Computing probabilities

This section covers 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.

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 compares two things:

• 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:

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

Think of probability as 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. When data is given as category counts, the grand total is the sum of all counts. Be careful with strict vs. inclusive inequalities when listing favorable outcomes: “less than $4$” excludes $4$, while “less than or equal to $4$” includes it.

When the sample space isn’t handed to you, you build it by enumeration - listing or counting every possible outcome.

Example: Rolling a pair of number cubes

You roll two fair six-sided number cubes. What is the probability that the sum of the two numbers rolled is $7$?

Each cube has $6$ faces, and the result of one cube doesn’t affect the other, so the sample space is every pair (first cube, second cube): $6\times 6=36$ equally likely outcomes.

Now list the pairs that sum to $7$:

$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$

That’s $6$ favorable outcomes out of $36$ total outcomes:

$$P(\text{sum}=7)=\frac{6}{36}=\frac{1}{6}$$

Answer: $\frac{1}{6}$

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

Reading two-way tables

A two-way table organizes counts for two categorical variables at once: rows represent one variable, columns represent the other, each cell holds a joint count, and the grand total represents the full sample space.

From a two-way table, you can compute three types of probabilities.

• Joint probability - the chance that both events occur together. Divide one cell count by the grand total:

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

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

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

• Conditional probability - the chance of one event given that another has occurred. The sample space shrinks to the relevant row or column total:

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

Joint and marginal probabilities use the grand total as the denominator; conditional probabilities use the row or column total for the given condition instead. The conditional probability formula works even when you aren’t reading from a table - if you’re given $P(A\cap B)$ and $P(B)$ directly, apply $P(A\mid B)=P(A\cap B)/P(B)$ the same way.

Common mistake: Using the grand total as the denominator for a conditional probability is one of the most frequent errors. When a question says “given that…” or “among those who…”, your denominator is the row or column total for that condition - not the overall total.

Example: Joint, marginal, and conditional probabilities

A café records how customers pay (Cash, Card, or Mobile) across three days.

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

If a table doesn’t give you row and column totals, add them up before computing any probabilities.

The marginal probability of paying by card uses the Card column total 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 - so the denominator is $60$, not $160$:

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

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 combining 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 individual 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 that both occur:

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

  For example, flipping two heads in a row with a fair coin: each flip is independent, so $P(\text{head then head})=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$.

Mutually exclusive ≠ independent. These two terms are easy to confuse, but they describe very different relationships. Mutually exclusive events cannot happen together - if $A$ occurs, $B$ is ruled out entirely. That means knowing $A$ occurred tells you $B$ did not occur, so the events are actually dependent. Independent events, by contrast, have no effect on each other at all. If $A$ and $B$ are both mutually exclusive and have nonzero probabilities, they cannot be independent.

Example: Mutually exclusive vs. non-mutually exclusive unions

You draw one card from a standard $52$-card deck.

Case 1 - mutually exclusive: Let $A$ = “draw a heart” and $B$ = “draw a spade.” A card cannot be both, so the events are mutually exclusive.

$$P(A)=\frac{13}{52},P(B)=\frac{13}{52}$$

$$P(A\cup B)=\frac{13}{52}+\frac{13}{52}=\frac{26}{52}=\frac{1}{2}$$

Case 2 - non-mutually exclusive: Let $A$ = “draw a heart” and $B$ = “draw a king.” The king of hearts is both, so the events can overlap.

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

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

The only difference between the two cases is whether you subtract the overlap. When events can share outcomes, count that overlap and remove it once.

Answer: Case 1: $\frac{1}{2}$; Case 2: $\frac{4}{13}$

With and without replacement

When you draw multiple items from a finite set, you need to consider 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 across draws.

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

Common mistake: Draws without replacement are dependent events - the outcome of the first draw changes the probabilities for the second. Don’t apply the independent-events multiplication rule $P(A\cap B)=P(A)\times P(B)$ here. Instead, multiply $P(\text{first draw})$ by the conditional probability $P(\text{second draw}\mid \text{first draw})$.

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: $0.3$