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Basic Concepts of Probability

Define probability including impossible and certain events.

Learning Objectives

After completing this section, you should be able to:

Define probability including impossible and certain events.
Calculate basic theoretical probabilities.
Calculate basic empirical probabilities.
Distinguish among theoretical, empirical, and subjective probability.
Calculate the probability of the complement of an event.

Introducing Probability

Uncertainty is, almost by definition, a nebulous concept. In order to put enough constraints on it that we can mathematically study it, we will focus on uncertainty strictly in the context of experiments. Recall that experiments are processes whose outcomes are unknown; the sample space for the experiment is the collection of all those possible outcomes. When we want to talk about the likelihood of particular outcomes, we sometimes group outcomes together; for example, in the Monopoly example at the beginning of this section, we were interested in the roll of 2 dice that might fall as a 4, 5, or 7. A grouping of outcomes that we’re interested in is called an event. In other words, an event is a subset of the sample space of an experiment; it often consists of the outcomes of interest to the experimenter.

Once we have defined the event that interests us, we can try to assess the likelihood of that event. We do that by assigning a number to each event (\(E\)) called the probability of that event (\(P(E)\)). The probability of an event is a number between 0 and 1 (inclusive). If the probability of an event is 0, then the event is impossible. On the other hand, an event with probability 1 is certain to occur. In general, the higher the probability of an event, the more likely it is that the event will occur.

Determining Certain and Impossible Events

Try it.

Consider an experiment that consists of rolling a single standard 6-sided die (with faces numbered 1-6). Decide if these probabilities are equal to zero, equal to one, or somewhere in between.

\(P(\text{roll a 4})\)
\(P(\text{roll a 7})\)
\(P(\text{roll a positive number})\)
\(P(\text{roll a}\ \frac{1}{3})\)
\(P(\text{roll an even number})\)
\(P(\text{roll a single-digit number})\)

Solution

Let's start by identifying the sample space. For one roll of this die, the possible outcomes are {1, 2, 3, 4, 5,6}. We can use that to assess these probabilities:

We see that 4 is in the sample space, so it’s possible that it will be the outcome. It’s not certain to be the outcome, though. So, \(0
Notice that 7 is not in the sample space. So, \(P(\text{roll a}\ 7)=0\).
Every outcome in the sample space is a positive number, so this event is certain. Thus, \(P(\text{roll a positive number})=1\).
Since \(\frac{1}{3}\) is not in the sample space, \(P(\text{roll a}\ \frac{1}{3})=0\).
Some outcomes in the sample space are even numbers (2, 4, and 6), but the others aren’t. So, \(0
Every outcome in the sample space is a single-digit number, so \(P(\text{roll a single-digit number})=1\).

Three Ways to Assign Probabilities

The probabilities of events that are certain or impossible are easy to assign; they’re just 1 or 0, respectively. What do we do about those in-between cases, for events that might or might not occur? There are three methods to assign probabilities that we can choose from. We’ll discuss them here, in order of reliability.

Condensed — the full section is in OpenStax Contemporary Mathematics.

New Probabilities from Old: Complements

One of the goals of the rest of this chapter is learning how to break down complicated probability calculations into easier probability calculations. We’ll look at the first of the tools we can use to accomplish this goal in this section; the rest will come later.

Given an event \(E\), the complement of \(E\) (denoted \(E'\)) is the collection of all of the outcomes that are not in \(E\). (This is language that is taken from set theory, which you can learn more about elsewhere in this text.) Since every outcome in the sample space either is or is not in \(E\), it follows that \(n(E)+n(E')=n(S)\). So, if the outcomes in \(S\) are equally likely, we can compute theoretical probabilities \(P(E)=\frac{n(E)}{n(S)}\) and \(P(E')=\frac{n(E')}{n(S)}\). Then, adding these last two equations, we get \[\]

\[\begin{array}{lll}P(E)+P(E') & = & \frac{n(E)}{n(S)}+\frac{n(E')}{n(S)} \\ & = & \frac{n(E)+n(E')}{n(S)} \\ & = & \frac{n(S)}{n(S)} \\ & = & 1\end{array}\]

Thus, if we subtract \(P(E')\) from both sides, we can conclude that \(P(E)=1-P(E')\). Though we performed this calculation under the assumption that the outcomes in \(S\) are all equally likely, the last equation is true in every situation.

How is this helpful? Sometimes it is easier to compute the probability that an event won’t happen than it is to compute the probability that it will. To apply this principle, it’s helpful to review some tricks for dealing with inequalities. If an event is defined in terms of an inequality, the complement will be defined in terms of the opposite inequality: Both the direction and the inclusivity will be reversed, as shown in the table below.

If \(E\) is defined with:	then \(E'\) is defined with:
\[<\]	\[\ge\]
\[\le\]	\[>\]
\[>\]	\[\le\]
\[\ge\]	\[<\]

Condensed — the full section is in OpenStax Contemporary Mathematics.

Key Concepts

The theoretical probability of an event is the ratio of the number of equally likely outcomes in the event to the number of equally likely outcomes in the sample space.
Empirical probabilities are computed by repeating the experiment many times, and then dividing the number of replications that result in the event of interest by the total number of replications.
Subjective probabilities are assigned based on subjective criteria, usually because the experiment can’t be repeated and the outcomes in the sample space are not equally likely.
The probability of the complement of an event is found by subtracting the probability of the event from one.

Formulas

For an experiment whose sample space \(S\) consists of equally likely outcomes, the theoretical probability of the event \(E\) is the ratio \(P(E)=\frac{n(E)}{n(S)}\) where \(n(E)\) and \(n(S)\) denote the number of outcomes in the event and in the sample space, respectively.
\(P(E)=1-P(E')\)

Practice (7)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Consider an experiment that consists of rolling a single standard 6-sided die (with faces numbered 1-6). Decide if these probabilities are equal to zero, equal to one, or somewhere in between.
1. \(P(\text{roll a 4})\)
2. \(P(\text{roll a 7})\)
3. \(P(\text{roll a positive number})\)
4. \(P(\text{roll a}\ \frac{1}{3})\)
5. \(P(\text{roll an even number})\)
6. \(P(\text{roll a single-digit number})\)
Lafunua jibu
Let's start by identifying the sample space. For one roll of this die, the possible outcomes are {1, 2, 3, 4, 5,6}. We can use that to assess these probabilities:
1. We see that 4 is in the sample space, so it’s possible that it will be the outcome. It’s not certain to be the outcome, though. So, \(0
2. Notice that 7 is not in the sample space. So, \(P(\text{roll a}\ 7)=0\).
3. Every outcome in the sample space is a positive number, so this event is certain. Thus, \(P(\text{roll a positive number})=1\).
4. Since \(\frac{1}{3}\) is not in the sample space, \(P(\text{roll a}\ \frac{1}{3})=0\).
5. Some outcomes in the sample space are even numbers (2, 4, and 6), but the others aren’t. So, \(0
6. Every outcome in the sample space is a single-digit number, so \(P(\text{roll a single-digit number})=1\).
Recall that a standard deck of cards consists of 52 unique cards which are labeled with a rank (the whole numbers from 2 to 10, plus J, Q, K, and A) and a suit (\(♣\), \(♢\), \(♡\), or \(♠\)). A standard deck is thoroughly shuffled, and you draw one card at random (so every card has an equal chance of being drawn). Find the theoretical probability of each of these events:
1. The card is \(10♠\).
2. The card is a \(♡\).
3. The card is a king (K).
Lafunua jibu
There are 52 cards in the deck, so the sample space for each of these experiments has 52 elements. That will be the denominator for each of our probabilities.
1. There is only one \(10♠\) in the deck, so this event only has one outcome in it. Thus, \(P(10♠)=\frac{1}{52}\).
2. There are 13 \(♡\text{s}\) in the deck, so \(P(♡)=\frac{13}{52}=\frac{1}{4}\).
3. There are 4 cards of each rank in the deck, so \(P(\text{K})=\frac{4}{52}=\frac{1}{13}\).
In the Basic Concepts of Probability, we were considering a Monopoly game where, if your sister rolled a sum of 4, 5, or 7 with 2 standard dice, you would win the game. What is the probability of this event? Use tables to determine your answer.

Lafunua jibu

We should think of this experiment as occurring in two stages: (1) one die roll, then (2) another die roll. Even though these two stages will usually occur simultaneously in practice, since they’re independent, it’s okay to treat them separately.
Step 1: Since we have two independent stages, let’s create a table (), which is probably the most efficient method for determining the sample space.
Now, each of the 36 ordered pairs in the table represent an equally likely outcome.
Step 2: To make our analysis easier, let’s replace each ordered pair with the sum ().
Step 3: Since the event we’re interested in is the one consisting of rolls of 4, 5, or 7. Let’s shade those in ().
Our event contains 13 outcomes, so the probability that your sister rolls a losing number is \(\frac{13}{36}\).
If you flip a fair coin 3 times, what is the probability of each event? Use a tree diagram to determine your answer
1. You flip exactly 2 heads.
2. You flip 2 consecutive heads at some point in the 3 flips.
3. All 3 flips show the same result.
Lafunua jibu
Let’s build a tree to identify the sample space ().
The sample space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, which has 8 elements.
1. Flipping exactly 2 heads occurs three times (HHT, HTH, THH), so the probability is \(\frac{3}{8}\).
2. Flipping 2 consecutive heads at some point in the experiment happens 3 times: HHH, HHT, THH. So, the probability is \(\frac{3}{8}\).
3. There are 2 outcomes that all show the same result: HHH and TTT. So, the probability is \(\frac{2}{8}=\frac{1}{4}\).
Assign an empirical probability to the following events:
1. Jose is on the basketball court practicing his shots from the free throw line. He made 47 out of his last 80 attempts. What is the probability he makes his next shot?
2. Amy is about to begin her morning commute. Over her last 60 commutes, she arrived at work 12 times in under half an hour. What is the probability that she arrives at work in 30 minutes or less?
3. Felix is playing Yahtzee with his sister. Felix won 14 of the last 20 games he played against her. How likely is he to win this game?
Lafunua jibu
1. Since Jose made 47 out of his last 80 attempts, assign this event an empirical probability of \(\frac{47}{80}\approx 59\%\).
2. Amy completed the commute in under 30 minutes in 12 of the last 60 commutes, so we can estimate her probability of making it in under 30 minutes this time at \(\frac{12}{60}=20\%\).
3. Since Felix has won 14 of the last 20 games, assign a probability for a win this time of \(\frac{14}{20}=70\%\).
Classify each of the following probabilities as theoretical, empirical, or subjective.
1. An eccentric billionaire is testing a brand new rocket system. He says there is a 15% chance of failure.
2. With 4 seconds to go in a close basketball playoff game, the home team need 3 points to tie up the game and send it to overtime. A TV commentator says that team captain should take the final 3-point shot, because he has a 38% chance of making it (greater than every other player on the team).
3. Felix is losing his Yahtzee game against his sister. He has one more chance to roll 2 dice; he’ll win the game if they both come up 4. The probability of this is about 2.8%.
Lafunua jibu
1. This experiment has never been run before, so the given probability is subjective.
2. Presumably, the commentator has access to each player’s performance statistics over the entire season. So, the given probability is likely empirical.
3. Rolling 2 dice results in a sample space with equally likely outcomes. This probability is theoretical. (We’ll learn how to calculate that probability later in this chapter.)
1. If you roll a standard 6-sided die, what is the probability that the result will be a number greater than one?
2. If you roll two standard 6-sided dice, what is the probability that the sum will be 10 or less?
3. If you flip a fair coin 3 times, what is the probability that at least one flip will come up tails?
Lafunua jibu
1. Here, the sample space is {1, 2, 3, 4, 5, 6}. It’s easy enough to see that the probability in question is \(\frac{5}{6}\), because there are 5 outcomes that fall into the event “roll a number greater than 1.” Let’s also apply our new formula to find that probability. Since \(E\) is defined using the inequality \(\text{roll}>1\), then \(E'\) is defined using \(\text{roll}\le 1\). Since there’s only one outcome (1) in \(E'\), we have \(P(E')=\frac{1}{6}\). Thus, \(P(E)=1-P(E')=\frac{5}{6}\).
2. In , we found the following table of equally likely outcomes for rolling 2 dice ():
  Here, the event \(E\) is defined by the inequality \(\text{sum}\le 10\). Thus, \(E'\) is defined by \(\text{sum}>10\). There are three outcomes in \(E'\): two 11s and one 12. Thus, \(P(E)=1-P(E')=1-\frac{3}{36}=\frac{11}{12}\).
3. In , we found the sample space for this experiment consisted of these equally likely outcomes: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Our event \(E\) is defined by \(\text{T}\ge 1\), so \(E'\) is defined by \(\text{T}<1\). The only outcome in \(E'\) is the first one on the list, where zero tails are flipped. So, \(P(E)=1-P(E')=1-\frac{1}{8}=\frac{7}{8}\).

Symbols used here

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\approx

approximately equal
Equal to the precision shown, not exactly.

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\bar{x},\ \mu

sample mean, population mean
Average of the data; average of the whole population.

\sigma,\ s,\ \sigma^2

standard deviation, sample s.d., variance
Typical distance from the mean; its square.

E[X],\ \operatorname{Var}(X)

expected value, variance
Probability-weighted average of X; its spread.

N(\mu, \sigma^2),\ z

normal distribution, z-score
The bell curve with mean μ and variance σ²; (x − μ)/σ.

How to: Basic Concepts of Probability

Define probability including impossible and certain events.
Calculate basic theoretical probabilities.
Calculate basic empirical probabilities.
Distinguish among theoretical, empirical, and subjective probability.
Calculate the probability of the complement of an event.
We see that 4 is in the sample space, so it’s possible that it will be the outcome. It’s not certain to be the outcome, though. So,
Notice that 7 is not in the sample space. So,
Every outcome in the sample space is a positive number, so this event is certain. Thus,

Questions people ask

Mean or median — which should I use?

Median when the data have outliers or a long tail (incomes, house prices); mean when the data are roughly symmetric and you want every value to count. Report both if they disagree — the gap is itself information.

What does a p-value actually say?

The probability of seeing data at least this extreme if the null hypothesis were true. It is not the probability that the null hypothesis is true.

Why divide by n − 1 for the sample variance?

The sample mean sits closer to the sample than the true mean does, so squared deviations from it are slightly too small on average; dividing by n − 1 instead of n corrects the bias.

Jaribu kufanya mambo yako mwenyewe

Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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