Series and Probability

🎲 Series and Probability

Part 1 of 5 — Series & Sigma Notation

Topics in This Part

🔑 Key Concept: A sequence is a list of numbers; a series is what you get when you add the terms of a sequence. This lesson has two big ideas — series (Parts 1–2) and probability (Parts 3–5) — and they meet in the world of counting and patterns.

Sequence vs. Series

A sequence is an ordered list of terms:

$3,\; 7,\; 11,\; 15,\; 19$

A series is the sum of those terms:

$3 + 7 + 11 + 15 + 19 = 55$

We label terms with subscripts: $a_1$ is the first term, $a_2$ the second, and $a_n$ the general (nth) term.

💡 Think of it this way: a sequence is the ingredients, a series is the total on the receipt.

Concept Check 🎯

Summation (Sigma) Notation

Writing long sums gets tedious, so we use the Greek capital letter sigma, $\Sigma$, to mean "add up":

$\sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \cdots + a_n$

The pieces:

• $k$ is the index — a counter.
• $k=1$ (bottom) is where the counter starts.
• $n$ (top) is where the counter ends.
• $a_k$ is the rule — plug each value of $k$ into it.

Worked Example

$\sum_{k=1}^{4} (2k+1) = (2\cdot 1+1) + (2\cdot 2+1) + (2\cdot 3+1) + (2\cdot 4+1)$

$= 3 + 5 + 7 + 9 = 24$

🔑 To expand a sigma: substitute each integer from the bottom number up to the top number, then add the results.

Expand & Evaluate 🧮

Find each sum.

1) $\displaystyle\sum_{k=1}^{3} 4k$ 2) $\displaystyle\sum_{k=1}^{4} (k^2)$ 3) $\displaystyle\sum_{k=1}^{5} (3k-2)$

Read the Sigma 🔽

Match each part of $\displaystyle\sum_{k=1}^{6}(3k-2)$ to its meaning.

Wrapping Up Part 1

You can now tell a sequence from a series and read, expand, and evaluate sigma notation.

Right now we are adding terms one by one. In Part 2 you'll learn formulas that add hundreds of terms in a single step.

🎲 Series and Probability

Part 2 of 5 — Arithmetic & Geometric Series

🔑 The Idea: Two of the most important series follow simple patterns. Arithmetic series add a constant each step; geometric series multiply by a constant. Each has a sum formula that beats adding by hand.

Arithmetic Series

An arithmetic sequence adds a fixed amount — the common difference $d$ — each step. Its series sum is:

$S_n = \frac{n}{2}\,(a_1 + a_n)$

In words: the number of terms, times the average of the first and last term.

Worked Example: $3 + 7 + 11 + \cdots$ (first 10 terms)

Here $a_1 = 3$ and $d = 4$. First find the 10th term with $a_n = a_1 + (n-1)d$:

$a_{10} = 3 + (10-1)(4) = 3 + 36 = 39$

Now apply the sum formula:

$S_{10} = \frac{10}{2}\,(3 + 39) = 5 \cdot 42 = 210$

💡 The famous trick: to add $1+2+3+\cdots+100$, pair the ends — $1+100=101$, and there are $50$ pairs, so $S = 50 \cdot 101 = 5050$.

Arithmetic Sums 🧮

Use $S_n = \dfrac{n}{2}(a_1 + a_n)$ (find $a_n$ first if needed).

1) Sum of the first $20$ terms of $5, 8, 11, \ldots$ ($a_1=5,\,d=3$) 2) Sum of the first $50$ positive even numbers: $2 + 4 + \cdots + 100$ 3) Sum: $1 + 2 + 3 + \cdots + 100$

Geometric Series

A geometric sequence multiplies by a fixed common ratio $r$ each step. The sum of the first $n$ terms is:

$S_n = a_1\,\frac{1 - r^{\,n}}{1 - r}, \qquad r \ne 1$

Worked Example: $2 + 6 + 18 + 54 + 162$ (first 5 terms)

Here $a_1 = 2$ and $r = 3$ (each term is $3\times$ the one before):

$S_5 = 2 \cdot \frac{1 - 3^5}{1 - 3} = 2 \cdot \frac{1 - 243}{-2} = 2 \cdot \frac{-242}{-2} = 2 \cdot 121 = 242$

✅ Check: $2+6+18+54+162 = 242$ ✓

Concept Check 🎯

Infinite Geometric Series

If $|r| < 1$, the terms shrink toward $0$ and the infinite sum converges to a finite number:

$S = \frac{a_1}{1 - r}, \qquad |r| < 1$

Worked Example: $8 + 4 + 2 + 1 + \cdots$

Here $a_1 = 8$ and $r = \frac{1}{2}$:

$S = \frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16$

Repeating Decimals

A repeating decimal is an infinite geometric series. $0.7777\ldots = \frac{7}{10} + \frac{7}{100} + \cdots$ with $a_1 = \frac{7}{10}$, $r = \frac{1}{10}$:

$S = \frac{7/10}{1 - 1/10} = \frac{7/10}{9/10} = \frac{7}{9}$

⚠️ The infinite formula only works when $|r| < 1$. If $|r| \ge 1$ the terms don't shrink, so the sum grows without bound.

Infinite Geometric Sums 🧮

Use $S = \dfrac{a_1}{1 - r}$. Enter a whole number, decimal, or fraction.

1) $27 + 9 + 3 + 1 + \cdots$ ($a_1 = 27,\, r = \frac{1}{3}$) 2) $4 + 2 + 1 + \frac{1}{2} + \cdots$ 3) The repeating decimal $0.3333\ldots$ as a fraction (form like 1/3)

🎲 Series and Probability

Part 3 of 5 — Counting Principles

🔑 Why counting first? Probability is just favorable outcomes ÷ total outcomes. To find those numbers you must be able to count outcomes fast. The counting tools here power every probability question in Parts 4 and 5.

The Fundamental Counting Principle

If one choice can be made $m$ ways and a second independent choice $n$ ways, then together there are:

$m \times n \text{ ways}$

This extends to any number of stages — just multiply.

Worked Example: Building an Outfit

You have $3$ shirts, $4$ pairs of pants, and $2$ pairs of shoes. The number of outfits is:

$3 \times 4 \times 2 = 24$

Worked Example: Lining Up

How many ways can $5$ people stand in a line? The first spot has $5$ choices, the next $4$, then $3$, $2$, $1$:

$5 \times 4 \times 3 \times 2 \times 1 = 120$

💡 The product $5 \times 4 \times 3 \times 2 \times 1$ is written $5!$ ("five factorial"). By definition $0! = 1$.

Count the Ways 🧮

1) A menu has $4$ appetizers, $6$ entrées, and $3$ desserts. How many 3-course meals? 2) Evaluate $6!$ ($= 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$). 3) A 4-digit PIN uses digits $0$–$9$ and digits may repeat. How many PINs? (enter the number)

Permutations vs. Combinations

The big question: does order matter?

• Permutation — 1st/2nd/3rd place in a race, a seating order, a password. Rearranging counts as different.
• Combination — a committee, a hand of cards, a pizza's toppings. Rearranging is the same group.

Worked Example: Race Medals (order matters → permutation)

From $5$ runners, how many ways to award gold, silver, bronze?

$P(5,3) = \frac{5!}{(5-3)!} = \frac{120}{2} = 60$

Worked Example: Committee (order doesn't matter → combination)

From $5$ people, how many $3$-person committees?

$C(5,3) = \frac{5!}{3!\,2!} = \frac{120}{6 \cdot 2} = 10$

🔑 Sanity check: there are always fewer combinations than permutations of the same $n$ and $k$, because a combination doesn't count the different orderings separately.

Permutation or Combination? 🎯

Compute Them 🧮

1) $P(6,2) = \dfrac{6!}{4!}$ 2) $C(10,3) = \dfrac{10!}{3!\,7!}$ 3) A pizza shop has $5$ toppings; choose any 2. How many topping pairs? (order doesn't matter)

🎲 Series and Probability

Part 4 of 5 — Probability Basics

🔑 The Big Formula: For equally likely outcomes, $P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}.$ Every probability is between $0$ (impossible) and $1$ (certain).

Theoretical Probability

Count the favorable outcomes, divide by the total. The counting tools from Part 3 do the heavy lifting.

Worked Example: One Die

Roll a fair $6$-sided die. $P(\text{rolling an even number})$? The evens are $2,4,6$ — three favorable out of six total:

$P(\text{even}) = \frac{3}{6} = \frac{1}{2}$

Worked Example: One Card

Draw from a standard $52$-card deck. There are $13$ hearts:

$P(\text{heart}) = \frac{13}{52} = \frac{1}{4}$

💡 A probability can be written as a fraction, a decimal, or a percent: $\frac{1}{4} = 0.25 = 25\%$.

Find the Probability 🧮

Give each answer as a fraction in lowest terms (e.g. 1/2).

1) Rolling a number greater than $4$ on a fair die. 2) Drawing a King from a $52$-card deck. 3) A bag has $4$ red and $6$ green marbles. $P(\text{green})$?

The Complement Rule

The complement of an event $A$ is "$A$ does not happen," written $A'$ or $A^c$. Since something either happens or doesn't:

$P(A') = 1 - P(A)$

Worked Example

If $P(\text{rain}) = 0.3$, then:

$P(\text{no rain}) = 1 - 0.3 = 0.7$

💡 When to use it: problems that say "at least one" are usually fastest through the complement. $P(\text{at least one}) = 1 - P(\text{none})$.

Worked Example: At Least One Head

Flip a coin twice. The only way to get no heads is TT, which has probability $\frac{1}{4}$:

$P(\text{at least one head}) = 1 - \frac{1}{4} = \frac{3}{4}$

The Addition Rule (Or)

For the probability that $A$ or $B$ happens:

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

You subtract the overlap so it isn't counted twice. If $A$ and $B$ can't both happen (mutually exclusive), the overlap is $0$ and it simplifies to $P(A) + P(B)$.

Worked Example: Heart or King

In a $52$-card deck: $13$ hearts, $4$ Kings, but the King of Hearts is in both:

$P(\text{heart or King}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$

⚠️ Forgetting to subtract the overlap is the #1 mistake. Always ask: can both happen at once?

Concept Check 🎯

Pick the Right Rule 🔽

🎲 Series and Probability

Part 5 of 5 — Compound Events & Mastery Check

You can sum series, count outcomes, and find single-event probabilities. The last skill: chaining events together with and.

The Multiplication Rule (And)

For two events happening in sequence, multiply:

$P(A \text{ and } B) = P(A) \cdot P(B \text{ given } A)$

The key question is whether the first event changes the second.

Independent Events (no effect)

A coin flip doesn't affect a die roll. $P(\text{heads and a }6)$:

$P = \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12}$

Dependent Events (the first changes the second)

A bag has $3$ red and $5$ blue marbles ($8$ total). Draw two without replacement. $P(\text{both red})$:

$P = \frac{3}{8} \cdot \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$

The second fraction is $\frac{2}{7}$ because after removing one red, only $2$ reds and $7$ marbles remain.

🔑 "And" → multiply. Just decide first: does removing the first item change the odds for the second? If yes, it's dependent and the second fraction shrinks.

Concept Check 🎯

Compound Probability 🧮

Give each answer as a fraction in lowest terms.

1) Flip $2$ coins: $P(\text{both heads})$. 2) Roll two dice: $P(\text{sum} = 7)$. (There are $6$ favorable pairs out of $36$.) 3) Bag of $5$ red, $3$ blue ($8$ total). Draw $2$ without replacement: $P(\text{both red})$.

Quick Reference — The Whole Lesson

⚠️ The two probability traps: forgetting the overlap in "or," and forgetting that "without replacement" makes the second fraction shrink.

Mixed Review 🔽

One quick decision for each scenario.

Exit Quiz ✅

Answer all three to finish the lesson.