Chapter 9: Probability & Statistics
Probability is the branch of mathematics that measures how likely an event is to occur. From predicting election outcomes and weather forecasts to calculating odds in games and assessing risk in business decisions, probability provides the mathematical framework for reasoning about uncertainty. In this chapter, we develop the foundational tools of counting, then apply them to compute probabilities for a wide variety of situations. We conclude by connecting the binomial theorem—a topic from algebra—to the binomial probability distribution, illustrating how algebraic structures have direct statistical applications.
9.1 Counting Principles
Before we can calculate probabilities, we need reliable methods for counting the number of possible outcomes. The most important tool is the Fundamental Counting Principle, which provides a systematic way to count the total number of outcomes when a process consists of multiple stages.
Fundamental Counting Principle
If a procedure can be broken into $k$ successive stages, where stage 1 can be performed in $n_1$ ways, stage 2 in $n_2$ ways, and so on through stage $k$ in $n_k$ ways, then the total number of ways to perform the entire procedure is:
$$n_1 \times n_2 \times n_3 \times \cdots \times n_k$$
Example 1: Outfit Choices
A student has 4 shirts, 3 pairs of pants, and 2 pairs of shoes. How many different outfits can the student create?
Solution: Each outfit requires choosing one item from each category. By the Fundamental Counting Principle:
$$4 \times 3 \times 2 = 24 \text{ outfits}$$
Tree Diagrams
A tree diagram is a visual tool that displays all possible outcomes of a multi-stage process. Each branch represents one possible choice at each stage, and the complete paths from root to leaves represent the full outcomes.
Example 2: Coin Flips
List all possible outcomes when flipping a coin three times.
Solution: We build a tree diagram. At each stage, the coin lands on Heads (H) or Tails (T).
There are $2 \times 2 \times 2 = 8$ total outcomes, confirming the Fundamental Counting Principle.
Tip: Tree diagrams become impractical for large problems (e.g., 10 coin flips yield 1024 paths). Use them for small cases to build intuition, and rely on counting formulas for larger problems.
Example 3: License Plates
A license plate consists of 3 letters followed by 4 digits. How many distinct plates are possible if repetition is allowed?
Solution: There are 26 choices for each letter position and 10 choices for each digit position:
$$26 \times 26 \times 26 \times 10 \times 10 \times 10 \times 10 = 26^3 \times 10^4 = 17{,}576 \times 10{,}000 = 175{,}760{,}000$$
9.2 Permutations
A permutation is an arrangement of objects where the order matters. Choosing a president, vice president, and treasurer from a group of 10 people is a permutation problem because the order of selection determines who gets which role.
Definition: Factorial
For any positive integer $n$, the factorial of $n$ is defined as:
$$n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$$
By convention, $0! = 1$.
Factorials grow extremely quickly: $5! = 120$, $10! = 3{,}628{,}800$, and $20!$ exceeds $2.4 \times 10^{18}$.
Permutation Formula
The number of permutations of $r$ objects chosen from $n$ distinct objects is:
$$P(n, r) = \frac{n!}{(n-r)!}$$
This counts the number of ordered arrangements of $r$ items selected from a pool of $n$.
Example 4: Race Finishers
In a race with 8 runners, how many ways can 1st, 2nd, and 3rd place be awarded?
Solution: We need the number of permutations of 3 items from 8:
$$P(8, 3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336$$
Permutations with Repetition
When some objects in the arrangement are identical, we must account for the fact that rearranging identical items does not produce a distinguishable new arrangement.
Permutations with Repeated Elements
The number of distinguishable permutations of $n$ objects, where there are $n_1$ identical objects of type 1, $n_2$ of type 2, ..., and $n_k$ of type $k$, is:
$$\frac{n!}{n_1! \cdot n_2! \cdot n_3! \cdots n_k!}$$
Example 5: Arranging Letters
How many distinguishable ways can the letters of MISSISSIPPI be arranged?
Solution: The word has 11 letters: M (1), I (4), S (4), P (2).
$$\frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!} = \frac{39{,}916{,}800}{1 \cdot 24 \cdot 24 \cdot 2} = \frac{39{,}916{,}800}{1{,}152} = 34{,}650$$
Example 6: Committee Officers
From a club of 12 members, how many ways can a president, vice president, secretary, and treasurer be chosen?
Solution: Each position is distinct, so order matters:
$$P(12, 4) = \frac{12!}{8!} = 12 \times 11 \times 10 \times 9 = 11{,}880$$
9.3 Combinations
A combination is a selection of objects where the order does not matter. Choosing 3 students from a class of 20 to form a committee (where all members have equal roles) is a combination problem. The key question to ask is: "Does rearranging the selection produce something different?" If no, use combinations.
Combination Formula
The number of combinations of $r$ objects chosen from $n$ distinct objects is:
$$C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$
The notation $\binom{n}{r}$ is read "$n$ choose $r$."
Notice the relationship between permutations and combinations: since every combination of $r$ objects can be arranged in $r!$ different ways, we have $P(n,r) = C(n,r) \times r!$, or equivalently $C(n,r) = P(n,r) / r!$.
Example 7: Choosing a Committee
From a group of 10 people, how many ways can a 4-person committee be formed?
Solution: Order does not matter (all committee members are equal):
$$\binom{10}{4} = \frac{10!}{4! \cdot 6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5{,}040}{24} = 210$$
Example 8: Lottery Numbers
A lottery requires choosing 6 numbers from 1 to 49. How many possible tickets exist?
Solution: The order in which numbers are drawn does not matter:
$$\binom{49}{6} = \frac{49!}{6! \cdot 43!} = \frac{49 \times 48 \times 47 \times 46 \times 45 \times 44}{720} = 13{,}983{,}816$$
This means the probability of winning with one ticket is approximately $1$ in $14$ million.
Pascal's Triangle and Combinations
Pascal's Triangle is a triangular array of numbers where each entry is the sum of the two entries directly above it. The entries in row $n$ (counting from row 0) are exactly the values $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$.
| Row 0: | 1 | |||||||||||||||||
| Row 1: | 1 | 1 | ||||||||||||||||
| Row 2: | 1 | 2 | 1 | |||||||||||||||
| Row 3: | 1 | 3 | 3 | 1 | ||||||||||||||
| Row 4: | 1 | 4 | 6 | 4 | 1 | |||||||||||||
| Row 5: | 1 | 5 | 10 | 10 | 5 | 1 | ||||||||||||
| Row 6: | 1 | 6 | 15 | 20 | 15 | 6 | 1 | |||||||||||
The identity underlying Pascal's Triangle is:
$$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$$
This recursive relationship, known as Pascal's Identity, states that choosing $r$ items from $n$ can be split into two cases: either a particular item is included (leaving $\binom{n-1}{r-1}$ ways to choose the rest) or it is excluded (leaving $\binom{n-1}{r}$ ways).
Permutation vs. Combination: Ask yourself: "Does order matter?" If arranging officers (president, VP, etc.), order matters — use $P(n,r)$. If choosing a committee with equal roles, order does not matter — use $C(n,r)$.
9.4 Probability Basics
With counting tools established, we can now define probability rigorously. Probability assigns a numerical value between 0 and 1 (inclusive) to events, where 0 means the event is impossible and 1 means it is certain.
Definition: Sample Space and Events
The sample space $S$ is the set of all possible outcomes of an experiment. An event $A$ is any subset of the sample space. The probability of event $A$ is:
$$P(A) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{|A|}{|S|}$$
This formula applies when all outcomes in $S$ are equally likely.
Key properties that every probability function must satisfy:
- For any event $A$: $0 \leq P(A) \leq 1$
- The certain event has probability 1: $P(S) = 1$
- The impossible event has probability 0: $P(\emptyset) = 0$
Complement Rule
The complement of event $A$, written $A'$ (or $\bar{A}$ or $A^c$), is the event that $A$ does not occur. The complement rule states:
$$P(A') = 1 - P(A)$$
This is often the easiest way to calculate a probability when the complementary event is simpler to count.
Example 9: Rolling a Die
A fair six-sided die is rolled. Find the probability of rolling (a) a 4, (b) an even number, (c) a number greater than 4.
Solution: The sample space is $S = \{1, 2, 3, 4, 5, 6\}$, with $|S| = 6$.
(a) $P(4) = \dfrac{1}{6}$
(b) Even numbers: $\{2, 4, 6\}$, so $P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$
(c) Greater than 4: $\{5, 6\}$, so $P(> 4) = \dfrac{2}{6} = \dfrac{1}{3}$
Example 10: Card Drawing
A card is drawn from a standard 52-card deck. What is the probability of not drawing a heart?
Solution: There are 13 hearts, so $P(\text{heart}) = \frac{13}{52} = \frac{1}{4}$. By the complement rule:
$$P(\text{not a heart}) = 1 - P(\text{heart}) = 1 - \frac{1}{4} = \frac{3}{4}$$
Example 11: At Least One
What is the probability of getting at least one head when flipping 3 fair coins?
Solution: "At least one head" is the complement of "no heads" (all tails). From our tree diagram in Example 2, there are $2^3 = 8$ total outcomes and exactly 1 outcome with all tails (TTT).
$$P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \frac{1}{8} = \frac{7}{8}$$
9.5 Compound Events
Many probability problems involve combining two or more events. Understanding how events relate to each other—whether they overlap, exclude each other, or depend on one another—is essential for computing compound probabilities.
Union: "A or B"
Addition Rule (General)
For any two events $A$ and $B$:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
We subtract $P(A \cap B)$ to avoid double-counting outcomes that belong to both events.
Definition: Mutually Exclusive Events
Two events $A$ and $B$ are mutually exclusive (or disjoint) if they cannot both occur: $A \cap B = \emptyset$. In this case, the addition rule simplifies to:
$$P(A \cup B) = P(A) + P(B)$$
Example 12: Cards — Union
A card is drawn from a standard deck. What is the probability of drawing a king or a heart?
Solution: Let $A$ = king and $B$ = heart. These events overlap (the king of hearts is both).
$P(A) = \frac{4}{52}$, $P(B) = \frac{13}{52}$, $P(A \cap B) = \frac{1}{52}$
$$P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$$
Intersection: "A and B"
Definition: Independent Events
Two events $A$ and $B$ are independent if the occurrence of one does not affect the probability of the other. Mathematically, $A$ and $B$ are independent if and only if:
$$P(A \cap B) = P(A) \cdot P(B)$$
Example 13: Independent Events
A coin is flipped and a die is rolled. What is the probability of getting heads and a 5?
Solution: The coin flip and die roll are independent events.
$$P(\text{H and 5}) = P(\text{H}) \cdot P(\text{5}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$
Conditional Probability
When events are not independent, the probability of one event may change based on whether another event has occurred. Conditional probability quantifies this dependence.
Conditional Probability Formula
The probability of event $A$ given that event $B$ has occurred is:
$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) \neq 0$$
Equivalently, the General Multiplication Rule states:
$$P(A \cap B) = P(B) \cdot P(A \mid B)$$
Example 14: Drawing Without Replacement
A bag contains 5 red and 3 blue marbles. Two marbles are drawn without replacement. What is the probability that both are red?
Solution: Let $A$ = first marble is red and $B$ = second marble is red.
$P(A) = \dfrac{5}{8}$. After removing one red marble, there are 4 red marbles among 7 total: $P(B \mid A) = \dfrac{4}{7}$.
$$P(A \cap B) = P(A) \cdot P(B \mid A) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$
Example 15: Conditional Probability from a Table
In a class of 30 students, 18 play sports and 12 are on the honor roll. Of those who play sports, 8 are on the honor roll. If a student is selected at random and is known to play sports, what is the probability the student is on the honor roll?
Solution: Let $S$ = plays sports, $H$ = honor roll. We need $P(H \mid S)$.
$$P(H \mid S) = \frac{P(H \cap S)}{P(S)} = \frac{8/30}{18/30} = \frac{8}{18} = \frac{4}{9}$$
Independence Test: Events $A$ and $B$ are independent if $P(A \mid B) = P(A)$. In Example 15, $P(H) = 12/30 = 2/5$ while $P(H \mid S) = 4/9 \neq 2/5$, so playing sports and being on the honor roll are not independent in this class.
9.6 Binomial Theorem & Probability
The Binomial Theorem from algebra and the binomial probability distribution from statistics are deeply connected. Both are built on the combination formula $\binom{n}{k}$, which is why the distribution is called "binomial."
The Binomial Theorem
Binomial Theorem
For any real numbers $a$ and $b$ and any non-negative integer $n$:
$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n$$
The coefficients $\binom{n}{k}$ are called binomial coefficients and correspond exactly to the entries in row $n$ of Pascal's Triangle.
Example 16: Expanding $(x + 2)^4$
Solution: Using the binomial theorem with $a = x$, $b = 2$, $n = 4$:
$$(x + 2)^4 = \binom{4}{0}x^4(2)^0 + \binom{4}{1}x^3(2)^1 + \binom{4}{2}x^2(2)^2 + \binom{4}{3}x^1(2)^3 + \binom{4}{4}x^0(2)^4$$
$$= 1 \cdot x^4 + 4 \cdot 2x^3 + 6 \cdot 4x^2 + 4 \cdot 8x + 1 \cdot 16$$
$$= x^4 + 8x^3 + 24x^2 + 32x + 16$$
Binomial Probability Distribution
A binomial experiment is a series of $n$ independent trials, each with exactly two outcomes (success or failure), where the probability of success $p$ is the same for every trial. The number of successes $X$ follows a binomial distribution.
Binomial Probability Formula
If $X$ is the number of successes in $n$ independent trials, each with probability of success $p$ and probability of failure $q = 1 - p$, then:
$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$
for $k = 0, 1, 2, \ldots, n$.
Notice the connection: if we expand $(p + q)^n$ using the Binomial Theorem, the term $\binom{n}{k} p^k q^{n-k}$ gives the probability of exactly $k$ successes. Since $p + q = 1$, the sum of all terms equals $1^n = 1$, confirming that the probabilities form a valid distribution.
Example 17: Free Throw Shooting
A basketball player makes 70% of her free throws. If she shoots 5 free throws, what is the probability that she makes exactly 3?
Solution: Here $n = 5$, $k = 3$, $p = 0.7$, and $q = 0.3$.
$$P(X = 3) = \binom{5}{3}(0.7)^3(0.3)^2 = 10 \times 0.343 \times 0.09 = 10 \times 0.03087 = 0.3087$$
The probability is approximately $30.87\%$.
Example 18: Quality Control
A factory produces items with a 5% defect rate. In a random sample of 10 items, what is the probability that (a) exactly 1 is defective, (b) at most 1 is defective?
Solution: Here $n = 10$ and $p = 0.05$ (probability of defect).
(a) $P(X = 1) = \binom{10}{1}(0.05)^1(0.95)^9 = 10 \times 0.05 \times 0.6302 \approx 0.3151$
(b) "At most 1" means $P(X = 0) + P(X = 1)$:
$P(X = 0) = \binom{10}{0}(0.05)^0(0.95)^{10} = 1 \times 1 \times 0.5987 \approx 0.5987$
$$P(X \leq 1) = 0.5987 + 0.3151 = 0.9138$$
There is about a 91.4% chance of finding at most 1 defective item.
Interactive: Binomial Distribution
Use the sliders below to adjust the number of trials $n$ and the probability of success $p$. Observe how the shape of the binomial distribution changes: as $p$ increases, the distribution shifts right; as $n$ increases, the distribution spreads and becomes more symmetric.
Key Properties of the Binomial Distribution:
Mean: $\mu = np$ Standard deviation: $\sigma = \sqrt{np(1-p)}$
For Example 17: $\mu = 5(0.7) = 3.5$ free throws made on average, with $\sigma = \sqrt{5(0.7)(0.3)} \approx 1.02$.
9.7 Practice Problems
Test your understanding with the following problems. Each covers a concept from this chapter. Try to solve each problem before revealing the solution.
Problem 1
A restaurant offers 5 appetizers, 8 entrees, and 4 desserts. How many different three-course meals (one appetizer, one entree, one dessert) are possible?
Show Solution
By the Fundamental Counting Principle:
$$5 \times 8 \times 4 = 160 \text{ meals}$$
$$\boxed{160}$$
Problem 2
How many ways can 6 books be arranged on a shelf?
Show Solution
This is a permutation of all 6 distinct objects:
$$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$$
$$\boxed{720}$$
Problem 3
How many distinguishable arrangements are there of the letters in BANANA?
Show Solution
BANANA has 6 letters: B (1), A (3), N (2).
$$\frac{6!}{1! \cdot 3! \cdot 2!} = \frac{720}{1 \cdot 6 \cdot 2} = \frac{720}{12} = 60$$
$$\boxed{60}$$
Problem 4
A committee of 5 is to be chosen from 9 men and 7 women. How many committees have exactly 3 men and 2 women?
Show Solution
Choose 3 men from 9 and 2 women from 7:
$$\binom{9}{3} \times \binom{7}{2} = \frac{9!}{3! \cdot 6!} \times \frac{7!}{2! \cdot 5!} = 84 \times 21 = 1{,}764$$
$$\boxed{1{,}764}$$
Problem 5
Two dice are rolled. What is the probability that the sum is 7?
Show Solution
The total sample space for two dice is $6 \times 6 = 36$ outcomes.
Outcomes that sum to 7: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — that is 6 outcomes.
$$P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$$
$$\boxed{\dfrac{1}{6}}$$
Problem 6
A jar contains 4 red, 6 green, and 5 blue marbles. If one marble is drawn at random, what is the probability of drawing a red or blue marble?
Show Solution
Total marbles: $4 + 6 + 5 = 15$. Red and blue are mutually exclusive events.
$$P(\text{red or blue}) = P(\text{red}) + P(\text{blue}) = \frac{4}{15} + \frac{5}{15} = \frac{9}{15} = \frac{3}{5}$$
$$\boxed{\dfrac{3}{5}}$$
Problem 7
A card is drawn from a standard deck. What is the probability it is a face card (J, Q, K) or a diamond?
Show Solution
Let $F$ = face card and $D$ = diamond. There are 12 face cards, 13 diamonds, and 3 face cards that are also diamonds (J, Q, K of diamonds).
$$P(F \cup D) = P(F) + P(D) - P(F \cap D) = \frac{12}{52} + \frac{13}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}$$
$$\boxed{\dfrac{11}{26}}$$
Problem 8
A bag has 8 white and 4 black balls. Two balls are drawn without replacement. What is the probability that both are white?
Show Solution
Using conditional probability (General Multiplication Rule):
$$P(\text{both white}) = P(W_1) \cdot P(W_2 \mid W_1) = \frac{8}{12} \times \frac{7}{11} = \frac{56}{132} = \frac{14}{33}$$
Alternative (combinations):
$$P = \frac{\binom{8}{2}}{\binom{12}{2}} = \frac{28}{66} = \frac{14}{33}$$
$$\boxed{\dfrac{14}{33}}$$
Problem 9
Expand $(2x - 3)^4$ using the Binomial Theorem.
Show Solution
Apply the Binomial Theorem with $a = 2x$, $b = -3$, $n = 4$:
$$(2x - 3)^4 = \sum_{k=0}^{4}\binom{4}{k}(2x)^{4-k}(-3)^k$$
$k=0$: $\binom{4}{0}(2x)^4(-3)^0 = 1 \cdot 16x^4 \cdot 1 = 16x^4$
$k=1$: $\binom{4}{1}(2x)^3(-3)^1 = 4 \cdot 8x^3 \cdot (-3) = -96x^3$
$k=2$: $\binom{4}{2}(2x)^2(-3)^2 = 6 \cdot 4x^2 \cdot 9 = 216x^2$
$k=3$: $\binom{4}{3}(2x)^1(-3)^3 = 4 \cdot 2x \cdot (-27) = -216x$
$k=4$: $\binom{4}{4}(2x)^0(-3)^4 = 1 \cdot 1 \cdot 81 = 81$
$$\boxed{16x^4 - 96x^3 + 216x^2 - 216x + 81}$$
Problem 10
A multiple-choice test has 8 questions, each with 4 choices. If a student guesses on every question, what is the probability of getting exactly 5 correct?
Show Solution
This is a binomial probability problem with $n = 8$, $k = 5$, $p = \frac{1}{4} = 0.25$.
$$P(X = 5) = \binom{8}{5}(0.25)^5(0.75)^3$$
$$= 56 \times (9.7656 \times 10^{-4}) \times 0.421875$$
$$= 56 \times 4.1199 \times 10^{-4}$$
$$\approx 0.02307$$
The probability is approximately $2.3\%$.
$$\boxed{\approx 0.0231}$$
Problem 11
In how many ways can a hand of 5 cards be dealt from a 52-card deck such that all 5 cards are of the same suit (a flush)?
Show Solution
There are 4 suits. For each suit, choose 5 cards from the 13 available:
$$4 \times \binom{13}{5} = 4 \times 1{,}287 = 5{,}148$$
The probability of a flush is:
$$P(\text{flush}) = \frac{5{,}148}{\binom{52}{5}} = \frac{5{,}148}{2{,}598{,}960} \approx 0.00198$$
(Note: this count includes straight flushes and royal flushes.)
$$\boxed{5{,}148 \text{ flush hands}}$$
Problem 12
A coin is biased so that $P(\text{heads}) = 0.6$. The coin is flipped 10 times. Find (a) the expected number of heads, (b) the standard deviation, and (c) $P(X = 7)$.
Show Solution
(a) Expected value: $\mu = np = 10(0.6) = 6$
(b) Standard deviation: $\sigma = \sqrt{np(1-p)} = \sqrt{10(0.6)(0.4)} = \sqrt{2.4} \approx 1.549$
(c) Binomial probability:
$$P(X = 7) = \binom{10}{7}(0.6)^7(0.4)^3 = 120 \times 0.0279936 \times 0.064 = 120 \times 0.001792 \approx 0.2150$$
$$\boxed{\mu = 6, \quad \sigma \approx 1.549, \quad P(X=7) \approx 0.215}$$