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MathChapter 9: Categorical Data and Probability
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Probability measures how likely an event is. SAT questions in this chapter focus on identifying the correct sample space, deciding whether one event changes another, and translating AND or OR language into the appropriate operation.

Outcomes, trials, events, and sample spaces

Core probability vocabulary
TermMeaningOriginal example
TrialOne performance of a random processSpin a fair spinner once
OutcomeOne possible result of a trialThe spinner lands on B
Sample spaceAll possible outcomes under consideration\(\{A,B,C,D\}\)
EventOne or more outcomes of interestLanding on A or C
Basic probability for equally likely outcomes
\[P(A)=\frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}\]

Use this ratio only when the listed outcomes are equally likely.

Worked example

Count favorable outcomes

A fair spinner has eight equal sectors labeled \(1\) through \(8\). What is the probability of spinning a factor of \(8\)?

  1. List favorable outcomes

    The factors shown are \(1,2,4,8\), so there are \(4\) favorable outcomes.

  2. Divide by the sample space

    There are \(8\) equally likely sectors, so \(P=4/8=1/2\).

The probability is \(\frac12\), or \(50\%\).

Independent and dependent events

Does the first event change the second?

Independent

The first result does not change the probability of the second. Standard draws with replacement are independent.

Dependent

The first result changes the remaining counts or probabilities. Standard draws without replacement are dependent.

Replacement determines dependence

A two-panel comparison shows how returning or keeping the first item changes the second draw.

What happens after the first draw?

Item returnedWith replacement: independent

The original category counts and total are restored before the second draw.

Example: The second denominator is the same as the first denominator.
Item kept outWithout replacement: dependent

The remaining category count and total depend on the first result.

Example: The second denominator is one less than the first denominator.

AND means follow one complete path

Independent AND rule
\[P(A\text{ and }B)=P(A)\cdot P(B)\]

Use the unchanged second probability when the events are independent.

Dependent AND rule
\[P(A\text{ and }B)=P(A)\cdot P(B\text{ following }A)\]

The second factor must reflect what remains after event \(A\).

Two draws with replacement

A container has 3 green tokens and 2 gold tokens. Each token is returned after selection, so every second-stage branch keeps the original probabilities.

First draw
  • Green\(\frac{3}{5}\)
    • Green \(\frac{3}{5}\)Green, then green\(\frac{3}{5}\cdot\frac{3}{5}=\frac{9}{25}\)
    • Gold \(\frac{2}{5}\)Green, then gold\(\frac{3}{5}\cdot\frac{2}{5}=\frac{6}{25}\)
  • Gold\(\frac{2}{5}\)
    • Green \(\frac{3}{5}\)Gold, then green\(\frac{2}{5}\cdot\frac{3}{5}=\frac{6}{25}\)
    • Gold \(\frac{2}{5}\)Gold, then gold\(\frac{2}{5}\cdot\frac{2}{5}=\frac{4}{25}\)
Worked example

Without replacement

A container holds \(5\) square tiles and \(3\) round tiles. Two tiles are drawn without replacement. What is the probability of square, then round?

  1. First draw

    The probability of square is \(5/8\).

  2. Update totals

    After removing a square, \(7\) tiles remain and all \(3\) round tiles remain.

  3. Multiply the path

    \(P(\text{square then round})=\frac58\cdot\frac37=\frac{15}{56}\).

The probability is \(\frac{15}{56}\).

OR for mutually exclusive events

Mutually exclusive events
Events that cannot occur on the same trial. One outcome cannot belong to both events.
Mutually exclusive OR rule
\[P(A\text{ or }B)=P(A)+P(B)\]

This chapter-level rule applies when \(A\) and \(B\) cannot occur together. For a finite equally likely sample space, direct counting of distinct favorable outcomes is also reliable.

Worked example

Add disjoint outcomes

A fair number cube is rolled. What is the probability of rolling a \(1\) or a \(6\)?

  1. Check overlap

    A single roll cannot be both \(1\) and \(6\), so the events are mutually exclusive.

  2. Add

    \(P(1\text{ or }6)=1/6+1/6=2/6=1/3\).

The probability is \(\frac13\).

Same-category multi-draw problems

Calculate each valid branch

  1. Name the branches

    For two categories, 'same category' means first-first or second-second.

  2. Multiply each AND path

    Update the second probability if sampling is without replacement.

  3. Add the disjoint paths

    The two complete paths cannot happen simultaneously, so add their probabilities.

Worked example

Two tiles of the same shape

A box has \(4\) triangles and \(2\) circles. Two shapes are drawn without replacement. What is the probability that they match?

  1. Triangle branch

    \(P(TT)=\frac46\cdot\frac35=\frac25\).

  2. Circle branch

    \(P(CC)=\frac26\cdot\frac15=\frac1{15}\).

  3. Add branches

    \(\frac25+\frac1{15}=\frac7{15}\).

The probability of matching shapes is \(\frac7{15}\).

Recognition guide

Probability language and the corresponding action
Prompt signalActionCheck
AND, then, followed byMultiply along one pathDoes the first event change the second?
OR between mutually exclusive eventsAdd the disjoint event probabilitiesCan both occur on one trial?
With replacementRestore original countsSecond denominator stays the same
Without replacementUpdate remaining countsSecond denominator decreases by one

Common mistakes and traps

  • Adding probabilities for an AND path instead of multiplying.
  • Multiplying mutually exclusive alternatives instead of adding them.
  • Treating no-replacement draws as independent.
  • Changing the second denominator after an item was replaced.
  • Reducing the total but forgetting to reduce the favorable category after drawing from it.
  • Counting overlapping favorable outcomes twice in an OR question.
  • Reporting a favorable-outcome count instead of dividing by the sample-space size.
Mini check

Replacement check

A jar has \(3\) mint and \(5\) peach tokens. Two are drawn without replacement. What is \(P(\text{mint then mint})\)?

  1. \(\frac38\cdot\frac38\)
  2. \(\frac38\cdot\frac27\)
  3. \(\frac38+\frac27\)
  4. \(\frac28\cdot\frac37\)
Show answer and explanation

Answer: \(\frac38\cdot\frac27=\frac3{28}\)

After one mint token is removed, \(2\) mint tokens remain among \(7\) total tokens.

Key takeaways

What to remember

  • For equally likely outcomes, probability is favorable outcomes divided by total outcomes.
  • Replacement restores the original composition; no replacement changes it.
  • Multiply probabilities along an AND path.
  • Add probabilities for mutually exclusive OR alternatives.
  • For same-category questions, calculate each valid branch and then add the branches.
Continue learning

Put these notes into practice

Apply the ideas with SAT-style questions, then reinforce key details with flashcards.