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How do you compute the probability of simple and compound events on the ACT?

Compute the probability of single events, complements, and compound events using the addition and multiplication rules, including independent and mutually exclusive events (Statistics and Probability).

An ACT Statistics answer on probability: the basic ratio of favorable to total outcomes, complements, the multiplication rule for independent events, and the addition rule for mutually exclusive events, with worked ACT-style questions and common traps.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. Basic probability
  3. Compound events: "and" versus "or"
  4. Using the complement for "at least one"
  5. Dependent events and drawing without replacement
  6. Why reading the wording matters
  7. Try this

What this topic is asking

Probability measures how likely an event is, as a number from 0 (impossible) to 1 (certain). The ACT tests the basic ratio of favorable to total outcomes, the complement rule, and compound events using the multiplication rule (independent events) and addition rule (mutually exclusive events).

Basic probability

Every probability starts from counting outcomes.

The complement rule is a powerful shortcut: when "at least one" is asked, it is often easier to compute the probability of none and subtract from 1.

Compound events: "and" versus "or"

The two compound rules are the heart of ACT probability.

Read the wording: "and" (both happen) usually means multiply; "or" (either happens) usually means add.

Using the complement for "at least one"

When a question asks for "at least one" success, the complement is "none", which is often a single multiplication. For example, the probability of at least one head in three coin flips is 1P(no heads)=1(12)3=118=781 - P(\text{no heads}) = 1 - \left(\frac{1}{2}\right)^{3} = 1 - \frac{1}{8} = \frac{7}{8}. Computing the "none" case and subtracting from 1 is much faster than adding up the probabilities of exactly one, two and three heads.

Dependent events and drawing without replacement

When outcomes do affect each other, the probabilities change between steps. Drawing two marbles without replacement is dependent: if a bag has 5 marbles with 2 red, the chance both draws are red is 25×14\frac{2}{5}\times \frac{1}{4}, because after removing one red, only 1 red remains among 4 marbles. Adjusting the second probability for the changed contents is the key skill here. With replacement, the draws are independent and the probability stays the same each time.

Why reading the wording matters

Most probability errors come from misreading "and" versus "or", or from forgetting that drawing without replacement changes the second probability. The reliable approach is to identify whether events are independent (multiply), mutually exclusive (add), or dependent (adjust the later probabilities), and to use the complement whenever "at least one" appears. Reducing the final fraction to lowest terms matches the answer choices the ACT provides.

Try this

Q1. A die is rolled. What is the probability of rolling a number greater than 4? [1 point]

  • Cue. Favorable outcomes are 5 and 6, so 26=13\frac{2}{6} = \frac{1}{3}.

Q2. Two independent events have probabilities 12\frac{1}{2} and 15\frac{1}{5}. What is the probability both occur? [1 point]

  • Cue. Multiply: 12×15=110\frac{1}{2}\times \frac{1}{5} = \frac{1}{10}.

Exam-style practice questions

Practice questions written in the style of ACT exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

ACT Math (style)1 marksA bag has 4 red, 3 blue and 5 green marbles. If one is drawn at random, what is the probability it is blue? (A) 14\frac{1}{4} (B) 312\frac{3}{12} (C) 13\frac{1}{3} (D) 512\frac{5}{12}
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The correct answer is (A), 14\frac{1}{4}.

Probability is favorable over total outcomes. There are 3 blue out of 4+3+5=124 + 3 + 5 = 12 marbles, so 312=14\frac{3}{12} = \frac{1}{4}. Choice (B) is correct unsimplified, but 14\frac{1}{4} is the reduced form, the expected answer.

ACT Math (style)1 marksA fair coin is flipped twice. What is the probability of getting heads both times? (A) 12\frac{1}{2} (B) 14\frac{1}{4} (C) 18\frac{1}{8} (D) 1
Show worked answer →

The correct answer is (B), 14\frac{1}{4}.

The two flips are independent, so multiply: P(H and H)=12×12=14P(\text{H and H}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. Choice (A) is the probability of one head; the second flip must also be accounted for.

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