Probability and Conditional Probability | Free SAT Math Course

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<title>Probability and Conditional Probability - Free SAT Math Lesson | The School of Mathematics</title>

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<h1>Probability and Conditional Probability: Free SAT Math Lesson</h1>

"Probability and conditional probability" is an officially named skill in College Board's Problem-Solving and Data Analysis domain, and it's tested through everything from simple single-event probability to reading two-way frequency tables. This free, complete lesson covers basic probability, independent events and mutually exclusive events, reading two-way tables to find conditional probability, and the complement rule for "at least one" problems. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback. Everything here is free, and you can keep practicing afterward with the full SAT Math Question Bank linked below.

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

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

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

<ol>

<li><a href="#basic">Basic Probability</a></li>

<li><a href="#independent">Independent and Mutually Exclusive Events</a></li>

<li><a href="#tables">Two-Way Tables and Conditional Probability</a></li>

<li><a href="#complement">The Complement Rule</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</h2>

<p>The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes:</p>

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

<p>Probability is always between 0 (impossible) and 1 (certain), and can be written as a fraction, decimal, or percent.</p>

<p><strong>Worked Example:</strong> A bag contains 5 red, 3 blue, and 2 green marbles. What is the probability of randomly selecting a blue marble?</p>

<p>There are 3 + 5 + 2 = 10 total marbles.</p>

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<p class="prompt">1. A standard six-sided die is rolled once. What is the probability of rolling a number greater than 4?</p>

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<p><span class="label">Explanation:</span> The numbers greater than 4 are 5 and 6, two outcomes out of six: <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>

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<p class="prompt">2. A jar contains 8 red, 5 blue, and 7 yellow candies. What is the probability of randomly selecting a yellow candy?</p>

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<p><span class="label">Explanation:</span> There are 8 + 5 + 7 = 20 candies total. P(yellow) = <span class="frac"><span class="num">7</span><span class="den">20</span></span>.</p>

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<p class="prompt">3. A standard deck of 52 cards is shuffled. What is the probability of drawing a heart?</p>

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<p><span class="label">Explanation:</span> There are 13 hearts out of 52 cards: <span class="frac"><span class="num">13</span><span class="den">52</span></span> = <span class="frac"><span class="num">1</span><span class="den">4</span></span>.</p>

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<p class="prompt">4. A spinner is divided into 8 equal sections numbered 1 through 8. What is the probability of landing on an even number?</p>

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<p><span class="label">Explanation:</span> The even numbers from 1 to 8 are 2, 4, 6, 8, four out of eight: <span class="frac"><span class="num">4</span><span class="den">8</span></span> = <span class="frac"><span class="num">1</span><span class="den">2</span></span>.</p>

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<h2 id="independent">2. Independent and Mutually Exclusive Events</h2>

<p>Two events are <strong>independent</strong> if the outcome of one doesn't affect the probability of the other. For independent events, multiply their individual probabilities to find the probability of both happening:</p>

<p>Two events are <strong>mutually exclusive</strong> if they cannot both happen at the same time. For mutually exclusive events, add their individual probabilities to find the probability of either happening:</p>

<p><strong>Worked Example:</strong> A coin is flipped and a die is rolled. What is the probability of getting heads AND rolling a 6?</p>

<p>These events are independent, one doesn't affect the other.</p>

<p class="step-math">P(heads and 6) = <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>

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<p class="prompt">1. A fair coin is flipped twice. What is the probability of getting heads both times?</p>

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<p><span class="label">Explanation:</span> The two flips are independent: <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">2</span></span> = <span class="frac"><span class="num">1</span><span class="den">4</span></span>.</p>

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<p class="prompt">2. A bag contains 4 red and 6 blue marbles. A marble is drawn, replaced, and a second marble is drawn. What is the probability both marbles are red?</p>

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<p><span class="label">Explanation:</span> Since the marble is replaced, each draw has P(red) = <span class="frac"><span class="num">4</span><span class="den">10</span></span> = <span class="frac"><span class="num">2</span><span class="den">5</span></span>, and the draws are independent: <span class="frac"><span class="num">2</span><span class="den">5</span></span> × <span class="frac"><span class="num">2</span><span class="den">5</span></span> = <span class="frac"><span class="num">4</span><span class="den">25</span></span>.</p>

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<p class="prompt">3. Two dice are rolled. What is the probability that the first die shows a 3 AND the second die shows an even number?</p>

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<p><span class="label">Explanation:</span> The two dice are independent: P(3) = <span class="frac"><span class="num">1</span><span class="den">6</span></span> and P(even) = <span class="frac"><span class="num">3</span><span class="den">6</span></span> = <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">2</span></span> = <span class="frac"><span class="num">1</span><span class="den">12</span></span>.</p>

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<p class="prompt">4. A single card is drawn from a standard deck. What is the probability that the card is a king OR a queen?</p>

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<p><span class="label">Explanation:</span> A card can't be both a king and a queen, so these events are mutually exclusive: <span class="frac"><span class="num">4</span><span class="den">52</span></span> + <span class="frac"><span class="num">4</span><span class="den">52</span></span> = <span class="frac"><span class="num">8</span><span class="den">52</span></span> = <span class="frac"><span class="num">2</span><span class="den">13</span></span>.</p>

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<p class="prompt">5. A single six-sided die is rolled. What is the probability of rolling a 2 OR a 5?</p>

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<p><span class="label">Explanation:</span> Rolling a 2 and rolling a 5 are mutually exclusive: <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">2</span><span class="den">6</span></span> = <span class="frac"><span class="num">1</span><span class="den">3</span></span>.</p>

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<h2 id="tables">3. Two-Way Tables and Conditional Probability</h2>

<p>A <strong>two-way table</strong> organizes data by two categories at once and is a common way the SAT presents information for conditional probability questions. A <strong>conditional probability</strong>, "the probability of A given B," restricts your attention to only the row or column matching the given condition, then asks what fraction of that restricted group also satisfies the first condition.</p>

<tr><th></th><th>Owns a Car</th><th>Doesn't Own a Car</th><th>Total</th></tr>

<tr><td><strong>Female</strong></td><td class="num">90</td><td class="num">60</td><td class="num">150</td></tr>

<tr><td><strong>Total</strong></td><td class="num">210</td><td class="num">90</td><td class="num">300</td></tr>

</table>

<p><strong>Worked Example:</strong> Using the table above, what is the probability that a randomly selected person owns a car, given that they are male?</p>

<p>"Given that they are male" restricts attention to the male row only, 150 people, of whom 120 own a car.</p>

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<p class="prompt">1. Using the table above, what is the probability that a randomly selected person is female, given that they own a car?</p>

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<p><span class="label">Explanation:</span> "Given that they own a car" restricts attention to the car-owner column, 210 people, of whom 90 are female: <span class="frac"><span class="num">90</span><span class="den">210</span></span> = <span class="frac"><span class="num">3</span><span class="den">7</span></span>.</p>

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<p class="prompt">2. Using the table above, what is the probability that a randomly selected person does not own a car, given that they are female?</p>

</div>

<p><span class="label">Explanation:</span> "Given that they are female" restricts attention to the female row, 150 people, of whom 60 don't own a car: <span class="frac"><span class="num">60</span><span class="den">150</span></span> = <span class="frac"><span class="num">2</span><span class="den">5</span></span>.</p>

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<p class="prompt">3. Using the table above, what is the overall probability that a randomly selected person owns a car (regardless of gender)?</p>

</div>

<p><span class="label">Explanation:</span> Overall, 210 out of 300 people own a car: <span class="frac"><span class="num">210</span><span class="den">300</span></span> = <span class="frac"><span class="num">7</span><span class="den">10</span></span>.</p>

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<p class="prompt">4. Using the table above, what is the probability that a randomly selected person is male?</p>

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<p><span class="label">Explanation:</span> There are 150 males out of 300 total people: <span class="frac"><span class="num">150</span><span class="den">300</span></span> = <span class="frac"><span class="num">1</span><span class="den">2</span></span>.</p>

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<p class="prompt">5. Using the table above, what is the probability that a randomly selected person is male AND owns a car?</p>

</div>

<p><span class="label">Explanation:</span> Unlike the conditional probability questions above, this asks for the joint count out of the grand total: <span class="frac"><span class="num">120</span><span class="den">300</span></span> = <span class="frac"><span class="num">2</span><span class="den">5</span></span>.</p>

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<h2 id="complement">4. The Complement Rule</h2>

<p>The <strong>complement</strong> of an event is everything that event does not include. Since an event and its complement together account for every possible outcome:</p>

<p>This rule is especially useful for "at least one" problems, which are often much easier to solve by finding the probability of the opposite, "none," and subtracting from 1.</p>

<p><strong>Worked Example:</strong> A fair coin is flipped 3 times. What is the probability of getting at least one heads?</p>

<p>It's easier to find the complement, getting zero heads (all tails), first.</p>

<p class="step-math">P(all tails) = <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">2</span></span> × <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">8</span></span></p>

<p class="step-math">P(at least one heads) = 1 − <span class="frac"><span class="num">1</span><span class="den">8</span></span> = <span class="frac"><span class="num">7</span><span class="den">8</span></span></p>

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<p class="prompt">1. The probability of an event occurring is <span class="frac"><span class="num">3</span><span class="den">8</span></span>. What is the probability that the event does NOT occur?</p>

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<p><span class="label">Explanation:</span> P(not A) = 1 − <span class="frac"><span class="num">3</span><span class="den">8</span></span> = <span class="frac"><span class="num">5</span><span class="den">8</span></span>.</p>

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<p class="prompt">2. A basketball player makes 70% of her free throws. What is the probability she misses a given free throw?</p>

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<p><span class="label">Explanation:</span> 100% − 70% = 30%.</p>

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<p class="prompt">3. A fair coin is flipped 3 times. What is the probability of getting AT LEAST one heads?</p>

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<p><span class="label">Explanation:</span> This is the worked example above: P(at least one heads) = 1 − P(all tails) = 1 − <span class="frac"><span class="num">1</span><span class="den">8</span></span> = <span class="frac"><span class="num">7</span><span class="den">8</span></span>.</p>

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<p class="prompt">4. A box contains 10 lightbulbs, 2 of which are defective. If one bulb is selected at random, what is the probability it is NOT defective?</p>

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<p><span class="label">Explanation:</span> P(defective) = <span class="frac"><span class="num">2</span><span class="den">10</span></span> = <span class="frac"><span class="num">1</span><span class="den">5</span></span>. P(not defective) = 1 − <span class="frac"><span class="num">1</span><span class="den">5</span></span> = <span class="frac"><span class="num">4</span><span class="den">5</span></span>.</p>

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

<li><strong>Adding probabilities for events that aren't mutually exclusive.</strong> The simple addition rule, P(A or B) = P(A) + P(B), only works when A and B can't both happen. If they can overlap, you need P(A) + P(B) − P(A and B) instead.</li>

<li><strong>Multiplying probabilities for events that aren't independent.</strong> If one event affects the probability of another, like drawing cards without replacement, you can't simply multiply the original probabilities together.</li>

<li><strong>Reading a conditional probability from the wrong row or column of a two-way table.</strong> "P(A given B)" means restrict to the B row or column first, then find what fraction of that group also satisfies A. Reversing the order gives a different, incorrect answer.</li>

<li><strong>Confusing "and" probability with conditional probability in a two-way table.</strong> P(A and B) divides the joint count by the grand total. P(A given B) divides the joint count by only B's row or column total. These use different denominators and give different answers.</li>

<li><strong>Overcomplicating an "at least one" problem instead of using the complement.</strong> Directly calculating "at least one success" often requires adding up several cases. Finding the complement, "zero successes," and subtracting from 1 is almost always faster.</li>

</ul>

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

<h3>How can I tell if two events are independent just from a word problem?</h3>

<p>Ask whether the outcome of one event changes the possible outcomes or probabilities of the other. Flipping a coin twice, or rolling two separate dice, are independent, since one result doesn't affect the other. Drawing cards from a deck without replacement is not independent, since removing a card changes what remains.</p>

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<h3>What's the fastest way to set up a conditional probability from a two-way table?</h3>

<p>Find the row or column that matches the "given" condition, treat that row or column's total as your new denominator, and use the specific cell that also matches the event you're asked about as your numerator.</p>

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<h3>Why does the complement rule work so well for "at least one" problems?</h3>

<p>"At least one" success is the opposite of "zero" successes, and there's exactly one way to get zero successes (every single trial fails), which is usually much simpler to calculate than adding up every way to get one success, two successes, three successes, and so on.</p>

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

<p>The <a href="https://theschoolofmathematics.com/quiz/sat-probability-conditional-probability-quiz">Probability & Conditional Probability quiz</a> in the SAT Math Question Bank includes additional original problems on this topic, along with quizzes covering every other Algebra, Advanced Math, Problem-Solving and Data Analysis, and Geometry and Trigonometry skill on the Digital SAT.</p>

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

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

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