In this section, you’ll learn how to compute the probability of simple events, work with unions of mutually exclusive and non-mutually exclusive events, calculate intersections for independent events, and read one-way and two-way tables to find joint, marginal, and conditional probabilities.

Definitions

Probability

The likelihood of an event occurring, expressed as a number between $0$ (impossible) and $1$ (certain).

Sample space

The complete set of all possible outcomes of an experiment or situation, often denoted by the symbol $S$. For example, rolling a fair die has the sample space $S=\{1,2,3,4,5,6\}$.

Event

A subset of outcomes from the sample space that we are interested in.

Joint probability

The probability that two events $A$ and $B$ occur together, denoted

$$P(A\cap B)=\frac{\text{count of outcomes in both }A\text{ and }B}{\text{total number of outcomes}}.$$

Marginal probability

The probability of a single event ignoring any other variables, found by summing row or column totals in a two-way table and dividing by the grand total.

Conditional probability

The probability of event $A$ given that event $B$ has occurred, denoted

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}.$$

Mutually exclusive events

Events that cannot occur at the same time. For $A$ and $B$,

$$P(A\cup B)=P(A)+P(B).$$

Non-mutually exclusive events

Events that can occur together. For $A$ and $B$,

$$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$

Independent events

Events whose outcomes do not affect each other. For $A$ and $B$,

$$P(A\cap B)=P(A)\times P(B).$$

Basic probability formula

Probability is a way to compare:

• how many outcomes satisfy a condition (the event), to
• how many outcomes are possible overall (the sample space).

So every probability question starts by identifying the sample space and the event of interest.

For any event $E$ in a finite sample space,

$$P(E)=\frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}.$$

This works because probability is a proportion. If an event occurs in half of all equally likely outcomes, its probability is $\frac{1}{2}$. If it occurs in one out of ten outcomes, its probability is $\frac{1}{10}$. Probabilities are always between $0$ and $1$, inclusive: $0$ means the event cannot occur, and $1$ means it must occur.

Many probability questions - especially on the Praxis - give information in tables instead of listing outcomes. The same proportion idea still applies; you just read the favorable count and the total count from the table.

Reading one-way tables

A one-way table organizes counts for a single categorical variable. Each entry is a count for one category, and the entire table represents the sample space.

To find a probability from a one-way table:

• Compute the grand total by summing all counts in the table.
• Identify the count corresponding to the event of interest.
• Divide that category count by the grand total.

This is the basic probability formula in table form: the category count is the number of favorable outcomes, and the grand total is the number of possible outcomes.

Example: Marble colors A bag contains:

Color	Count
Red	$3$
Blue	$2$
Green	$5$

What is $P(\text{green})$?

Reading two-way tables

A two-way table organizes counts for two categorical variables at the same time. Each cell shows how often a specific combination occurs. That lets you compute probabilities for single events and for relationships between events.

In a two-way table:

• Rows typically represent categories of one variable.
• Columns represent categories of a second variable.
• Cell entries show joint counts.
• Row and column totals summarize overall frequencies and form the basis for marginal probabilities.
• The grand total represents the full sample space.

A generic two-way table looks like this:

	Event $B$	Not $B$	Total
Event $A$	$n_{AB}$	$n_{A\bar{B}}$	$n_A$
Not $A$	$n_{\bar{A}B}$	$n_{\bar{A}\bar{B}}$	$n_{\bar{A}}$
Total	$n_B$	$n_{\bar{B}}$	$N$

From this table, you can compute three different types of probabilities.

• Joint probability measures the chance that both events occur together. Use one cell count divided by the grand total:

  $$P(A\cap B)=\frac{n_{AB}}{N}.$$

• Marginal probability measures the chance of a single event, ignoring the other variable. Use a row or column total divided by the grand total:

  $$P(A)=\frac{n_A}{N},P(B)=\frac{n_B}{N}.$$

• Conditional probability measures the chance of one event given that another event has occurred. Here, the “total possible outcomes” becomes the relevant row or column total:

  $$P(A\mid B)=\frac{n_{AB}}{n_B}.$$

The main difference is what you treat as the sample space:

• Joint probabilities use the entire table.
• Marginal probabilities use the entire table but focus on one variable.
• Conditional probabilities restrict attention to outcomes that satisfy the condition.

Example: Joint, marginal, and conditional probabilities A café records how customers pay (Cash, Card, or Mobile) across three days.

Day	Cash	Card	Mobile
Monday	$25$	$15$	$10$
Tuesday	$20$	$25$	$5$
Wednesday	$15$	$30$	$15$

Sometimes a problem won’t provide totals. In that case, compute the row totals, column totals, and the grand total before finding probabilities.

First, compute the totals for each row, each column, and the grand total. The row totals and column totals both add to $160$, so the table is consistent.

Day	Cash	Card	Mobile	Total
Monday	$25$	$15$	$10$	$50$
Tuesday	$20$	$25$	$5$	$50$
Wednesday	$15$	$30$	$15$	$60$
Total	$60$	$70$	$30$	$160$

The marginal probability of paying by card uses the column total for Card divided by the grand total:

$$P(\text{Card})=\frac{70}{160}=0.4375$$

The joint probability of a customer paying in cash on Tuesday uses the Tuesday-Cash cell divided by the grand total:

$$P(\text{Tuesday and Cash})=\frac{20}{160}=0.125$$

The conditional probability of paying with mobile given it is Wednesday restricts the sample space to the Wednesday row:

$$P(\text{Mobile}\mid \text{Wednesday})=\frac{15}{60}=0.25$$

Interpretation: About $43.75\%$ of all customers pay by card overall (marginal).

The chance of randomly selecting a Tuesday cash transaction is $12.5\%$ (joint).

If it is known to be Wednesday, there is a $25\%$ chance the customer paid using mobile (conditional).

Answer: $P(\text{Card})=0.4375,P(\text{Tuesday and Cash})=0.125,P(\text{Mobile}\mid \text{Wednesday})=0.25$

After you can compute probabilities from sample spaces and tables, the next step is combining events. Here, it matters whether events can overlap and whether one event changes the probability of the other.

Unions and intersections of events

When you combine events, two common questions come up:

• Are you finding the probability that at least one of two events occurs (a union)?
• Or are you finding the probability that both events occur (an intersection)?

The correct rule depends on whether the events can occur together and whether they affect each other.

• Union, mutually exclusive If two events cannot occur at the same time, there is no overlap. The probability that at least one occurs is the sum of their probabilities:

  $$P(A\cup B)=P(A)+P(B).$$

• Union, non-mutually exclusive If two events can occur together, adding $P(A)$ and $P(B)$ counts the overlap twice. Subtract $P(A\cap B)$ once to correct for that:

  $$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$

• Intersection, independent If two events are independent, the occurrence of one does not change the probability of the other. Multiply their probabilities to get the probability both occur:

  $$P(A\cap B)=P(A)\times P(B).$$

Example: Mutually exclusive union Two fair dice are rolled once. Let $A$ = “sum is $2$,” and $B$ = “sum is $12$.” What is $P(A\cup B)$?

These two sums cannot occur at the same time, so the events are mutually exclusive.

$$>P(A)=\frac{1}{36},P(B)=\frac{1}{36}$$

$$>P(A\cup B)=\frac{1}{36}+\frac{1}{36}=\frac{2}{36}=\frac{1}{18}$$

Answer: $\frac{1}{18}$

Example: Non-mutually exclusive union You draw one card from a standard $52$-card deck. Let $A$ = “heart” and $B$ = “king.” What is $P(A\cup B)$?

These events are not mutually exclusive because the king of hearts is both a heart and a king.

• $P(A)=\frac{13}{52}$
• $P(B)=\frac{4}{52}$
• $P(A\cap B)=\frac{1}{52}$

$$>P(A\cup B)=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$$

Answer: $\frac{4}{13}$

With and without replacement

When you draw multiple items from a finite set, decide whether the first draw changes the second. That depends on whether you put the item back.

• With replacement means the item is returned before the next draw. The sample space stays the same, so the probabilities stay the same.

• Without replacement means the item is not returned. The sample space shrinks, so the probabilities change after each draw.

Example: Without replacement A box contains $3$ red and $2$ blue balls ($5$ total). You draw two balls without replacement. What is the probability both are red?

On the first draw, there are $3$ red balls out of $5$ total:

$$>P({\text{red}}_1)=\frac{3}{5}.$$

After drawing a red ball and not replacing it, there are $2$ red balls left out of $4$ total:

$$>P({\text{red}}_2\mid {\text{red}}_1)=\frac{2}{4}=\frac{1}{2}.$$

Because both events must occur, multiply the probabilities:

$$>P(\text{both red})=\frac{3}{5}\times \frac{1}{2}=\frac{3}{10}=\mathrm{0.3.}$$

Answer: $0.3$

Independent intersections

Sometimes one event has no effect on another. When events are independent, the probability that both occur is the product of their individual probabilities.

Example: Two-flip intersection What is the probability of flipping two heads in a row with a fair coin?

Each flip is independent, and the probability of heads on any single flip is $\frac{1}{2}$. Therefore,

$$>P(\text{head then head})=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}=\mathrm{0.25.}$$

Answer: $0.25$

Example: Pair of cards Draw two cards without replacement. What is $P(\text{both face cards})$?

There are $12$ face cards in a $52$-card deck. On the first draw,

$$>P({\text{face}}_1)=\frac{12}{52}.$$

After drawing a face card and not replacing it, there are $11$ face cards left out of $51$ total cards:

$$>P({\text{face}}_2\mid {\text{face}}_1)=\frac{11}{51}.$$

Multiply to find the joint probability:

$$>\frac{12}{52}\times \frac{11}{51}=\frac{132}{2652}\approx \mathrm{0.0498.}$$

Answer: $\approx 0.0498$