SAT Math Lesson Plan 13: Probability and Combinatorics

Lesson Summary

This 45-minute lesson introduces key probability and combinatorics concepts tested on the SAT. Approximately 5–10% of SAT Math questions cover these topics, including basic probability, permutations, combinations, and conditional probability. Students will learn how to calculate probabilities, distinguish between different types of counting problems, and interpret results in real-world contexts. The lesson includes a Warm-Up, six worked examples, a Review section with guided practice, and a 10-question multiple-choice quiz with answer explanations. Aligned with Common Core standards, this lesson builds essential skills in event analysis and logical reasoning.

Lesson Objectives

Define and calculate basic probability.
Differentiate between permutations and combinations and apply appropriate formulas.
Understand and apply conditional probability.
Solve real-world problems involving probability and counting principles.

Common Core Standards

CCSS.MATH.CONTENT.HSS.CP.A.1: Describe events as subsets of a sample space using characteristics of outcomes.
CCSS.MATH.CONTENT.HSS.CP.A.3: Understand the conditional probability of event A given event B as P(A | B) and interpret it in real-world situations.
CCSS.MATH.CONTENT.HSS.CP.B.9: Use permutations and combinations to solve counting problems.

Prerequisite Skills

Understanding of fractions, percentages, and ratios.
Ability to perform basic algebraic calculations.
Familiarity with factorials and exponents.

Key Vocabulary

Probability: A measure of how likely an event is to occur, given as a number between 0 and 1.
Sample Space: The set of all possible outcomes in a probability experiment.
Permutation: An arrangement of objects where order does matter.
Combination: A selection of objects where order does not matter.
Factorial (n!): The product of all positive integers up to a given number n.
Conditional Probability: The probability of an event occurring given that another event has already occurred.

Warm Up

This activity introduces the idea of probability as a ratio of favorable outcomes to possible outcomes. Students should recall how to calculate basic fractions and percentages before solving probability questions.

Problem: A bag contains 3 red marbles, 2 blue marbles, and 5 green marbles. What is the probability of randomly selecting a red marble?

Step 1:

Count the total number of marbles: \[ 3 + 2 + 5 = 10 \]

Step 2:

Count the favorable outcomes (red marbles): 3

Step 3:

Set up the probability as a fraction: \[ P(\text{Red}) = \frac{3}{10} \]

Step 4:

Convert to decimal or percent if needed: \[ \frac{3}{10} = 0.3 = 30\% \]

Answer: The probability of selecting a red marble is \( \frac{3}{10} \) or 30%.

Self-Study Tip: Always verify whether you are being asked for a probability as a fraction, decimal, or percent on the SAT. Any of the three may be correct, depending on the question.

Teach

Probability and combinatorics questions test your ability to count, calculate, and reason logically about possible outcomes. These problems often appear in word problems or data-based scenarios, and they typically test:

Basic probability (favorable outcomes over total outcomes)
Multi-stage events (compound probability)
Permutations and combinations (counting problems)
Conditional probability (based on prior outcomes)

In this section, we’ll explore each of these concepts through seven worked-out examples. You'll see how to identify the type of problem, apply the correct formula, and solve efficiently—skills that will help you recognize and master SAT probability questions.

Example 1: Basic Probability

This example reinforces the Warm-Up and introduces basic terminology.

Problem: A standard six-sided die is rolled. What is the probability of rolling an even number?

Step 1:

List the sample space: {1, 2, 3, 4, 5, 6}

Step 2:

Identify favorable outcomes (even numbers): {2, 4, 6}

Step 3:

\[ P(\text{Even}) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{3}{6} = \frac{1}{2} \]

Answer: The probability is \( \frac{1}{2} \).

Example 2: Compound Probability (Independent Events)

This example explores multi-step probability with independent outcomes.

Problem: A coin is flipped and a die is rolled. What is the probability of getting heads and a 6?

Step 1:

Probability of heads: \( \frac{1}{2} \)
Probability of rolling a 6: \( \frac{1}{6} \)

Step 2:

Since these are independent events: \[ P(\text{Heads and 6}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \]

Answer: The probability is \( \frac{1}{12} \).

Example 3: Compound Probability (Dependent Events)

This example demonstrates how probabilities change after an item is removed from a set.

Problem: A bag contains 4 red and 6 blue marbles. Two marbles are drawn without replacement. What is the probability that both are red?

Step 1:

First red marble: \( \frac{4}{10} \)

Step 2:

One red removed, 3 red remain, and 9 marbles total: \[ \frac{3}{9} \]

Step 3:

\[ P(\text{Red and Red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15} \]

Answer: The probability is \( \frac{2}{15} \).

Example 4: Permutations

Use permutations when order matters.

Problem: How many different ways can 3 students be selected from a group of 5 to win 1st, 2nd, and 3rd place in a contest?

Step 1:

Use the permutation formula: \[ P(n, r) = \frac{n!}{(n - r)!} \]

Step 2:

Plug in \( n = 5 \), \( r = 3 \): \[ P(5, 3) = \frac{5!}{(5 - 3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{120}{2} = 60 \]

Answer: There are 60 different ways to assign the three places.

Self-Study Tip: Use permutations when the positions or roles (like 1st, 2nd, 3rd) are distinct.

Example 5: Combinations

Use combinations when order does not matter.

Problem: From a club of 6 members, how many ways can a 3-person committee be formed?

Step 1:

Use the combination formula: \[ C(n, r) = \frac{n!}{r!(n - r)!} \]

Step 2:

\[ C(6, 3) = \frac{6!}{3!(6 - 3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = \frac{120}{6} = 20 \]

Answer: There are 20 ways to form the committee.

Self-Study Tip: Use combinations when the group has no specific roles or positions.

Example 6: Conditional Probability

This example shows how to calculate the probability of an event given that another event has already occurred.

Problem: A survey shows that 60 students play sports. Of those, 36 also play a musical instrument. What is the probability that a student plays an instrument given that they play sports?

Step 1:

This is a conditional probability: \[ P(\text{Instrument | Sports}) = \frac{P(\text{Instrument and Sports})}{P(\text{Sports})} \]

Step 2:

\[ P(\text{Instrument | Sports}) = \frac{36}{60} = \frac{3}{5} \]

Answer: The conditional probability is \( \frac{3}{5} \).

Self-Study Tip: Read carefully to identify the "given that" part—this tells you which group to focus on for your denominator.

Example 7: Counting from a Visual Set

This example reinforces how to determine probability by directly counting favorable and total outcomes from a visual.

Problem: The image below shows a box with 12 blocks labeled with letters. If one block is selected at random, what is the probability that the block has a vowel?

Step 1:

Count the total number of blocks: 12

Step 2:

Count the favorable outcomes (vowels): A, E, I → 3 blocks

Step 3:

\[ P(\text{vowel}) = \frac{3}{12} = \frac{1}{4} \]

Answer: The probability of selecting a vowel is \( \frac{1}{4} \).

Self-Study Tip: Always double check the question to see if you're being asked for a count or a probability. Many SAT visuals involve sets of labeled or patterned objects where the question requires careful visual inspection.

Review

This lesson covered key probability and combinatorics concepts that commonly appear on the SAT Math section. You learned how to calculate basic and compound probabilities, differentiate between permutations and combinations, and apply conditional probability in context. These concepts are essential for interpreting real-world situations involving logic and counting, and they frequently appear in SAT word problems, survey questions, and statistical scenarios.

Here’s a summary of what was covered:

Calculating the probability of single and multi-step events
Understanding the difference between independent and dependent events
Applying permutations when order matters
Using combinations when order does not matter
Calculating conditional probabilities using contextual information

Example 1: Basic and Compound Probability

Problem: A bag contains 4 white balls and 6 black balls. One ball is drawn, replaced, and then another ball is drawn. What is the probability both balls are black?

Step 1:

Since the ball is replaced, the events are independent.

Step 2:

\[ P(\text{Black}) = \frac{6}{10} = \frac{3}{5} \] \[ P(\text{Black and Black}) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} \]

Answer: The probability is \( \frac{9}{25} \).

Example 2: Permutations

Problem: In how many ways can the letters A, B, and C be arranged?

Step 1:

Since all letters are used and order matters: \[ 3! = 3 \times 2 \times 1 = 6 \]

Answer: There are 6 possible arrangements.

Example 3: Conditional Probability

Problem: Out of 40 students, 25 are in the art club and 10 are in both the art club and science club. What is the probability that a student is in the science club given that they are in the art club?

Step 1:

Use the conditional probability formula: \[ P(\text{Science | Art}) = \frac{P(\text{Art and Science})}{P(\text{Art})} = \frac{10}{25} = \frac{2}{5} \]

Answer: The conditional probability is \( \frac{2}{5} \).

Example 4: Visual Probability with Number Tiles

Problem: The image below shows 10 number tiles placed face-up. If one tile is selected at random, what is the probability that it shows an even number?

Step 1:

List the even numbers from 1 to 10: 2, 4, 6, 8, 10 → 5 favorable outcomes

Step 2:

Total number of tiles: 10

Step 3:

\[ P(\text{even}) = \frac{5}{10} = \frac{1}{2} \]

Answer: The probability of selecting a tile with an even number is \( \frac{1}{2} \).

Multimedia Resources

For additional support with probability and combinatorics—including videos, tutorials, and practice exercises—explore this collection of instructional resources from Media4Math that align with this lesson.

https://www.media4math.com/LessonPlans/SupportResourcesSATMathLesson13

Quiz

Directions: Choose the best answer for each question. Show your work on a separate sheet if needed.

A jar contains 5 red balls, 3 green balls, and 2 yellow balls. What is the probability of selecting a green ball?
a) \( \frac{1}{5} \)
b) \( \frac{3}{10} \)
c) \( \frac{1}{3} \)
d) \( \frac{2}{5} \)
What is the probability of flipping two coins and getting heads on both?
a) \( \frac{1}{2} \)
b) \( \frac{1}{4} \)
c) \( \frac{1}{3} \)
d) \( \frac{3}{4} \)
How many ways can 4 students line up for a photo?
a) 10
b) 12
c) 16
d) 24
From a group of 8 people, how many ways can a 3-person committee be formed?
a) 56
b) 24
c) 336
d) 64
If a number is chosen at random from 1 to 20, what is the probability of choosing a multiple of 4?
a) \( \frac{1}{4} \)
b) \( \frac{1}{5} \)
c) \( \frac{1}{3} \)
d) \( \frac{3}{10} \)
A deck has 10 cards numbered 1 to 10. If one card is drawn at random, what is the probability of drawing a number less than 7?
a) \( \frac{3}{5} \)
b) \( \frac{2}{5} \)
c) \( \frac{4}{5} \)
d) \( \frac{1}{2} \)
Two cards are drawn from a standard deck without replacement. What is the probability that both cards are hearts?
a) \( \frac{1}{4} \)
b) \( \frac{13}{52} \times \frac{12}{51} \)
c) \( \frac{1}{13} \)
d) \( \frac{13}{52} \times \frac{13}{51} \)
In how many ways can the letters in the word "MATH" be arranged?
a) 12
b) 16
c) 24
d) 30
A class has 30 students. 18 play soccer and 12 play basketball. If 7 students play both, what is the probability that a randomly chosen student plays soccer given that they play basketball?
a) \( \frac{7}{12} \)
b) \( \frac{5}{12} \)
c) \( \frac{7}{18} \)
d) \( \frac{11}{30} \)
What is the probability of rolling a number greater than 4 on a standard die?
a) \( \frac{1}{2} \)
b) \( \frac{1}{3} \)
c) \( \frac{1}{6} \)
d) \( \frac{2}{3} \)

Answer Key

Answer: b) \( \frac{3}{10} \)
Total balls = 5 + 3 + 2 = 10. Favorable = 3 green. \[ P(\text{green}) = \frac{3}{10} \]
Answer: b) \( \frac{1}{4} \)
\[ P(\text{HH}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]
Answer: d) 24
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]
Answer: a) 56
\[ C(8, 3) = \frac{8!}{3!(5)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
Answer: a) \( \frac{1}{4} \)
Multiples of 4 from 1–20: {4, 8, 12, 16, 20} = 5 values \[ P = \frac{5}{20} = \frac{1}{4} \]
Answer: a) \( \frac{3}{5} \)
Numbers less than 7: {1–6} → 6 numbers \[ \frac{6}{10} = \frac{3}{5} \]
Answer: b) \( \frac{13}{52} \times \frac{12}{51} \)
First card heart = \( \frac{13}{52} \), then 12 hearts left out of 51 \[ P = \frac{13}{52} \times \frac{12}{51} \]
Answer: c) 24
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]
Answer: a) \( \frac{7}{12} \)
Conditional probability: \[ P(\text{Soccer | Basketball}) = \frac{\text{Both}}{\text{Basketball}} = \frac{7}{12} \]
Answer: a) \( \frac{1}{3} \)
Greater than 4: {5, 6} → 2 outcomes \[ \frac{2}{6} = \frac{1}{3} \]