability exactly (5 heads in 10 tosses).

So an estimate from more trials is the more reliable one.

Coin-toss records

Count Buffon (1707 to 1788) tossed a coin 4040 times and got 2048 heads: $\frac{2048}{4040} \approx 0.507\text{.}$

John Kerrich (1903 to 1985) tossed a coin 10 000 times and got 5067 heads: $0.5067$ .

Karl Pearson (1857 to 1936) tossed a coin 24 000 times and got 12 012 heads: $0.5005$ .

Example 3: Which estimate to trust?

Two pupils each toss the same coin to estimate the probability of a head. Ali tosses it 10 times and gets 7 heads. Bala tosses it 1000 times and gets 520 heads. (a) Find each relative frequency. (b) Whose estimate is more reliable, and why?

Solution

(a) Ali: $\frac{7}{10} = 0.7$ . Bala: $\frac{520}{1000} = 0.52$ .

(b) Bala's, because it comes from far more trials (1000 against 10). With 10 tosses, 7 heads happens quite often even with a fair coin.

(a) $0.7$ and $0.52$ ; (b) Bala's, as it is based on many more trials.

When the Formula Cannot Be Used

The formula of lesson 13.2 needs equally likely outcomes. A drawing pin dropped on a table lands point up or point down, but the two are not equally likely, and no count of outcomes tells us its probability. The same is true of a biased coin or die. In such cases an experiment with many trials is how we estimate the probability.

Neither kind of probability is always better: when the outcomes are equally likely, the theoretical probability is the exact value and an experiment only estimates it; when they are not, the experiment is our way to an estimate.

A survey is different. A pupil chosen at random from a school is equally likely to be any of its pupils, so the formula would work if we knew how many pupils of the whole school travel by MRT, say. When only some of the pupils are asked, we do not know the counts for the whole school, and the relative frequency in the survey is our estimate of the probability.

Example 4: A drawing pin

A drawing pin is dropped 400 times and lands point up 248 times. (a) Estimate the probability that it lands point up. (b) Estimate the probability that it lands point down. (c) About how many times would it land point up in 1000 drops?

Solution

(a) Relative frequency $= \frac{248}{400} = 0.62$ .

(b) Point down $400 - 248 = 152$ times: $\frac{152}{400} = 0.38$ (or $1 - 0.62 = 0.38$ ).

(a) about $0.62$ ; (b) about $0.38$ ; (c) about 620 times.

Example 5: Fair or biased?

A coin is tossed 100 times and lands head 87 times. Do you think the coin is fair? Explain.

Solution

If the coin were fair, $P(\text{head}) = \frac{1}{2}$ , and in 100 tosses we would expect about $\frac{1}{2} \times 100 = 50$ heads: a few more or a few less is usual.

87 heads is far from about 50. The relative frequency $\frac{87}{100} = 0.87$ suggests the coin lands head much more often than tail.

We cannot be certain, as chance can give odd results, but 87 heads in 100 would be extremely rare with a fair coin.

The coin is very likely biased; 0.87 is our estimate of its probability of a head.

Using an Estimate

Example 6: How many red balls?

A bag holds 20 balls, some red and the rest white. Mei draws a ball at random, notes its colour and puts it back, 200 times. She gets a red ball 150 times. Estimate the number of red balls in the bag.

Solution

Because each ball is put back, every draw is from the same 20 balls.

Relative frequency of red $= \frac{150}{200} = 0.75$ , an estimate of $P(\text{red})$ .

If $P(\text{red}) = 0.75$ , then red balls $= 0.75 \times 20 = 15$ .

About 15 red balls.

Example 7: How many sixes?

A fair die is rolled 300 times. About how many sixes would you expect?

Solution

$P(6) = \frac{1}{6}$ , so about $\frac{1}{6}$ of the rolls give a six.

$\frac{1}{6} \times 300 = 50$ .

The actual number will usually be near 50 but not exactly 50.

About 50 sixes.

Common Mistakes

Expecting an experiment to give the theoretical value exactly. 28 heads in 50 tosses of a fair coin is a quite ordinary result.
Saying the relative frequency gets closer to the probability with every trial. It tends to settle, not to improve at each step.
Trusting an estimate from a few trials over one from many because it happens to be closer to a value you expected.
Using $\frac{1}{2}$ for a drawing pin or a biased coin. Two outcomes are not the same as two equally likely outcomes.
Dividing by the wrong number: relative frequency is the number of times divided by the number of trials.

Practice

Try each question first, then open the solution to check your working. Give decimals to 3 decimal places where they do not end sooner.

Question 1

Marcus rolls a fair die 60 times and gets a six 12 times. (a) What is the theoretical probability of a six? (b) Find the relative frequency of a six. (c) How far apart are they?

ShowHide solution

(a) $\frac{1}{6} \approx 0.167$ .

(b) $\frac{12}{60} = \frac{1}{5} = 0.2\text{.}$

(a) $\frac{1}{6} \approx 0.167$ ; (b) $0.2$ ; (c) about $0.033$ .

Question 2

Three experiments are done with a fair die to estimate the probability of a six: 30 rolls, 300 rolls and 3000 rolls. Complete the table, say which experiment gives the most reliable estimate, and why.

Rolls	Sixes	Relative frequency
30	8
300	55
3000	512

ShowHide solution

The column reads: $\frac{8}{30} = 0.2666\ldots \approx 0.267\text{,}$ $\frac{55}{300} = 0.1833\ldots \approx 0.183\text{,}$ $\frac{512}{3000} = 0.1706\ldots \approx 0.171\text{.}$

The experiment with 3000 rolls has by far the most trials, so its estimate is the most reliable. Its result is also the closest to $\frac{1}{6} \approx 0.167$ , which agrees with what we expect from more trials.

0.267, 0.183, 0.171; the experiment with 3000 rolls, because it has the most trials.

Question 3

In a survey of 250 pupils of a school, 85 said they travel to school by MRT. (a) Estimate the probability that a pupil of the school, chosen at random, travels by MRT. (b) The school has 1400 pupils. Estimate how many travel by MRT.

ShowHide solution

Only 250 of the pupils were asked, so the survey gives an estimate, not the exact probability.

(a) $\frac{85}{250} = 0.34$ .

(b) $0.34 \times 1400 = 476$ .

(a) about $0.34$ ; (b) about 476 pupils.

Question 4

A die is rolled 500 times with these results. (a) Find the relative frequency of a six. (b) Do you think the die is fair? Explain. (c) Estimate how many sixes there would be in the next 50 rolls.

Number	Frequency
1	62
2	75
3	80
4	78
5	85
6	120

ShowHide solution

Check: $62 + 75 + 80 + 78 + 85 + 120 = 500\text{.}$

(a) $\frac{120}{500} = 0.24$ .

(b) With a fair die we would expect about $\frac{1}{6} \times 500 \approx 83$ of each number. 120 sixes is far more than that, and 62 ones far fewer. The die is very likely biased, though we cannot be certain.

(a) $0.24$ ; (b) probably not fair: 120 sixes against about 83 expected; (c) about 12.

Question 5

A bag holds 30 counters, some blue and the rest yellow. A counter is taken at random, its colour noted and put back, 120 times. A blue counter is taken 40 times. Estimate the number of blue counters in the bag.

ShowHide solution

Relative frequency of blue $= \frac{40}{120} = \frac{1}{3}$ .

Blue counters $\approx \frac{1}{3} \times 30 = 10$ .

About 10 blue counters.

Question 6

Kumar drops a drawing pin and records how often it lands point up, writing the totals so far after every 100 drops. Find the relative frequency at each point, plot them against the number of drops, and give the best estimate of the probability that the pin lands point up.

Number of drops	Times point up
100	68
200	122
300	186
400	252
500	310

ShowHide solution

$\frac{68}{100} = 0.68$ , $\frac{122}{200} = 0.61$ , $\frac{186}{300} = 0.62$ , $\frac{252}{400} = 0.63$ , $\frac{310}{500} = 0.62$ .

There is no theoretical value to draw: the two outcomes are not equally likely. The relative frequency settles near 0.62, and the estimate from the most drops (500) is $0.62$ .

0.68, 0.61, 0.62, 0.63, 0.62; the best estimate is about $0.62$ .

Question 7

For a class game, Hana's group makes a spinner with 4 equal sectors coloured red, blue, green and yellow. She spins it 400 times and it lands on red 160 times. Is the spinner fair? Explain.

ShowHide solution

If it were fair, $P(\text{red}) = \frac{1}{4}$ , and about $\frac{1}{4} \times 400 = 100$ of the 400 spins would land on red.

160 is far above 100: the relative frequency $\frac{160}{400} = 0.4$ is well above 0.25 after many spins.

Very likely not fair: red came up 160 times against about 100 expected.

Question 8

Ahmad rolls a fair die 120 times. About how many times would you expect a number greater than 4?

ShowHide solution

Greater than 4: 5 and 6. $P = \frac{2}{6} = \frac{1}{3}$ .

$\frac{1}{3} \times 120 = 40$ .

About 40 times.

Question 9

True or false? Explain each answer. (a) If a fair coin is tossed 100 times, it will land head exactly 50 times. (b) Each time more trials are added, the relative frequency gets closer to the theoretical probability. (c) A theoretical probability is always more accurate than an experimental one.

ShowHide solution

(a) False. About 50 heads is likely, but 47 or 53 are just as ordinary; exactly 50 is not certain.

(b) False. The relative frequency tends to settle as the trials grow, but not at every step: in Example 2 it was 0.02 from 0.5 after 100 tosses and 0.025 after 200.

(c) False. When the outcomes are equally likely, the theoretical probability is exact. When they are not (a drawing pin, a biased die), there is no theoretical value from counting, and an experiment with many trials gives the estimate.

(a) false; (b) false; (c) false.

Key Points

A theoretical probability is worked out from equally likely outcomes; an experimental probability comes from trials.
Relative frequency $=$ number of times the event happens $\div$ number of trials. It estimates the probability.
As the trials grow, the relative frequency tends to settle near the probability, though not closer at every step.
An estimate from more trials is more reliable.
When the outcomes are not equally likely (a drawing pin, a biased die), or the counts are not known (a survey of some of the pupils of a school), an experiment or a survey with many trials is how we estimate the probability.
A result far from what a fair object would give suggests the object is biased, though we cannot be certain.
In $n$ trials, an event of probability $p$ happens about $p \times n$ times.