Probability | Free ACT Math Course | The School of Mathematics

<!DOCTYPE html>

<head>

<title>ACT Math: Probability | The School of Mathematics</title>

<style>

:root {

--blue: #4285f4;

--green: #34a853;

--red: #ea4335;

--gray-bg: #f8f9fa;

--border: #e0e0e0;

}

* { box-sizing: border-box; }

body {

font-family: -apple-system, BlinkMacSystemFont, "Segoe UI", Roboto, Helvetica, Arial, sans-serif;

background: #ffffff;

color: #202124;

line-height: 1.65;

margin: 0;

padding: 0;

}

.wrap { max-width: 820px; margin: 0 auto; padding: 32px 20px 80px; }

h1 { font-size: 2rem; line-height: 1.25; margin-bottom: 10px; }

h2 {

font-size: 1.5rem;

margin-top: 48px;

padding-top: 8px;

border-top: 2px solid var(--border);

}

h3 { font-size: 1.15rem; margin-bottom: 6px; }

.intro { font-size: 1.05rem; color: #3c4043; }

.cta-group {

display: flex;

gap: 12px;

flex-wrap: wrap;

margin: 22px 0 32px;

}

.cta-btn {

flex: 1 1 220px;

text-align: center;

text-decoration: none;

font-weight: 700;

font-size: 1rem;

padding: 15px 18px;

border-radius: 10px;

}

.cta-primary { background: var(--blue); color: #fff; }

.cta-primary:hover { background: #3367d6; }

.cta-secondary { background: #fff; color: var(--blue); border: 2px solid var(--blue); }

.cta-secondary:hover { background: #eef3fd; }

nav.toc {

background: var(--gray-bg);

border: 1px solid var(--border);

border-radius: 10px;

padding: 18px 22px;

margin-bottom: 36px;

}

nav.toc h2 { margin-top: 0; border-top: none; font-size: 1.1rem; }

nav.toc ol { margin: 0; padding-left: 22px; }

nav.toc a { color: var(--blue); text-decoration: none; }

nav.toc a:hover { text-decoration: underline; }

table.ref {

width: 100%;

border-collapse: collapse;

margin: 18px 0 26px;

font-size: 0.95rem;

}

table.ref th, table.ref td {

border: 1px solid var(--border);

padding: 10px 12px;

text-align: left;

vertical-align: top;

}

table.ref th { background: var(--gray-bg); }

.example {

background: var(--gray-bg);

border-left: 4px solid var(--blue);

border-radius: 6px;

padding: 16px 20px;

margin: 18px 0;

}

.example p { margin: 6px 0; }

.step-math { font-family: "Cambria Math", Georgia, serif; }

.frac {

display: inline-flex;

flex-direction: column;

vertical-align: middle;

text-align: center;

font-size: 0.85em;

line-height: 1;

margin: 0 2px;

position: relative;

top: 0.1em;

}

.frac .num { border-bottom: 1.3px solid #202124; padding: 0 4px 1px; }

.frac .den { padding: 1px 4px 0; }

.problem {

border: 1px solid var(--border);

border-radius: 10px;

padding: 18px 22px 20px;

margin: 18px 0;

}

.problem .prompt { font-weight: 600; margin-bottom: 12px; }

.options { display: flex; flex-direction: column; gap: 8px; margin: 0 0 4px; }

.option-btn {

text-align: left;

padding: 10px 14px;

border: 1px solid #ccc;

border-radius: 6px;

background: #fff;

cursor: pointer;

font-size: 0.98rem;

font-family: inherit;

}

.option-btn:hover:not(:disabled) { background: #f1f3f4; }

.option-btn.correct { background: #e6f4ea; border-color: var(--green); font-weight: 600; }

.option-btn.incorrect { background: #fce8e6; border-color: var(--red); }

.option-btn:disabled { cursor: default; }

.explanation {

margin-top: 0;

padding: 0 16px;

background: var(--gray-bg);

border-left: 4px solid var(--blue);

border-radius: 4px;

max-height: 0;

opacity: 0;

overflow: hidden;

transition: max-height 0.35s ease, opacity 0.35s ease, margin-top 0.35s ease, padding 0.35s ease;

}

.explanation.show {

max-height: 600px;

opacity: 1;

margin-top: 14px;

padding: 14px 16px;

}

.explanation p { margin: 6px 0; }

.explanation .label { font-weight: 700; }

.mistake-list li { margin-bottom: 10px; }

.faq-item { margin-bottom: 18px; }

.faq-item h3 { margin-bottom: 4px; }

footer.cta-final { margin-top: 56px; }

</style>

</head>

<body>

<h1>ACT Math: Probability</h1>

Probability questions on the ACT reward recognizing the right method (add or multiply, permutation or combination) more than heavy computation. This free, complete lesson covers basic probability and the complement rule, independent and dependent events, the fundamental counting principle, permutations and combinations, and expected value. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback, in the same four-answer-choice, calculator-allowed format as the current ACT. Everything here is free, and you can keep practicing afterward with the full ACT Math Question Bank linked below.

</p>

<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/act-probability-quiz-1">Practice Probability Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/ACT-Math-Qbank">Explore the Full ACT Math Qbank</a>

</div>

<h2>What's covered in this lesson</h2>

<ol>

<li><a href="#basic">Basic Probability and the Complement Rule</a></li>

<li><a href="#events">Independent and Dependent Events</a></li>

<li><a href="#counting">The Fundamental Counting Principle</a></li>

<li><a href="#perm-comb">Permutations and Combinations</a></li>

<li><a href="#expected-value">Expected Value</a></li>

<li><a href="#mistakes">Common Mistakes to Avoid</a></li>

<li><a href="#faq">Frequently Asked Questions</a></li>

</ol>

</nav>

<h2 id="basic">1. Basic Probability and the Complement Rule</h2>

<p class="step-math" style="text-align:center; font-size:1.1rem;">P(event) = <span class="frac"><span class="num">favorable outcomes</span><span class="den">total outcomes</span></span></p>

<p>The <strong>complement rule</strong>: the probability an event does NOT happen equals 1 minus the probability it does.</p>

<p><strong>Worked Example:</strong> A bag has 5 red, 3 blue, and 2 green marbles. Find P(red).</p>

<p>Total marbles: 5 + 3 + 2 = 10.</p>

</div>

<p class="prompt">1. A bag has 4 red and 6 blue marbles. Find P(blue).</p>

</div>

<p><span class="label">Explanation:</span> Total marbles: 10. P(blue) = <span class="frac"><span class="num">6</span><span class="den">10</span></span> = <span class="frac"><span class="num">3</span><span class="den">5</span></span>.</p>

</div>

<p class="prompt">2. A standard six-sided die is rolled once. What is P(rolling a number greater than 4)?</p>

</div>

<p><span class="label">Explanation:</span> Numbers greater than 4: {5, 6}, so 2 favorable outcomes out of 6: <span class="frac"><span class="num">2</span><span class="den">6</span></span> = <span class="frac"><span class="num">1</span><span class="den">3</span></span>.</p>

</div>

<p class="prompt">3. The probability of rain tomorrow is 0.35. What is the probability it does NOT rain?</p>

</div>

<p><span class="label">Explanation:</span> 1 − 0.35 = 0.65.</p>

</div>

<p class="prompt">4. A jar has 12 candies, and the probability of picking a red one is <span class="frac"><span class="num">1</span><span class="den">3</span></span>. How many red candies are in the jar?</p>

</div>

<p><span class="label">Explanation:</span> 12 × <span class="frac"><span class="num">1</span><span class="den">3</span></span> = 4.</p>

</div>

<h2 id="events">2. Independent and Dependent Events</h2>

<p>Events are <strong>independent</strong> if one doesn't affect the other's probability (like separate coin flips, or drawing "with replacement"). Events are <strong>dependent</strong> if one changes the probability of the next (like drawing "without replacement"). Either way, to find P(A and B), multiply the two probabilities, recalculating the second probability if the events are dependent. Use addition instead when you want P(A or B) for outcomes that can't happen together.</p>

<p><strong>Worked Example:</strong> A coin is flipped and a six-sided die is rolled. Find P(heads AND rolling a 4).</p>

<p>These are independent events.</p>

<p class="step-math">P(heads) × P(4) = <span class="frac"><span class="num">1</span><span class="den">2</span></span> × <span class="frac"><span class="num">1</span><span class="den">6</span></span> = <span class="frac"><span class="num">1</span><span class="den">12</span></span></p>

</div>

<p class="prompt">1. Two six-sided dice are rolled. What is P(both show a 6)?</p>

</div>

<p><span class="label">Explanation:</span> These are independent events: <span class="frac"><span class="num">1</span><span class="den">6</span></span> × <span class="frac"><span class="num">1</span><span class="den">6</span></span> = <span class="frac"><span class="num">1</span><span class="den">36</span></span>.</p>

</div>

<p class="prompt">2. A bag has 5 red and 3 blue marbles (8 total). One marble is drawn, replaced, and a second is drawn. Find P(both red).</p>

</div>

<p><span class="label">Explanation:</span> With replacement, these are independent: <span class="frac"><span class="num">5</span><span class="den">8</span></span> × <span class="frac"><span class="num">5</span><span class="den">8</span></span> = <span class="frac"><span class="num">25</span><span class="den">64</span></span>.</p>

</div>

<p class="prompt">3. The same bag (5 red, 3 blue, 8 total) has one marble drawn and NOT replaced, then a second is drawn. Find P(both red).</p>

</div>

<p><span class="label">Explanation:</span> Without replacement, these are dependent: <span class="frac"><span class="num">5</span><span class="den">8</span></span> × <span class="frac"><span class="num">4</span><span class="den">7</span></span> = <span class="frac"><span class="num">20</span><span class="den">56</span></span> = <span class="frac"><span class="num">5</span><span class="den">14</span></span>.</p>

</div>

<p class="prompt">4. A deck of 52 cards has 4 aces. Two cards are drawn without replacement. What is P(first card is an ace AND second card is an ace)?</p>

</div>

<p><span class="label">Explanation:</span> <span class="frac"><span class="num">4</span><span class="den">52</span></span> × <span class="frac"><span class="num">3</span><span class="den">51</span></span> = <span class="frac"><span class="num">12</span><span class="den">2652</span></span> = <span class="frac"><span class="num">1</span><span class="den">221</span></span>.</p>

</div>

<p class="prompt">5. Is rolling a die twice in a row an example of independent or dependent events?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pb-5',true)">A) Independent</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pb-5',false)">B) Dependent</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pb-5',false)">C) Neither, this isn't a valid probability scenario</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pb-5',false)">D) It depends on the specific numbers rolled</button>

</div>

<p><span class="label">Explanation:</span> Each roll of the die has no effect on the outcome of any other roll, making them independent.</p>

</div>

<h2 id="counting">3. The Fundamental Counting Principle</h2>

<p>When a process has multiple independent stages, multiply the number of choices at each stage to find the total number of possible outcomes.</p>

<p><strong>Worked Example:</strong> A restaurant offers 4 appetizers, 6 entrees, and 3 desserts. How many different 3-course meals are possible?</p>

</div>

<p class="prompt">1. A license plate has 3 letters followed by 4 digits. If letters and digits can repeat, how many different license plates are possible?</p>

</div>

<p><span class="label">Explanation:</span> 26 × 26 × 26 × 10 × 10 × 10 × 10 = 17,576 × 10,000 = 175,760,000.</p>

</div>

<p class="prompt">2. A store has 5 shirt styles, 4 pants styles, and 2 shoe styles. How many different complete outfits (one of each) can be made?</p>

</div>

<p><span class="label">Explanation:</span> 5 × 4 × 2 = 40.</p>

</div>

<p class="prompt">3. How many different 4-digit codes can be made using digits 0-9, if digits can repeat?</p>

</div>

<p><span class="label">Explanation:</span> 10 × 10 × 10 × 10 = 10,000.</p>

</div>

<p class="prompt">4. How many different 4-digit codes can be made using digits 0-9, if NO digit can repeat?</p>

</div>

<p><span class="label">Explanation:</span> With no repeats, each choice reduces the options by one: 10 × 9 × 8 × 7 = 5,040.</p>

</div>

<h2 id="perm-comb">4. Permutations and Combinations</h2>

<p>A <strong>permutation</strong> counts arrangements where order matters (like assigning distinct roles). A <strong>combination</strong> counts selections where order does NOT matter (like choosing a group with no distinct roles).</p>

<p><strong>Worked Example:</strong> In how many ways can 5 runners finish 1st, 2nd, and 3rd place?</p>

<p>Order matters here (this is a permutation):</p>

</div>

<p class="prompt">1. A club has 8 members. In how many ways can a president, vice president, and secretary (3 distinct roles) be chosen?</p>

</div>

<p><span class="label">Explanation:</span> Since the roles are distinct, order matters: 8 × 7 × 6 = 336.</p>

</div>

<p class="prompt">2. A club has 8 members. In how many ways can a 3-person committee (no distinct roles) be chosen?</p>

</div>

<p><span class="label">Explanation:</span> Since a committee has no distinct roles, order doesn't matter. Take the permutation (336) and divide by the number of ways to arrange 3 people (3! = 6): 336 ÷ 6 = 56.</p>

</div>

<p class="prompt">3. Does choosing a team captain and co-captain from 10 players require a permutation or a combination?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pd-3',true)">A) Permutation</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-3',false)">B) Combination</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-3',false)">C) Neither</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-3',false)">D) Both give the same answer</button>

</div>

<p><span class="label">Explanation:</span> Captain and co-captain are distinct roles, so which player fills which role matters, this is a permutation.</p>

</div>

<p class="prompt">4. How many ways can you choose 2 pizza toppings from a menu of 6 toppings, if order doesn't matter?</p>

</div>

<p><span class="label">Explanation:</span> The permutation would be 6 × 5 = 30. Since order doesn't matter, divide by 2! = 2: 30 ÷ 2 = 15.</p>

</div>

<h2 id="expected-value">5. Expected Value</h2>

<p><strong>Expected value</strong> is a weighted average of all possible outcomes, multiply each outcome's value by its probability, then add up the results.</p>

<p><strong>Worked Example:</strong> A game costs $2 to play. You win $10 with probability 0.1, and win nothing otherwise. Find the expected net value of playing.</p>

<p>Expected winnings: 0.1(10) + 0.9(0) = 1.</p>

<p class="step-math">Net expected value: 1 − 2 = −1 (an average loss of $1 per play)</p>

</div>

<p class="prompt">1. A spinner has 4 equal sections: win $8 (1 section), win $2 (1 section), and win $0 (2 sections). What is the expected value of one spin?</p>

</div>

<p><span class="label">Explanation:</span> <span class="frac"><span class="num">1</span><span class="den">4</span></span>(8) + <span class="frac"><span class="num">1</span><span class="den">4</span></span>(2) + <span class="frac"><span class="num">2</span><span class="den">4</span></span>(0) = 2 + 0.5 + 0 = $2.50.</p>

</div>

<p class="prompt">2. A raffle has 100 tickets. One ticket wins $200, and the rest win nothing. If a ticket costs $3, what is the expected net value of buying one ticket?</p>

</div>

<p><span class="label">Explanation:</span> Expected winnings: <span class="frac"><span class="num">1</span><span class="den">100</span></span>(200) = $2. Net: 2 − 3 = −$1.</p>

</div>

<p class="prompt">3. A die is rolled. You win $6 if it lands on 6, and lose $1 otherwise. What is the expected value?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pe-3',true)">A) <span class="frac"><span class="num">1</span><span class="den">6</span></span> dollars</button>

</div>

<p><span class="label">Explanation:</span> <span class="frac"><span class="num">1</span><span class="den">6</span></span>(6) + <span class="frac"><span class="num">5</span><span class="den">6</span></span>(−1) = 1 − <span class="frac"><span class="num">5</span><span class="den">6</span></span> = <span class="frac"><span class="num">1</span><span class="den">6</span></span>.</p>

</div>

<p class="prompt">4. An insurance policy pays $50,000 if a rare event occurs (probability 0.001), and $0 otherwise. What is the expected payout?</p>

</div>

<p><span class="label">Explanation:</span> 0.001 × 50,000 = $50.</p>

</div>

<h2 id="mistakes">Common Mistakes to Avoid</h2>

<li><strong>Adding probabilities when you should multiply, or vice versa.</strong> Use multiplication for "and" (both events happening together); use addition for "or" between outcomes that can't both happen.</li>

<li><strong>Forgetting to adjust the denominator for dependent events.</strong> "Without replacement" means the total number of outcomes shrinks for the second event, recalculate the probability using the new total.</li>

<li><strong>Confusing permutations and combinations.</strong> If the roles or positions are distinguishable (order matters), use a permutation. If you're just selecting a group with no distinct roles (order doesn't matter), use a combination.</li>

<li><strong>Forgetting to subtract the cost when finding a net expected value.</strong> Expected value calculates the average payout; if there's a cost to play, subtract it separately at the end.</li>

<li><strong>Treating a "greater than" or "at least" condition as excluding the boundary value incorrectly.</strong> Carefully check whether the condition (like "more than 4" versus "4 or more") includes the boundary number itself before counting favorable outcomes.</li>

</ul>

<h2 id="faq">Frequently Asked Questions</h2>

<h3>How can I quickly tell if I need a permutation or a combination?</h3>

<p>Ask whether swapping two selected items would create a meaningfully different outcome. If yes (like assigning a president versus a vice president), order matters, use a permutation. If no (like just picking a group of people), order doesn't matter, use a combination.</p>

</div>

<h3>Do I need to memorize the permutation and combination formulas for the ACT?</h3>

<p>They're useful, but many ACT counting problems can be solved just as quickly with the fundamental counting principle or the "permutation, then divide by the redundant arrangements" approach shown in this lesson, without needing to plug into a formula.</p>

</div>

<h3>What does a negative expected value actually mean?</h3>

<p>It means that, on average over many repeated trials, you'd expect to lose money (or whatever quantity is being measured) rather than gain it, even though any single outcome could still be a win.</p>

</div>

<h3>Where can I practice more problems like these?</h3>

<p>The <a href="https://theschoolofmathematics.com/quiz/act-probability-quiz-1">Probability quizzes</a> in the ACT Math Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested on the ACT Math section.</p>

</div>

<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/act-probability-quiz-1">Practice Probability Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/ACT-Math-Qbank">Explore the Full ACT Math Qbank</a>

</div>

</footer>

</div>

function checkAnswer(btn, problemId, isCorrect) {

var problem = document.getElementById(problemId);

var buttons = problem.querySelectorAll('.option-btn');

buttons.forEach(function(b) { b.disabled = true; });

if (isCorrect) {

btn.classList.add('correct');

} else {

btn.classList.add('incorrect');

var correctBtn = problem.querySelector('.option-btn[data-correct="true"]');

if (correctBtn) { correctBtn.classList.add('correct'); }

}

var exp = problem.querySelector('.explanation');

if (exp) { exp.classList.add('show'); }

}

</script>

</body>

</html>