Probability and Counting — ACT Math

Introduction

Probability questions look deceptively simple until you hit 'at least one' or 'given that' — those two phrases signal the hardest probability questions on the ACT.

Probability and counting questions appear 3-6 times per ACT, mostly in the medium-hard range (questions 30-55). Counting/combinations questions are rare but high-difficulty when they appear.

By the end of this lesson you will be able to:

Calculate theoretical probability and use complement, AND/OR rules
Distinguish between permutations and combinations and apply the correct formula
Solve conditional probability and 'at least one' problems

You'll calculate the probability that at least one of two events occurs using P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

The Concept

The Core Rule

P(event) = favorable outcomes / total outcomes. Complement: P(not A) = 1 − P(A). Addition rule: P(A or B) = P(A) + P(B) − P(A and B). Multiplication rule: P(A and B) = P(A) × P(B|A). Independent events: P(A and B) = P(A) × P(B). Permutation: nPr = n!/(n−r)!. Combination: nCr = n!/((n−r)!r!).

How the ACT tests this

Probability from a table or two-way frequency table
Asking for the number of ways to arrange or select items (permutation vs combination)
Asking for 'at least one' probability (best solved with complement)

Basic Probability Rules

P(event) = (# favorable outcomes)/(# total outcomes). All probabilities are between 0 and 1. P(A or B) = P(A) + P(B) − P(A and B). For mutually exclusive events, P(A and B) = 0, so P(A or B) = P(A) + P(B).

Complement shortcut: P(at least one) = 1 − P(none)
For two independent events: P(both) = P(A) × P(B)
Expected value = sum of (outcome × probability) for all outcomes

Counting Principles

Fundamental Counting Principle: if one event has m outcomes and another has n outcomes, together they have m × n outcomes. Permutations (order matters): nPr = n!/(n−r)!. Combinations (order doesn't matter): nCr = n!/((n−r)!r!).

Arrange n objects in a line: n! ways
Choose r from n without order: nCr = nPr / r!
Combinations with restriction: subtract arrangements that violate the restriction

Conditional Probability

P(A|B) = P(A and B) / P(B) — the probability of A given that B has already occurred. For two-way tables: P(A|B) = (cell count for A and B) / (row or column total for B).

Given a two-way table, use the restricted row/column total as the denominator
Independent events: P(A|B) = P(A) — knowing B gives no info about A
Conditional probability changes the sample space to only outcomes where B occurred

Your strategy

For 'at least one' problems, use complement: 1 − P(none of the events happen)
Decide permutation vs combination: if order matters → permutation; if not → combination
For two-way table probability, underline which row/column forms the restricted sample space
Draw a tree diagram or list outcomes for small sample spaces to avoid formula errors

Worked Examples

Easy Example 1 Using The Number Of Non-blue Marbles (7) As The Denominator Instead Of Total Marbles (10)

A bag contains 4 red, 3 blue, and 3 green marbles. One marble is drawn at random. What is the probability of drawing a blue marble?

A. 1/10
B. 3/10 (Correct answer)
C. 4/10
D. 3/7
E. 1/3

Step 1

Total marbles: 4 + 3 + 3 = 10

Step 2

Favorable outcomes (blue): 3

Step 3

P(blue) = 3/10

Correct answer: B

Why B is correct

Correct: 3/10

Why other options are wrong

A: 1/10 would be the probability of one specific marble out of 10

C: 4/10 is the probability of red, not blue

D: Used 7 non-blue marbles as denominator: wrong total

E: 1/3 assumes equal groups of 3; ignores that red has 4

⚠ Trap: Using the number of non-blue marbles (7) as the denominator instead of total marbles (10)

Medium Example 2 Using Combinations (5C3 = 10) When Order Matters, Or Using 5³ When Letters Can't Repeat

How many different 3-letter arrangements can be made from the letters A, B, C, D, E if each letter is used at most once and order matters?

A. 10
B. 20
C. 60 (Correct answer)
D. 120
E. 125

Step 1

Order matters → Permutation: 5P3 = 5!/(5−3)! = 5!/2!

Step 2

5!/2! = (5 × 4 × 3 × 2 × 1)/(2 × 1) = 120/2 = 60

Step 3

Answer: 60

Correct answer: C

Why C is correct

Correct: 5 × 4 × 3 = 60

Why other options are wrong

A: 5C3 = 10: combination where order doesn't matter — wrong approach

B: Arithmetic error

D: 5! = 120: all five letters arranged; but we only select 3

E: 5³ = 125: allows repeated letters — problem says at most once

⚠ Trap: Using combinations (5C3 = 10) when order matters, or using 5³ when letters can't repeat

Hard Example 3 Calculating P(exactly 1 Head) = 3/8 Instead Of P(at Least 1 Head) = 7/8

A coin is flipped 3 times. What is the probability of getting at least 1 head?

A. 1/8
B. 3/8
C. 6/8
D. 7/8 (Correct answer)
E. 1

Step 1

Use complement: P(at least 1 head) = 1 − P(no heads)

Step 2

P(no heads) = P(all tails) = (1/2)³ = 1/8

Step 3

P(at least 1 head) = 1 − 1/8 = 7/8

Correct answer: D

Why D is correct

Correct: 1 − 1/8 = 7/8

Why other options are wrong

A: This is P(all tails) = 1/8 — the complement, not the answer

B: P(exactly 1 head) = 3/8; only counts one-head scenarios

C: P(exactly 1 head) + P(exactly 2 heads) = 3/8 + 3/8 = 6/8; misses P(all 3 heads)

E: P = 1 would mean it's certain; tails-tails-tails is possible

⚠ Trap: Calculating P(exactly 1 head) = 3/8 instead of P(at least 1 head) = 7/8

Strategy Tips

Always use complement for 'at least one' problems — it is almost always faster
Order matters → permutation (nPr); order doesn't matter → combination (nCr)
For two-way table problems, write the denominator before the numerator
List out all outcomes for simple probability spaces (2-4 items) to avoid formula errors
Picking numbers: test small cases (2 coins, not 3) to verify your counting logic

Common pitfalls

Using combinations when the problem requires permutations (order matters)

Forgetting to subtract P(A and B) in the addition rule for non-mutually-exclusive events

Confusing 'at least one' with 'exactly one'

Basic probability: 30-45 seconds. Complement method: 60 seconds. Permutation/combination: 60-90 seconds. If a counting problem requires more than 2 minutes, use the Fundamental Counting Principle slot-by-slot.

Summary

P(at least one) = 1 − P(none) is the fastest approach for complement problems
Permutation (order matters): nPr = n!/(n−r)!; Combination (order doesn't): nCr = n!/((n−r)!r!)
Conditional probability restricts the sample space to the given condition

A password is 4 digits, each from 1-9, with no repeated digits. How many possible passwords are there? Identify whether this is a permutation or combination, then calculate.

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