Chapter 11 · Probability

Making Predictions With Probability

Use a known probability or data from many trials to predict about how often an event will happen, and see why such predictions are approximate, not guaranteed.

18 min read

You already know how to find an experimental probability. You repeat a chance experiment, which is an activity whose result you cannot know ahead of time, such as spinning a spinner. Each run of the experiment, such as one spin, is a trial. The result, or group of results, you are watching for, such as landing on blue, is an event. The experimental probability is the number of trials in which the event happened divided by the total number of trials.

In this lesson you turn that around and use a probability to look ahead. A probability is a number from $0$ to $1$ that tells how likely an event is. You will answer two kinds of question:

You are given the probability of an event. About how often should the event happen in many trials?
You only have data from past trials. About how often should the event happen in future trials?

Both questions use the same reasoning, and both give predictions that are approximate, not guaranteed.

Start with what a probability tells you about a long run of trials. The relative frequency of an event is the fraction of the trials in which the event happened: the number of times it happened divided by the number of trials.

Here is the link. Over many trials, the relative frequency of an event tends to settle close to its probability. So if an event has probability $0.2$ , then in a long run of trials it should happen in about $0.2$ , or two tenths, of them.

The word many matters. In just $5$ trials, an event with probability $0.2$ might happen $0$ times or $3$ times. That gives a relative frequency of $0$ or $0.6$ , far from $0.2$ . The link becomes dependable only as the number of trials grows, and even then it is approximate, not guaranteed.

The table shows what about two tenths of the trials means for different numbers of trials.

Predictions for an event with probability 0.2 over many trials

Number of trials	Predicted relative frequency	Two tenths of the trials	Predicted count
$100$	about $0.2$	$0.2 \times 100$	about $20$
$200$	about $0.2$	$0.2 \times 200$	about $40$
$400$	about $0.2$	$0.2 \times 400$	about $80$
$800$	about $0.2$	$0.2 \times 800$	about $160$

Look down the table. The predicted relative frequency stays at about $0.2$ , while the predicted count grows. Each time the number of trials doubles, the predicted count doubles too. The predicted count is proportional to the number of trials: it is always the same number, $0.2$ , times the number of trials. That fixed multiplier, the probability, is the constant of proportionality.

Keep the word about in mind. These are approximate predictions for many trials, not guarantees, so expect real counts to be near them, and only sometimes exactly equal to them.

Now write the same reasoning with symbols. Let $n$ be the number of trials, a positive whole number. Let $p$ be the probability of the event, a number from $0$ to $1$ . Let $c$ be the number of trials in which the event happens. The relative frequency is $\frac{c}{n}$ , and over many trials it tends to be close to $p$ . You can write this as $\frac{c}{n} \approx p$ , where the symbol $\approx$ means is approximately equal to.

Multiplying both sides by $n$ undoes the division by $n$ and leaves the count alone on the left: $c \approx p \times n$ . This predicted value is called an expected count. It is the number of times you expect the event to happen over many trials. Treat it as an approximate target, not a guarantee. The real count will usually be a little off, where a little means small compared with $n$ .

Key concept

A probability predicts a long-run count

Over many trials, the relative frequency of an event tends to be close to its probability. So you can predict about how many times the event will happen: multiply the probability $p$ by the number of trials $n$ . The product is the expected count, your prediction for the real count $c$ .

If the probability is given, use it for $p$ .
If you only have data, use the experimental probability from many past trials as your estimate of $p$ , and predict for trials run in the same way.

The expected count is approximate, not a guarantee. Expect the real count to be near it, not exactly on it, and do not rely on it for only a few trials.

c \approx p \times n

Worked example

From a given probability to a predicted count

The probability that a spinner lands on blue is $\frac{3}{8}$ . You will spin it $400$ times. Predict the approximate relative frequency of blue, and about how many of the $400$ spins should land on blue.

Name the quantities. The probability of blue, p, is 3/8. The number of trials, n, is 400 spins. Let c stand for the number of those 400 spins that land on blue. This is the count you want to predict.
$p = \frac{3}{8}, \quad n = 400$
Predict the relative frequency, which is the count divided by the number of trials. Over many trials the relative frequency tends to be close to the probability, and 400 spins is many trials. So expect a relative frequency of about 3/8. Dividing 3 by 8 gives 0.375 as a decimal. This is approximate, not exact.
$\frac{c}{400} \approx \frac{3}{8} = 0.375$
Turn the relative frequency into a count. Multiply both sides by 400, the number of spins. This undoes the division by 400 and leaves c alone on the left, so the count is about three eighths of the 400 spins.
$c \approx \frac{3}{8} \times 400$
Do the arithmetic. One eighth of 400 is 400 divided by 8, which is 50. Three eighths is 3 times as much.
$400 \div 8 = 50, \qquad 3 \times 50 = 150$
Say what the number means. 150 is an expected count over many spins, not a promise. The real count will usually land near 150, a little above or a little below, and only sometimes exactly on it.

Answer

Over $400$ spins, expect the relative frequency of blue to be about $\frac{3}{8}$ , so the spinner should land on blue about $150$ times. This is an approximate expected count, not a guaranteed result.

Sometimes nobody can give you the probability. How likely is a paper cup to land on its side when you toss it? There is no simple way to reason that out from the shape of the cup. You have to toss it and count.

This is where experimental probability helps. It is the relative frequency of an event in trials that have already happened. If there were many of those trials, that relative frequency is likely to be close to the probability. So you can use it as your estimate of $p$ and multiply as before: the predicted count is about the experimental probability times the number of future trials.

Two conditions come with this prediction. The data must come from many trials, because a relative frequency from only a few trials can be far from the probability. And the future trials must be run in the same way as the past ones, because a different experiment can have a different probability. Even then, the result is an approximate expected count, not a guarantee.

The table shows the results of tossing one paper cup $200$ times.

Results of 200 tosses of one paper cup

How the cup landed	Number of tosses
On its side	$146$
Upside down	$34$
Right side up	$20$
Total	$200$

Worked example

From past data to a prediction

A paper cup was tossed $200$ times. It landed on its side $146$ times, upside down $34$ times and right side up $20$ times. If you toss the same cup $500$ more times in the same way, about how many times should you expect it to land on its side?

Find the experimental probability of the event 'lands on its side'. It is the relative frequency in the past trials: the number of times the event happened, 146, divided by the number of trials, 200. The other two counts are not needed for this event. Dividing both 146 and 200 by 2 gives 73/100, which is 0.73.
$\frac{146}{200} = \frac{73}{100} = 0.73$
Use 0.73 as the estimate of the probability, p. This is reasonable because 200 tosses is many trials, so the relative frequency is likely to be close to the probability, and the new tosses use the same cup in the same way.
$p \approx 0.73$
Set up the prediction. Let c stand for the number of the 500 new tosses that land on the side. Over many trials the event should happen in about the fraction p of them, so multiply the estimated probability by the number of new trials.
$c \approx 0.73 \times 500$
Do the arithmetic. Because 500 is 5 times 100, first multiply 0.73 by 100 to get 73, and then multiply 73 by 5.
$0.73 \times 100 = 73, \qquad 73 \times 5 = 365$
Check the size of the answer. 0.73 is a little less than 3/4, and 3/4 of 500 is 375, so a prediction a little under 375 makes sense. This checks the arithmetic only. The real count is still left to chance and is only expected to be near 365.
$\frac{3}{4} \times 500 = 375$

Answer

Based on the $200$ past tosses, you should expect the cup to land on its side about $365$ times in $500$ more tosses. This is an approximate expected count, not a guarantee.

Worked example

When the prediction is not a whole number

In a test, $210$ of $250$ seeds from one batch sprouted. A gardener will plant $330$ seeds from the same batch in the same conditions. About how many of the $330$ seeds should the gardener expect to sprout?

Find the experimental probability that a seed sprouts. Each seed is one trial, and the event 'sprouts' happened in 210 of the 250 trials. Dividing both numbers by 5 gives 42/50, and multiplying both by 2 gives 84/100, which is 0.84.
$\frac{210}{250} = \frac{42}{50} = \frac{84}{100} = 0.84$
Use 0.84 as the estimate of the probability. The conditions hold: 250 seeds is many trials, and the new seeds come from the same batch and grow in the same conditions. Let c stand for the number of the 330 new seeds that sprout, and multiply as before.
$c \approx 0.84 \times 330$
Do the arithmetic by splitting 330 into 300 and 30. This is allowed because multiplying a sum gives the same result as multiplying each part and adding.
$0.84 \times 300 = 252, \qquad 0.84 \times 30 = 25.2$
Add the two parts to finish the multiplication.
$252 + 25.2 = 277.2$
Interpret the result. A count of seeds must be a whole number, so 277.2 cannot be the real count. That is no problem, because the prediction is approximate anyway. Round to the nearest whole number and report about 277.
$277.2 \approx 277$

Answer

The gardener should expect about $277$ of the $330$ seeds to sprout. This is an approximate expected count based on the test, not a guarantee, so the real number could be somewhat higher or lower.

Here is the idea to take with you. Over many trials, the relative frequency of an event tends to stay close to its probability. So you can predict a count by multiplying the probability by the number of trials. If the probability is given, use it. If it is not, use the experimental probability from many past trials, and predict only for trials run in the same way.

Every such prediction is an approximate expected count, not a guarantee. Say about, and do not trust the method for only a few trials.

Next, you will compare experimental probability with theoretical probability, which is the probability you find by reasoning about the possible results of a trial instead of collecting data.

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Comparing Experimental and Theoretical Probability

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