GED Math • Quantitative Problem Solving

Determine Probability and Counting Combinations

Learn how to calculate the likelihood of events and count possible groupings using practical, real-world methods.

A multi-subject study notebook with math and science symbols for GED preparation
Inside this lesson 10 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

Humans have always tried to predict the future — from ancient dice games to modern insurance calculations. The mathematics of probability grew out of a simple but powerful question: when something uncertain happens, can we measure how likely each outcome is? Alongside probability, the study of combinations developed to answer a related question: how many different ways can we select items from a group? Together, these two ideas form the backbone of decision-making in fields ranging from medicine to finance to everyday life.

1654 • The Birth of Probability

French mathematicians Blaise Pascal and Pierre de Fermat exchanged letters about gambling problems, creating the first formal theory of probability.

1713 • Bernoulli's Theorem

Jacob Bernoulli published the Law of Large Numbers, proving that as you repeat an experiment many times, the observed results approach the true probability.

1812 • Laplace's Systematic Framework

Pierre-Simon Laplace published a comprehensive book on probability, defining it as the ratio of favorable outcomes to total outcomes — the formula you will use on the GED.

Modern Era • Everyday Applications

Today, probability and counting methods drive weather forecasts, medical testing, quality control in manufacturing, lottery odds, and data science.

The central question this lesson addresses is straightforward: given a set of possible outcomes, how do you calculate the chance that a specific event happens, and how do you count the number of ways to choose items from a group? On the GED, you will encounter these ideas in real-world scenarios like drawing cards, selecting committee members, or interpreting data about risk.

SECTION 2

Core Principles & Definitions

Before diving into formulas, let's establish the key ideas that every probability and combination problem builds on. Understanding these definitions clearly will make the math much easier to follow.

1

Probability

A number between 0 and 1 (or 0% and 100%) that measures how likely an event is to happen. A probability of 0 means impossible; a probability of 1 means certain.
2

Outcome

One specific result from an experiment or situation. For example, rolling a 4 on a die is one outcome. The set of all possible outcomes is called the sample space.
3

Favorable Outcomes

The outcomes that match the event you care about. If you want to roll an even number on a die, the favorable outcomes are 2, 4, and 6 — that's 3 favorable outcomes.
4

Combination

A selection of items from a group where the order does not matter. Choosing players A, B, C for a team is the same combination as choosing C, A, B.
5

Factorial (n!)

The product of all whole numbers from n down to 1. For example, 4! = 4 × 3 × 2 × 1 = 24. By definition, 0! = 1. Factorials are the building blocks of counting formulas.
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KEY TAKEAWAY

Think of probability like a weather forecast. When the forecaster says there is a 70% chance of rain, that number comes from counting how many times similar conditions produced rain out of all similar weather patterns. Probability works the same way: you compare the number of ways something can happen to the total number of things that could happen.

SECTION 3

Visual Explanation — Probability at a Glance

The Probability Scale: 0 to 10Impossible0.5Equally Likely1CertainExample: Drawing from a Bag of 10 MarblesSample Space10 total marblesProbability CalculationsP(Red)= 4/10 = 2/5 = 0.4P(Blue)= 3/10 = 0.3P(Green)= 3/10 = 0.3Total= 0.4 + 0.3 + 0.3 = 1.0 ✓
The probability scale runs from 0 (impossible) to 1 (certain). The marble example shows how you calculate probability: divide the number of favorable marbles by the total number of marbles. Notice that the probabilities of all outcomes always add up to 1.

The diagram above illustrates two essential ideas. First, probability always falls on a scale from 0 to 1 — you can also express it as a fraction, decimal, or percentage. Second, to find the probability of drawing a specific color, you count how many marbles of that color exist (favorable outcomes) and divide by the total number of marbles (total outcomes). On the GED, this is the core calculation you will perform again and again, whether the context involves marbles, cards, survey data, or any other scenario.

SECTION 4

Mathematical Framework

The GED provides a formula sheet during the test, so you do not need to memorize these formulas — but you do need to understand what each part means and when to apply them. Let's walk through the key formulas one at a time.

BASIC PROBABILITY

P(event)=numberoffavorableoutcomes÷totalnumberofoutcomesP(event) = number of favorable outcomes ÷ total number of outcomes

P(event) is the probability of the event occurring. The result is always a number from 0 to 1. Multiply by 100 to convert to a percentage.

FACTORIAL

n!=n×(n−1)×(n−2)×…×2×1n! = n × (n − 1) × (n − 2) × … × 2 × 1

n is a non-negative whole number. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. Special case: 0! = 1 by definition. Factorials count the number of ways to arrange n items in a line.

COMBINATION FORMULA

C(n,r)=n!÷[r!×(n−r)!]C(n, r) = n! ÷ [r! × (n − r)!]

n = total number of items to choose from. r = number of items you are selecting. Use this when order does NOT matter — such as choosing team members, lottery numbers, or pizza toppings.

PERMUTATION FORMULA (FOR COMPARISON)

P(n,r)=n!÷(n−r)!P(n, r) = n! ÷ (n − r)!

Use this when order DOES matter — such as assigning 1st, 2nd, and 3rd place. Permutations always give a larger number than combinations because every different arrangement counts separately.

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GED Test Tip

The GED formula sheet includes the combination formula. When you see a problem, ask yourself: Does the order of selection matter? If you are choosing a committee (order doesn't matter), use combinations. If you are ranking finishers in a race (order matters), use permutations.

SECTION 5

Combinations vs. Permutations — Knowing the Difference

One of the most common mistakes on the GED is mixing up combinations and permutations. The difference comes down to one word: order. If switching the arrangement creates a different result, order matters and you use a permutation. If switching the arrangement gives you the same group, order does not matter and you use a combination.

Combination vs. Permutation: Choosing 2 from {A, B, C}COMBINATIONOrder does NOT matter(e.g., choosing team members){A, B} = {B, A}Same group → counted ONCE{A, C} = {C, A}Same group → counted ONCE{B, C} = {C, B}Same group → counted ONCEC(3,2) = 33! ÷ (2! × 1!) = 33 unique groupsPERMUTATIONOrder DOES matter(e.g., assigning 1st & 2nd place)(A, B)(B, A)(A, C)(C, A)(B, C)(C, B)P(3,2) = 63! ÷ 1! = 66 unique arrangements
When choosing 2 items from the set {A, B, C}, combinations count only 3 unique groups (since {A, B} = {B, A}), while permutations count 6 arrangements (since the order (A, B) ≠ (B, A)). Permutations always produce a number equal to or greater than the corresponding combination.
Clue in the ProblemOrder Matters?Use…
"Choose," "select," "pick," "committee," "group"NoCombination
"Arrange," "rank," "1st/2nd/3rd," "password," "sequence"YesPermutation

Quick decision table: how to tell if a problem uses combinations or permutations

SECTION 6

Worked Example — Lottery Committee Problem

Let's work through a complete problem that combines both counting and probability — the kind of multi-step question you might see on the GED.

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Problem

A company has 10 employees. Management wants to select a safety committee of 3 people. How many different committees are possible, and what is the probability that a specific group of 3 friends — Maria, James, and Tanya — are all selected?

Step-by-Step Solution

  1. 1

    Step 1 — Identify the type of counting problem

    We are selecting a committee, which means the order of selection does not matter. Choosing Maria-James-Tanya is the same committee as Tanya-Maria-James. This tells us to use the combination formula.
  2. 2

    Step 2 — Identify n and r

    We have n = 10 employees to choose from, and we are selecting r = 3 for the committee.
    n = 10, r = 3
  3. 3

    Step 3 — Substitute into the combination formula

    C(10, 3) = 10! ÷ [3! × (10 − 3)!] = 10! ÷ [3! × 7!]. Now simplify: 10! ÷ 7! = 10 × 9 × 8 (the remaining factors cancel). So we get (10 × 9 × 8) ÷ (3 × 2 × 1).
  4. 4

    Step 4 — Calculate

    Numerator: 10 × 9 × 8 = 720. Denominator: 3 × 2 × 1 = 6. Divide: 720 ÷ 6 = 120.
    There are 120 possible committees.
  5. 5

    Step 5 — Find the probability

    There is only 1 committee that consists of Maria, James, and Tanya. The total number of possible committees is 120. Using the probability formula: P = favorable outcomes ÷ total outcomes = 1 ÷ 120.
    P(Maria, James, Tanya) = 1/120 ≈ 0.0083, or about 0.83%

SECTION 7

Common Pitfalls & Strengths of These Methods

Common MistakeWhy It's WrongCorrect Approach
Using permutations when order doesn't matterOvercounts groups — treats {A, B} and {B, A} as differentAsk: "Does switching the order create a new result?" If no, use combinations.
Writing probability greater than 1Probability can never exceed 1 (100%)If your answer is greater than 1, you likely flipped the fraction — check numerator vs. denominator.
Forgetting that 0! = 1Leads to division by zero errors in formulasMemorize: 0! = 1 by definition. It makes the formula work when r = n.
Not simplifying factorials before multiplyingCreates enormous numbers that are hard to work withCancel common factorial factors first. For C(10,3), write 10×9×8 ÷ 3! instead of computing 10! fully.

Frequent errors and how to avoid them on the GED

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KEY TAKEAWAY

Think of combinations like choosing toppings for a pizza — it doesn't matter if you add peppers first and then mushrooms, or mushrooms first and then peppers. The pizza is the same. Permutations are like a locker combination (ironically named!) — entering 1-2-3 gives a different result than 3-2-1. Always ask: does the order of my choices create a different outcome?

SECTION 8

Connections to Advanced Topics

The probability and combination skills you are learning now form the foundation for many advanced topics you may encounter in college or in your career. Understanding where these ideas lead can motivate your study and show you how useful this knowledge really is.

GED Level ConceptAdvanced ExtensionReal-World Use
Basic probability (favorable ÷ total)Conditional probability — how one event affects anotherMedical testing: what is the chance you are actually sick given a positive test?
Counting combinations C(n, r)Binomial theorem — expanding expressions like (a + b)ⁿFinance: modeling the probability of investment gains and losses
Probability of simple eventsExpected value — average outcome over many trialsInsurance: calculating premiums based on average claim costs

How GED probability concepts connect to college-level mathematics

You don't need to learn any of these advanced ideas for the GED test. But knowing they exist shows that the skills you're building now — counting outcomes, computing probabilities, and reasoning about chance — are genuine building blocks for higher education and career growth. Master the basics here, and these more complex topics will feel much more approachable when you encounter them.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A bag contains 5 red chips and 3 blue chips. You draw one chip at random. Which of the following best describes the probability of drawing a blue chip?
  • A. 3/5, because there are 3 blue chips and 5 red chips
  • B. 3/8, because there are 3 favorable outcomes out of 8 total chips
  • C. 5/8, because you are more likely to draw a red chip
  • D. 1/3, because blue is one of the two possible colors
Show answer (B)
The correct answer is B. Probability = favorable outcomes ÷ total outcomes. There are 3 blue chips (favorable) and 5 + 3 = 8 total chips, so P(blue) = 3/8. Choice A incorrectly uses only the red chips as the denominator instead of all chips. Choice C gives the probability of drawing a red chip (5/8), not a blue one. Choice D incorrectly divides 1 by the number of colors rather than counting individual chips.

Question 2: BASIC CALCULATION

A club has 8 members. They need to choose a subcommittee of 2 members. How many different subcommittees are possible?
  • A. 16
  • B. 56
  • C. 28
  • D. 64
Show answer (C)
The correct answer is C. Since the order of selection does not matter for a committee, use the combination formula: C(8, 2) = 8! ÷ [2! × 6!] = (8 × 7) ÷ (2 × 1) = 56 ÷ 2 = 28. Choice A (16) comes from simply multiplying 8 × 2, which is not the correct operation. Choice B (56) is the permutation P(8, 2) = 8 × 7, which overcounts by treating each pair twice. Choice D (64) is 8², which would apply to a different type of problem entirely.

Question 3: INTERMEDIATE

A standard deck of 52 playing cards contains 4 aces. If you draw 2 cards at random without replacement, what is the probability that both cards are aces?
  • A. 1/221
  • B. 1/169
  • C. 1/13
  • D. 6/52
Show answer (A)
The correct answer is A. Find the number of ways to choose 2 aces from 4: C(4, 2) = 6. Find the total ways to choose 2 cards from 52: C(52, 2) = (52 × 51) ÷ 2 = 1,326. Probability = 6 ÷ 1,326 = 1/221. Choice B (1/169) incorrectly computes (1/13) × (1/13), treating draws as independent with replacement. Choice C (1/13) is the probability of drawing a single ace on one draw. Choice D (6/52) incorrectly places the favorable combinations over the total cards instead of total combinations.

Question 4: APPLIED

A hiring manager has 12 applicants for 4 identical internship positions. She calculates the number of ways to fill the positions. After one applicant withdraws, she recalculates. How many fewer possible groups of interns exist after the withdrawal?
  • A. 165
  • B. 495
  • C. 330
  • D. 120
Show answer (A)
The correct answer is A. With 12 applicants: C(12, 4) = (12 × 11 × 10 × 9) ÷ (4 × 3 × 2 × 1) = 11,880 ÷ 24 = 495. With 11 applicants: C(11, 4) = (11 × 10 × 9 × 8) ÷ (4 × 3 × 2 × 1) = 7,920 ÷ 24 = 330. The difference is 495 − 330 = 165. Choice B (495) is the original count, not the difference. Choice C (330) is the new count after the withdrawal, not the difference. Choice D (120) has no correct basis in this calculation.

Question 5: CRITICAL THINKING

3 pts
A teacher wants to form a study group of 3 students from a class of 7 boys and 5 girls. The group must include at least 1 girl. How many different study groups are possible? Show your work.
Show answer
Rubric (3 points): Point 1: Correctly identifies the total number of groups of 3 from 12 students. C(12, 3) = (12 × 11 × 10) ÷ (3 × 2 × 1) = 220. Point 2: Correctly identifies the number of all-boy groups (the groups with NO girls). C(7, 3) = (7 × 6 × 5) ÷ (3 × 2 × 1) = 35. Point 3: Uses the complement strategy to subtract: groups with at least 1 girl = total groups − all-boy groups = 220 − 35 = 185. The final answer is 185 different study groups. Alternative correct method: Add the cases — (exactly 1 girl) + (exactly 2 girls) + (exactly 3 girls) = C(5,1)×C(7,2) + C(5,2)×C(7,1) + C(5,3)×C(7,0) = 5×21 + 10×7 + 10×1 = 105 + 70 + 10 = 185. Full credit for either method.

SUMMARY

Lesson Summary

Summary

In this lesson, you learned how to calculate probability by dividing the number of favorable outcomes by the total number of outcomes. Probability always ranges from 0 (impossible) to 1 (certain) and can be expressed as a fraction, decimal, or percentage. You also learned to use factorials (n!) as the building block for counting formulas.

The combination formula C(n, r) = n! ÷ [r! × (n − r)!] counts selections where order does not matter — committees, groups, and teams. The permutation formula counts arrangements where order matters. On the GED, read carefully for keywords like "choose" or "select" (combinations) versus "arrange" or "rank" (permutations). Remember that the GED provides these formulas on the formula sheet — your job is to identify which formula to use, plug in the correct values, and simplify by canceling factorial factors before computing.

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