Probability: Outcomes, Experiments and Equally Likely Events

A coin flip has a result you cannot know in advance. Probability gives you a way to describe its chance. The same language works for rolling a die, spinning a spinner, or choosing a counter.

Name what happens

A probability experiment is an action whose result depends on chance. One attempt is a trial. A possible result is an outcome. A standard six-sided die, also called a number cube, has six outcomes: 1, 2, 3, 4, 5, and 6.

An event groups the outcomes you are interested in. For example, “roll an even number” contains 2, 4, and 6. If the die lands on 4, the outcome is 4 and the event has happened. An event can contain more than one outcome.

Read the scale

A probability measures how likely an event is. Write it as $P(\text{event})$. Read $P(\text{even})$ as “the probability of rolling an even number.” Every probability lies from 0 to 1, including both ends.

An even chance means the event is as likely to happen as to fail. It is also called a “50-50 chance” or “evens.” For the even-number event, three of the six faces work, giving $3\mathrm{/}6=1\mathrm{/}2$. For a 6, only one face works, giving $1\mathrm{/}6\approx 0.17$. That belongs much nearer 0 on the scale.

The farther right a probability sits, the more likely its event. “Likely” still leaves room for failure. A probability of 0.9 means success about nine times in ten over many trials. A probability of 0.1 means success about once in ten; it is small, but possible.

Fair, biased, and random

A fair die gives every face the same chance. A biased die favors some faces. For example, extra weight can make one face appear more often. A fair coin gives heads and tails equal chances. On a fair spinner, equal-sized sections have equal chances; a color covering several sections can have a larger chance than another color.

Choosing an item at random gives each item the same chance of selection. You can know the chance of picking a particular item without knowing which item will come out.

Check the scale when calculating. Values such as −0.3, 1.2, 7/5, and 120% cannot describe a probability. A chance of 1 already includes every possible result.

A probability can look different without changing its value. A quarter, 0.25, and 25% all name the same chance. Connecting these forms lets you compare events on one scale.

Change the form

For a fraction, divide the top number by the bottom number. To change a decimal to a percent, multiply by 100:

To go back from a percent, divide by 100. So 37.5% is 0.375. Percent means “out of 100,” and the percent probability scale runs from 0% to 100%.

You can also simplify a fraction first. For example, $6\mathrm{/}10=3\mathrm{/}5=0.6=60\mathrm{\%}$. Both fractions are correct. Simplifying changes the numbers used to write the chance, but keeps their ratio the same.

Keep exact values while working

Some decimals repeat forever. For one third, $1\mathrm{/}3=0.333\dots$. You might report this as about 0.33 or 33.3%. For one sixth, $1\mathrm{/}6=0.1666\dots$, about 0.17 or 16.7%.

Use the exact fraction during a calculation. Round at the end when a decimal or percent is requested. Two rounded values can look equal even when their exact probabilities differ.

Read “out of” carefully

“1 in 4,” “1 out of 4,” and “a 25% chance” all mean $1\mathrm{/}4$. The bottom number counts the whole group. If 3 out of 10 students wear glasses, the probability of choosing one of them at random is $3\mathrm{/}10$. The other 7 students are part of that total, so $3\mathrm{/}7$ uses the wrong whole.

The phrase “1 to 4” compares one favorable result with four unfavorable results. That makes five results altogether, giving $1\mathrm{/}5$. It is different from “1 out of 4.”

Put unlike forms in order

Suppose event A has probability 0.45, event B has probability $2\mathrm{/}5$, and event C has probability 42%. Change them to decimals:

• A: 0.45.
• B: $2÷5=0.40$.
• C: $42÷100=0.42$.

The order from least likely to most likely is B, C, A. On a probability line, B sits left of C, and C sits left of A.

Do not compare just the top numbers of two fractions. For example, $3\mathrm{/}8=0.375$ is smaller than $2\mathrm{/}5=0.4$. Although 3 is larger than 2, eighths are smaller pieces than fifths.

To mark 0.25 on a line from 0 to 1, go a quarter of the way across. Mark 0.8 four fifths of the way across. Equal probabilities, such as 0.8 and 80%, belong at the same point.

Before finding a probability, decide what counts as a result. A list of every possible outcome is called the sample space. An event picks out some of those outcomes.

Make a complete list

Write a sample space in braces. For a coin flip, it is {heads, tails}. For one roll of a standard die, it is {1, 2, 3, 4, 5, 6}. List numbers in order so that gaps are easy to spot.

A spinner with eight sections numbered 1 through 8 has sample space {1, 2, 3, 4, 5, 6, 7, 8}. There are eight outcomes. The list includes every possible result, even when your event needs only a few of them.

Pick the outcomes for an event

The favorable outcomes are the ones that make an event happen. “Favorable” means they match the event; the result need not be something you want.

For a die, “roll a prime number” contains {2, 3, 5}. Remember that 1 is not prime. “Roll a number greater than 4” contains {5, 6}. The sample space stays the same, but these events choose different parts of it.

An event can contain:

• One outcome, such as “roll a 3.”
• Several outcomes, such as “roll an odd number.”
• Every outcome, such as “roll a whole number from 1 to 6.”
• No outcomes, such as “roll a 9.”

The last two are certain and impossible events.

A list does not give equal chances

Equally likely outcomes have the same chance. A fair die gives that chance to each face. Writing down three color names does not make those colors equally likely.

A bag holds 5 red, 3 blue, and 2 green counters. One counter is chosen at random. The possible colors are {red, blue, green}, but red has more counters than either other color.

To see the equal chances, give each counter a label: R1 through R5, B1 through B3, and G1 through G2. Those ten individual counters are ten equally likely outcomes. The event “red” contains five of them. Color names group the outcomes; they do not count the individual chances.

Repeated letters work the same way. The word “MISSISSIPPI” has 11 letter positions. It contains 1 M, 4 I’s, 4 S’s, and 2 P’s. Picking a position at random gives each position the same chance. The four letter names {M, I, S, P} do not have equal chances, because some names appear in more positions.

When every outcome has the same chance, count the outcomes that work and compare them with the whole sample space. This gives a theoretical probability: a probability calculated from a model of the situation.

Use the counting rule

First check that the outcomes have equal chances. Then count the whole set and the favorable part. Write the fraction, and simplify or convert it as needed.

On one roll of a fair die, the numbers greater than 4 are 5 and 6. Two faces work out of six:

The prime numbers are 2, 3, and 5, so their probability is $3\mathrm{/}6=1\mathrm{/}2$. No favorable outcomes gives 0. If every outcome works, the favorable and total counts are equal, giving 1.

Count objects, not color names

In a bag with 5 red, 3 blue, and 2 green counters, there are ten counters altogether. Each counter has the same chance of being chosen.

For “red or green,” either color works. There are $5+2=7$ favorable counters, so the probability is $7\mathrm{/}10$. Using three as the denominator would count color names instead of the objects being picked.

The same rule applies to people. A class has 28 students, including 15 girls. A random choice gives one particular student, Jane, probability $1\mathrm{/}28$. The event “choose a girl” includes 15 students, so its probability is $15\mathrm{/}28\approx 0.54$.

For a position chosen from “MISSISSIPPI,” 4 of the 11 positions contain S, 2 contain P, and 1 contains M. Their probabilities are $4\mathrm{/}11$, $2\mathrm{/}11$, and $1\mathrm{/}11$. The only vowel is I, appearing four times, so the probability of a vowel is $4\mathrm{/}11$.

Update the whole when it changes

Add two blue counters to the bag above. It now has five blue counters and twelve counters in total:

The denominator changes as well as the numerator. Keeping ten underneath would treat new counters as favorable without including them in the whole bag.

Unequal groups need different treatment

A spinner can be half red, a quarter blue, and a quarter yellow. It has three colors, but red’s chance is $1\mathrm{/}2$ because red covers half the circle. Equal-sized sections have equal chances; unequal color areas do not.

Likewise, naming two events does not make each a half chance. A fair die either shows a 6 or does not, but those events contain one face and five faces. Their probabilities are $1\mathrm{/}6$ and $5\mathrm{/}6$.

A probability model gives a probability to every possible outcome.

Equal and unequal models

A fair die has a uniform model: each of its six outcomes has probability $1\mathrm{/}6$. A spinner with unequal color areas has a model that is not uniform. Its probabilities can come from the areas. A 90° section takes $90\mathrm{/}360=1\mathrm{/}4$ of the circle, giving a probability of $1\mathrm{/}4$.

Model values can also come from experimental data. Whatever their source, each value must lie from 0 to 1, and the values for all outcomes must total 1.

Find a missing probability

A spinner’s colors have these probabilities:

Add the known values: $0.35+0.20+0.15=0.70$. The missing part of the whole is $1-0.70=0.30$, so green has probability 0.30.

A list such as 0.5, 0.3, and 0.3 fails this check. It totals 1.1, which is more than the whole chance.

A model may use a letter for a missing value. Suppose a team has win probability $2x$, tie probability $x$, and loss probability 0.25. The three results cover every possibility:

Combine the unknown parts and subtract 0.25:

The probabilities are therefore 0.50, 0.25, and 0.25. Substituting them into the table gives a total of 1.

Add events that cannot happen together

Two events are mutually exclusive when they cannot both happen on the same trial. On a single die roll, “roll a 2” and “roll a 5” are mutually exclusive. On one spin, red and blue are mutually exclusive, so:

All the individual outcomes of an experiment are mutually exclusive. Together they are exhaustive, meaning that they cover every possibility. This is why their probabilities add to 1.

Watch for shared outcomes

“Roll an even number” and “roll a number greater than 3” can both happen: 4 and 6 belong to both events. Adding their two half chances would count these faces twice.

List the outcomes that fit either description: {2, 4, 5, 6}. Four of the six faces work, so the probability is $4\mathrm{/}6=2\mathrm{/}3$.

Sometimes it is easier to calculate the chance that an event fails. The complement of an event contains every outcome outside that event.

Include everything outside the event

If A is “roll a 6,” then “not A” is “roll 1, 2, 3, 4, or 5.” The complement is also written A′, read “A prime,” or Aᶜ. Its probability can be written $P(\text{not }A)$, $P(A^{\mathrm{\prime }})$, or $P(A^c)$.

In any trial, exactly one of A and “not A” happens. They cannot happen together, and there is no outcome outside the pair. Their probabilities therefore add to 1:

Subtract the event’s probability to find its complement:

Subtract from the whole

For a fraction, write 1 with the same denominator. If $P(A)=2\mathrm{/}9$, then:

With a decimal, subtract from 1. If $P(A)=0.27$, then $P(\text{not }A)=0.73$. With a percent, subtract from 100%. A 65% chance of rain gives a 35% chance of no rain during that forecast period.

The rule also works for unequal chances. A player who makes a free throw with probability 0.78 misses with probability $1-0.78=0.22$. Making and missing cover every result of the throw.

Choose the shorter count

A bag has 5 red, 3 blue, and 2 green counters. For “not green,” you could count red and blue to get eight favorable counters. You could also subtract green’s chance:

Both routes give the same value. The complement is useful when one small group is easier to count than all the other groups.

Keep the boundary

The complement of “greater than 4” is “less than or equal to 4.” The number 4 belongs in the complement because it is not greater than 4. On a die, that complement is {1, 2, 3, 4}.

Two events can be mutually exclusive without being complements. On a four-color spinner, “red” and “blue” cannot happen together, but green and yellow remain possible. With probabilities red 0.35, blue 0.20, green 0.30, and yellow 0.15, the complement of red includes all three other colors. Its probability is $1-0.35=0.65$. Red and blue alone total only 0.55.

A coin has a simple familiar model. An object such as a paper cup may land in several ways with unequal chances. Repeated trials help you estimate those chances from what actually happens.

Record every trial

Choose clear outcome categories before starting. Carry out each trial in the same way, using the same object and method. Record results as they happen so that none is forgotten. Tally marks can help, and a frequency table gathers the final counts.

The frequency of an outcome is how many times it happens. Its relative frequency compares that count with the number of trials:

Used to estimate a probability, this is also called experimental probability.

Work from a table

In a recorded experiment, a thumbtack landed point up 32 times and point down 18 times in 50 drops. If you carry out a sharp-object experiment, have an adult supervise and keep hands clear of the landing area.

The estimated probability of point up is 0.64, or 64%. Dividing by 18 would compare the two outcome counts; the required denominator is all 50 trials.

The frequencies total 50, and the relative frequencies total 1. These totals help you spot a missed or repeated record. Another run of 50 drops could give a different count, such as 30 or 35 point-up results, so 0.64 is an estimate.

Use records from everyday events

If 47 of 60 seeds sprout, their relative frequency is $47\mathrm{/}60\approx 0.78$. This estimates the chance that a seed of this kind sprouts under similar conditions.

A bus that was late on 9 of the last 40 school days has relative frequency $9\mathrm{/}40=0.225$, about 23%. Those records are useful for estimating future lateness while the route and timetable stay much the same. A new route could change the chance.

Simulate a situation

A simulation uses another chance process to imitate a situation. Suppose a player makes 70% of free throws. Generate equally likely random digits from 0 through 9, and assign them before starting:

• Digits 0 through 6: a make.
• Digits 7 through 9: a miss.

Seven of the ten digits mean “make,” matching the probability 0.7. Each generated digit represents one throw. A calculator or computer can generate many digits quickly. Count the simulated makes and divide by the total throws to find the simulated relative frequency.

A fair coin can produce seven heads in ten flips. Another person may get four heads in ten. Short experiments vary a lot, even when the coin and method are the same.

Watch the proportion settle

One run of fair-coin flips gave these running totals:

The proportion moved toward the theoretical probability of 0.5. This long-run tendency is called the Law of Large Numbers: as trials increase, relative frequency tends to get closer to the probability.

It can move away for a while. A further head after the first ten flips would make the fraction $8\mathrm{/}11$, farther from 0.5 than 0.7 is. The law describes the pattern over many trials, not a rule for each new point.

Read a running graph

Put the number of trials on the horizontal axis and relative frequency on the vertical axis. After each trial, plot the updated fraction. Early points can jump sharply. Later points tend to vary less and settle near one value.

For a fair coin, that value is near 0.5. For another object, the value where the line settles gives an estimate of its probability. An estimate of 0.52 from 200 careful flips is more reliable than 0.7 from ten comparable flips because it uses many more results.

Combine counts, not group averages

Suppose group A records 8 successes in 20 trials. Group B records 40 successes in 80 trials. To combine the results, add both counts and both totals:

The two group proportions are 0.4 and 0.5. Averaging them gives 0.45, which gives the small group the same weight as the large one. Pooling counts gives each individual trial the same weight. That is the right comparison when the groups repeated the same experiment.

A streak does not change the next chance

After five heads in a row, a fair coin still has probability $1\mathrm{/}2$ of heads on the next flip. The coin has no memory, and tails is not “due.” The same idea holds for fresh rolls of a die.

Later trials help the proportion settle because early results become a smaller share of the growing total. The coin does not need to produce a matching streak of tails to correct the earlier heads.

A model predicts chances. An experiment gives observed proportions. Comparing the two helps you decide whether the model fits the object and the way it is used.

Compare the same event

For a fair die, the theoretical probability of a 6 is $1\mathrm{/}6\approx 0.167$. Suppose a die shows 6 on 80 of 300 rolls. Its relative frequency is $80\mathrm{/}300\approx 0.267$. A fair die would be expected to give about 50 sixes in 300 rolls.

The gap is large for that many trials. It is strong evidence that the die favors 6. If the count were 53 instead, the relative frequency would be $53\mathrm{/}300\approx 0.177$. That small gap is ordinary chance variation and gives no reason to suspect bias.

Read the gap and the trial count together

Four sixes in twelve rolls gives about 0.33. This is farther from 0.167, but twelve rolls are few. Short runs can be uneven by chance, so this is only weak evidence of bias.

A large gap that remains after many careful trials is stronger evidence against a model. Experimental evidence supports a judgment about fairness; even a large experiment leaves some uncertainty.

Check possible causes of a mismatch:

• Natural chance variation, especially with few trials.
• A biased object.
• A method that favors some results.
• A missed result or another recording error.

For example, always dropping a coin the same way from a very small height may stop it behaving like an ordinary coin flip. Repeating a poor method more times keeps that problem in the data.

Build a model from data

For an unfamiliar object, you may have no known theoretical model. A tossed paper cup lands on its side 62 times, open end down 30 times, and open end up 8 times in 100 trials. Divide each count by 100:

These values total 1. The model is not uniform: three possible outcomes have very different estimated chances.

Data can also test a familiar guess. A penny spun on a table lands heads up 31 times in 100 spins. Its relative frequency, 0.31, is far from 0.5 for 100 trials. Heads and tails do not appear equally likely for this spinning method.

A known theoretical model can have equal or unequal probabilities. Use it to calculate chances; use many careful trials to check its fit. When no known model gives the chances, use relative frequencies to estimate them.

If an event has probability 0.4, you expect it in about four tenths of many repeated trials. The expected frequency turns that proportion into a predicted count.

Multiply probability by trials

A fair die is rolled 600 times. The event “3 or 6” has two favorable faces out of six, so $P(3\text{ or }6)=2\mathrm{/}6=1\mathrm{/}3$. The expected frequency is:

This predicts about 200 rolls showing a 3 or a 6. The actual count will vary. The remaining faces have an expected total of $600-200=400$ rolls. Expected frequencies for all outcomes add to the number of trials.

A decimal probability works the same way. If red has probability 0.35 on a spinner, the expected count in 40 spins is $0.35\times 40=14$.

You can also use a probability estimated from data. With the thumbtack estimate of 0.64 for point up, the predicted count in 200 drops is $0.64\times 200=128$. The prediction depends on that experimental estimate and on repeating the same kind of drop.

Interpret a fractional count

In 50 fair-die rolls, the expected number of sixes is:

A real experiment has a whole-number count. The value 8.3 describes the average count over many repeats of the entire 50-roll experiment. For a rough prediction of one run, say “about 8 sixes.”

Expected frequency is a count, so it can exceed 1. A probability cannot. Multiplying a small chance by many trials can produce a large expected count.

Work backward

You can find a missing number of trials from an expected count. If an outcome has probability 0.15 and is expected 45 times:

To find a probability, divide the expected count by the trials. An expected frequency of 30 in 120 trials gives $30\mathrm{/}120=0.25$.

Leave room for chance variation

For 100 flips of a fair coin, the expected number of heads is 50. Counts such as 46 and 55 are ordinary results. Exactly 50 heads occurs in only about 8% of such 100-flip experiments.

Probability helps you judge whether game rules are fair and compare choices with uncertain results. It also helps you separate a small chance from an impossible event.

Check the rules of a game

A fair chance game gives every player the same probability of winning. A fair die alone does not make the game fair.

Suppose A wins on 1 or 2, while B wins on 3, 4, 5, or 6. Their chances are $2\mathrm{/}6=1\mathrm{/}3$ and $4\mathrm{/}6=2\mathrm{/}3$. In 60 rounds, their expected wins are 20 and 40. A might lead after a few rounds, but B still has the larger winning chance.

Giving A the faces 1, 2, and 3 and B the faces 4, 5, and 6 makes both chances $3\mathrm{/}6=1\mathrm{/}2$. Now the rules are fair.

Find the chance of a guess

A question with four options and one correct answer gives a random guess probability $1\mathrm{/}4$ of being correct. Guessing on 20 such questions gives an expected score of $20\times 1\mathrm{/}4=5$ correct answers.

Ruling out one wrong answer leaves three possibilities. Choosing at random among them gives probability $1\mathrm{/}3$ of being correct. You have improved the chance, but an individual guess can still be wrong.

Compare unwanted events

A risk is the chance of an unwanted event. A risk of “1 in 200” means:

It is equal to 5 in 1000. Among 600 people facing this risk, the expected number affected is $600\mathrm{/}200=3$.

To compare “1 in 200” with “3 in 500,” use the same form. Their decimals are 0.005 and 0.006, so 3 in 500 is the larger risk. The larger denominator alone does not decide the comparison.

Consider the consequences

Two events can have the same chance but very different effects. A 1% chance of a scratch and a 1% chance of a serious injury call for different decisions. The severity of an injury can make protection worthwhile even when the chance is small.

A weather forecast also describes a chance. A 30% rain forecast gives a 30% chance of measurable rain at your location during the stated period. Across many comparable forecasts that match the true chances well, rain occurs about three times in ten. It does not mean rain lasts for 30% of the day. A dry day is possible under that forecast.

Holding 5 of 400 raffle tickets gives a winning probability of $5\mathrm{/}400=1\mathrm{/}80$. You are unlikely to win, yet a person holding just one ticket can win. Probability guides a decision without telling you which result the next trial will produce.