Introduction to Probability

Intuition

You already use the language of chance every day. A forecast says there is a good chance of rain. A coin flip decides who kicks off, and a board game turns on the roll of a die. Each of these is a guess about how likely something is. Probability is the part of math that pins that likeliness down to a single number. The idea is tidy: when the possible results are all equally likely, count how many results you want. Then divide that count by how many results there are in total. That total is exactly the kind of count you learned to find in the last lesson, which is why counting comes right before probability. By the end you will be able to say not just that heads is “fifty-fifty” but that its probability is $\frac{1}{2}$ . You will also be able to handle dice, spinners, and bags of marbles the same way.

Learning goals

Place a probability on the scale from $0$ to $1$
Divide favorable outcomes by equally likely total outcomes
State the equally-likely assumption a counting argument depends on
Apply the complement rule, since all outcome probabilities add to one
Distinguish experimental probability from theoretical, and say how trials close the gap
Count a two-stage experiment's outcomes first, then apply the same formula

A number from 0 to 1

Probability is measured on a fixed scale from $0$ to $1$ . A probability of $0$ means the event is impossible, and a probability of $1$ means it is certain. Everything else falls in between, and the larger the number, the more likely the event. A probability of $\frac{1}{2}$ sits exactly in the middle and means the event is as likely to happen as not, the “fifty-fifty” of everyday speech.

Because a probability is just a number between $0$ and $1$ , you can write it three equivalent ways. The three forms are a fraction, a decimal, and a percent from $0\%$ to $100\%$ . A one-in-four chance is $\frac{1}{4}$ , which is $0.25$ , which is $25\%$ , all naming the same spot on the scale. Use whichever form is clearest for the problem. Be ready to convert between them, since these are the fractions, decimals, and percents from earlier chapters doing familiar work.

The probability scale runs from 0 (impossible) to 1 (certain). The same value can be written as a fraction, a decimal, or a percent, so an even chance is one-half, 0.5, or 50 percent.

No probability is ever negative or larger than $1$ , because an event cannot happen less often than never or more often than always. If a calculation hands you $1.3$ or $-0.2$ , you have made a mistake somewhere.

Outcomes, and what equally likely means

To compute a probability we have to be careful about the setup. An experiment is any action with an uncertain result, such as flipping a coin or rolling a die. Each possible result is an outcome, and the full list of outcomes is the sample space. An event is the part of the sample space you care about, a chosen collection of outcomes. Rolling a die has the sample space $\{1, 2, 3, 4, 5, 6\}$ , and “roll an even number” is the event $\{2, 4, 6\}$ .

A fair die has six equally likely faces. The three shaded faces are the even numbers 2, 4, and 6, so the event 'roll an even number' covers 3 of the 6 outcomes.

The method of this lesson rests on one assumption: that the outcomes are equally likely, meaning each one has the same chance as any other. A fair coin, a fair die, a spinner with equal sectors, and a marble drawn at random are the standard ways of promising this. The words fair, equal, and at random are not decoration; they are what let you find a probability just by counting.

Equally likely is an assumption you must check, not something that is always true. Tomorrow’s weather is either rain or no rain, but those two outcomes are usually not equally likely. So the chance of rain is not automatically $\frac{1}{2}$ . A bent coin might favor one side, and a spinner whose sectors have different sizes lands more often on the big ones. When the outcomes are not equally likely, counting them is not enough, and you need data or extra information instead.

The probability of an event

Now the payoff. Suppose an experiment has $N$ equally likely outcomes. Something must happen, so the chances of all $N$ outcomes together make up the whole, and we name that whole $1$ (certain). Since the outcomes are equal, split that one whole into $N$ equal shares, so each single outcome has probability $\frac{1}{N}$ .

An event is just some of those outcomes. If the event contains $k$ of them, it collects $k$ of the equal shares, so its probability is $k \times \frac{1}{N} = \frac{k}{N}$ . That is the central formula of the lesson:

P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{total number of equally likely outcomes}}.

“Favorable” is only a name for the outcomes in your event; it does not mean good, just counted. The denominator is the size of the whole sample space, which is exactly the total you practiced counting in the previous lesson.

Worked example 1 The probability of tails

A fair coin is flipped once. What is the probability that it lands tails?

A flip has two equally likely outcomes, heads and tails, so the sample space has $2$ outcomes. Exactly one of them, tails, is favorable.

P(\text{tails}) = \frac{1}{2} = 0.5 = 50\%.

The word “fair” is what lets us count this way. It promises the two sides are equally likely, so each one takes half of the whole.

Worked example 2 The probability of an even roll

A fair die is rolled once. What is the probability of rolling an even number?

The sample space is the six faces $\{1, 2, 3, 4, 5, 6\}$ , all equally likely, so the total is $6$ . The event “even” is the set $\{2, 4, 6\}$ , the three faces shaded in the die figure above, so there are $3$ favorable outcomes.

P(\text{even}) = \frac{3}{6} = \frac{1}{2} = 0.5 = 50\%.

Always reduce the fraction. Here $\frac{3}{6}$ and $\frac{1}{2}$ are the same number, and $\frac{1}{2}$ is the simplest way to write it.

Check your understanding

A fair die is rolled once. What is the probability of rolling a $3$ ?

\frac{1}{6}

\frac{1}{3}

\frac{1}{2}

\frac{5}{6}

More single-event probabilities

The formula does not care whether the experiment is a coin, a die, a spinner, or a bag of marbles. The work is always the same: count the favorable outcomes, count the total equally likely outcomes, and divide. Here it is on a spinner and on a random draw.

A fair spinner with 8 equal sectors. Because the sectors are equal, each is equally likely. Three sectors are shaded, so the probability of landing on a shaded sector is 3 out of 8.

Worked example 3 Reading a probability off a spinner

The spinner above has $8$ equal sectors, and $3$ of them are shaded. If you spin it once, what is the probability of landing on a shaded sector?

Because the sectors are equal, the $8$ sectors are equally likely, so the sample space has $8$ outcomes. The shaded sectors are the $3$ favorable ones.

P(\text{shaded}) = \frac{3}{8} = 0.375 = 37.5\%.

Equal sectors are what make this work. If the shaded part were one wide sector and the rest were narrow, the outcomes would not be equally likely. In that case you could not just count sectors.

An experiment with exactly $100$ equally likely outcomes is worth a closer look. The reason is that the counting and the three ways of writing a probability all show up in one picture. Picture a drum of $100$ raffle tickets, mixed and drawn at random. Each square below is one ticket, and shading a square marks it as a favorable outcome. So the count of shaded squares is the numerator, while the $100$ squares are always the denominator.

Counting favorable outcomes out of a hundred

Squares shaded

25 squares of 100 outcomes are favorable. So the probability is 25/100, or 1/4 in lowest terms. As a percent, 25%. As a decimal, 0.25.

One hundred equally likely outcomes drawn as a hundred-square. Shade the outcomes that count as favorable, and read the probability as a count, a percent, a fraction, and a decimal.

Give yourself some targets. Shade $25$ squares and one probability is reported three ways at once, as $\frac{25}{100}$ , as $0.25$ , and as $25\%$ , all naming the one-in-four chance. Shade none and you are at $0$ , the impossible end of the scale; shade all $100$ and you are at $1$ , certain. Then find the shading where the favorable squares and the ones left unshaded are equally likely. That shading lands you on the even chance in the middle.

Worked example 4 Drawing a marble at random

A bag holds $3$ red, $5$ blue, and $2$ green marbles, all the same size and weight. You draw one marble without looking. What is the probability it is red?

Drawing without looking makes the marbles equally likely, so the total is the whole bag. Count it with the addition principle from the last lesson:

3 + 5 + 2 = 10 \text{ marbles}.

The favorable outcomes are the $3$ red marbles, so

P(\text{red}) = \frac{3}{10} = 0.3 = 30\%.

The same draw gives $P(\text{blue}) = \frac{5}{10} = \frac{1}{2}$ and $P(\text{green}) = \frac{2}{10} = \frac{1}{5}$ , and notice the three probabilities add to $\frac{3}{10} + \frac{5}{10} + \frac{2}{10} = 1$ , the whole bag.

Why the probabilities add to one

That last observation, that the probabilities of all the outcomes add to $1$ , is not a coincidence of that bag. It is always true, and it is worth a careful statement, because it is also the engine behind a shortcut called the complement rule.

The outcome probabilities add to $1$ , and $P(\text{not } A) = 1 - P(A)$ #

Take an experiment with $N$ equally likely outcomes, so each outcome has probability $\frac{1}{N}$ . Add up the probabilities of all $N$ of them. There are $N$ shares, each equal to $\frac{1}{N}$ , so the total is

\underbrace{\frac{1}{N} + \frac{1}{N} + \cdots + \frac{1}{N}}_{N \text{ terms}} = N \times \frac{1}{N} = 1.

The whole sample space has probability $1$ , which is just the statement that something is certain to happen.

Now take any event $A$ built from $k$ of these outcomes, so $P(A) = \frac{k}{N}$ . Its complement, written “not $A$ ”, is everything else in the sample space, the other $N - k$ outcomes, so $P(\text{not } A) = \frac{N - k}{N}$ . Adding the two,

P(A) + P(\text{not } A) = \frac{k}{N} + \frac{N - k}{N} = \frac{N}{N} = 1,

because every outcome is either in $A$ or not in $A$ , never both and never neither. Subtracting $P(A)$ from each side gives the complement rule:

P(\text{not } A) = 1 - P(A).

The same counting also shows every probability lands on the scale. Since $k$ is at least $0$ , the fraction $\frac{k}{N}$ is never less than $0$ , the value it takes when $k = 0$ , an impossible event. Since $k$ is at most $N$ , that same fraction $\frac{k}{N}$ is never more than $1$ , the value it takes when $k = N$ , a certain event.

The complement rule

The complement rule earns its own line because it often saves work. When the outcomes you want are awkward to count but the leftover outcomes are easy, count the leftovers and subtract from $1$ . Problems that ask for “at least one” of something are the classic case, since “at least one” is the complement of “none”.

Worked example 5 Using the complement rule

Using the same bag of $10$ marbles ( $3$ red, $5$ blue, $2$ green), what is the probability of drawing a marble that is not red?

You could count the favorable outcomes directly. Not red means blue or green, which is $5 + 2 = 7$ marbles, so $\frac{7}{10}$ . The complement rule reaches the same answer faster, starting from $P(\text{red}) = \frac{3}{10}$ :

P(\text{not red}) = 1 - P(\text{red}) = 1 - \frac{3}{10} = \frac{7}{10} = 0.7 = 70\%.

Both routes agree, which is the whole point. A marble is either red or not red, so those two probabilities must fill up the one whole.

Check your understanding

A fair die is rolled once. What is the probability of not rolling a $3$ ?

\frac{1}{6}

\frac{1}{3}

\frac{5}{6}

\frac{1}{2}

Theoretical and experimental probability

Everything so far has been theoretical probability: you reason about equally likely outcomes and compute a fraction before ever running the experiment. There is a second kind. Experimental probability, also called relative frequency, comes from actually doing the experiment many times and recording what happens:

P_{\text{exp}}(\text{event}) = \frac{\text{number of times the event happened}}{\text{number of trials}}.

If you flip a real coin $20$ times and get $11$ heads, the experimental probability of heads is $\frac{11}{20} = 0.55$ . That happens even though the theoretical value is $\frac{1}{2}$ . The two are different numbers, and that is expected, because a short run of flips rarely splits exactly evenly.

Here is the key link between them. As you take more and more trials, the experimental probability tends to settle closer and closer to the theoretical probability. Twenty flips might give $55\%$ heads, but ten thousand flips will almost certainly land very near $50\%$ . This pattern, that relative frequency homes in on the true probability as the number of trials grows, is called the law of large numbers. It is why an insurer can predict the long-run rate of claims precisely even though any single policy is uncertain. It is also the bridge that makes a theoretical $\frac{1}{2}$ something you can actually watch happen.

Worked example 6 Experimental probability of heads

A coin is flipped $50$ times and lands heads $27$ times. What is the experimental probability of heads from this data, and how does it compare with the theoretical value?

Experimental probability divides the number of times the event happened by the number of trials.

P_{\text{exp}}(\text{heads}) = \frac{27}{50} = 0.54 = 54\%.

The theoretical probability for a fair coin is $\frac{1}{2} = 50\%$ . The experimental value is close but not exactly equal, which is normal for only $50$ flips. Flip the same coin ten thousand times and the fraction of heads would almost certainly sit much nearer to $50\%$ .

A gentle step to two-stage experiments

Single events are the heart of probability, but the same formula stretches to experiments built from two stages, as long as you keep counting carefully. Rolling two dice, flipping two coins, or spinning twice are all sequences of stages, and the multiplication principle counts their total outcomes for you. The probability formula does not change at all: it is still favorable outcomes over total equally likely outcomes. Only the counting gets a little bigger.

Rolling two dice gives 6 times 6 = 36 equally likely outcomes, one per cell. The 6 shaded cells down the diagonal are the doubles, where both dice match, so the probability of a double is 6 out of 36.

Worked example 7 Rolling a double with two dice

Two fair dice, one red and one blue, are rolled. What is the probability that they show the same number, a double?

First count the total. The red die has $6$ results and the blue die has $6$ . The roll is a sequence of two stages, so the multiplication principle gives

6 \times 6 = 36

equally likely outcomes, one per cell of the grid above. The doubles are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$ , which is $6$ favorable outcomes, the shaded diagonal.

P(\text{double}) = \frac{6}{36} = \frac{1}{6} \approx 0.167 \approx 16.7\%.

We found the denominator by counting, exactly the tool from the previous lesson, and the outcomes stay equally likely because each die is fair.

Check your understanding

Two fair coins are flipped. What is the probability that both land heads?

\frac{1}{2}

\frac{1}{4}

\frac{1}{3}

\frac{3}{4}

Looking back, and what comes next

This is the last lesson of pre-algebra, and probability is a fitting place to stop, because it quietly uses almost everything you have built. The probabilities are fractions you simplify, decimals you convert, and percents you read; the totals come from the counting principles of the last lesson. The whole idea of measuring chance rests on the plain arithmetic you started with. From here the road branches into more counting and probability, into algebra, and into geometry, each leaning on these same foundations. You now have them.

Common mistakes

Counting outcomes that are not equally likely. The formula divides favorable by total only when the outcomes are equally likely. “Rain or no rain” is two outcomes, but that does not make the chance of rain $\frac{1}{2}$ . Check for “fair”, “equal”, or “at random” before you count.
Forgetting to simplify, or mixing forms. Decide on a form and reduce. Write $\frac{3}{6}$ as $\frac{1}{2}$ , and make sure a percent answer is the same number as the fraction, not a different one.
A probability outside $0$ to $1$ . Probabilities never leave the scale. If you compute a value above $1$ or below $0$ , something is wrong, so recheck the counts.
Confusing an event with its complement. The rule is $P(\text{not } A) = 1 - P(A)$ : subtract from $1$ , do not flip the fraction. The complement of $\frac{1}{6}$ is $\frac{5}{6}$ , not $\frac{6}{1}$ or $\frac{1}{5}$ .
Miscounting the total number of outcomes. The denominator is the full sample space. For a draw, count every item in the bag; for two stages, multiply the stage counts. A second draw made without replacement leaves one fewer item to choose from, so the total shrinks.
Expecting experiment to match theory exactly. A short run of trials will not hit the theoretical value on the nose. Only as the number of trials grows do experimental and theoretical probabilities close in.

Key takeaways

Probability measures likelihood on a scale from $0$ (impossible) to $1$ (certain), and $\frac{1}{2}$ is an even chance. Any probability can be written as a fraction, a decimal, or a percent from $0\%$ to $100\%$ .
When outcomes are equally likely, $P(\text{event}) = \dfrac{\text{favorable outcomes}}{\text{total equally likely outcomes}}$ , and the denominator is the count you find with the counting principles.
“Equally likely” is an assumption (a fair coin, a fair die, equal sectors, a random draw). Without it, you cannot find the probability by counting alone.
All the outcome probabilities add to $1$ , which gives the complement rule $P(\text{not } A) = 1 - P(A)$ , handy when the opposite event is easier to count.
Experimental probability is $\dfrac{\text{times it happened}}{\text{number of trials}}$ , and by the law of large numbers it settles toward the theoretical value as the number of trials grows.
For a two-stage experiment, count the total with the multiplication principle and apply the same formula; two dice, for example, give $6 \times 6 = 36$ equally likely outcomes.
This completes pre-algebra. Probability rests on the fractions, decimals, percents, and counting you have been building all along.

Introduction to Probability

Learning goals

A number from 0 to 1

Outcomes, and what equally likely means

The probability of an event

More single-event probabilities

Why the probabilities add to one

The complement rule

Theoretical and experimental probability

A gentle step to two-stage experiments

Looking back, and what comes next

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)