Sequences, Series, and Probability
From the Math curriculum
Sequences, Series, and Probability
TL;DR
You'll learn about ordered lists of numbers (sequences) and their sums (series), including how they behave. We'll also cover probability, which helps you understand the likelihood of events happening. Together, these tools let you predict outcomes and analyze data.
1. The Mental Model
Think of sequences as recipes where each step gives you a new number based on the last. Series are like adding up all the ingredients in that recipe. Probability is your way of guessing how likely a specific recipe will turn out a certain way.
2. The Core Material
Sequences
A sequence is just an ordered list of numbers. Each number in the list is called a term.
- Arithmetic Sequence: You get the next term by adding a constant value (the common difference,
d) to the previous term.- Example: 2, 5, 8, 11, ... (here,
d= 3) - The
n-th term:a_n = a_1 + (n-1)d(wherea_1is the first term)
- Example: 2, 5, 8, 11, ... (here,
- Geometric Sequence: You get the next term by multiplying the previous term by a constant value (the common ratio,
r).- Example: 3, 6, 12, 24, ... (here,
r= 2) - The
n-th term:a_n = a_1 * r^(n-1)
- Example: 3, 6, 12, 24, ... (here,
Series
A series is the sum of the terms in a sequence.
- Arithmetic Series Sum: The sum of the first
nterms (S_n) of an arithmetic sequence.S_n = n/2 * (a_1 + a_n)orS_n = n/2 * (2a_1 + (n-1)d)
- Geometric Series Sum: The sum of the first
nterms (S_n) of a geometric sequence.S_n = a_1 * (1 - r^n) / (1 - r)(whereris not 1)
- Infinite Geometric Series: If the common ratio
ris between -1 and 1 (i.e.,|r| < 1), the sum of an infinite geometric series converges to a finite value.S_infinity = a_1 / (1 - r)
Probability Basics

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Probability measures how likely an event is to occur. It's always a number between 0 (impossible) and 1 (certain).
- Formula:
P(Event) = (Number of favorable outcomes) / (Total number of possible outcomes) - Sample Space: The set of all possible outcomes.
- Event: A specific outcome or set of outcomes.
- Complement: The probability that an event doesn't happen is
P(not Event) = 1 - P(Event). - Independent Events: Two events are independent if the outcome of one doesn't affect the outcome of the other.
P(A and B) = P(A) * P(B)
- Dependent Events: The outcome of one event affects the outcome of the other.
P(A and B) = P(A) * P(B|A)(whereP(B|A)is the probability of B given A)
- Mutually Exclusive Events: Two events can't happen at the same time.
P(A or B) = P(A) + P(B)
- Non-Mutually Exclusive Events: They can happen at the same time.
P(A or B) = P(A) + P(B) - P(A and B)
Counting Principles (Permutations & Combinations)

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These help you figure out the total number of possible outcomes.
- Permutation: The number of ways to arrange items where order matters.
P(n, k) = n! / (n-k)!(arrangingkitems fromntotal)n!(n factorial) isn * (n-1) * ... * 1
- Combination: The number of ways to choose items where order doesn't matter.
C(n, k) = n! / (k! * (n-k)!)(choosingkitems fromntotal)
graph TD
A["Sequences & Series"] --> B["Sequences"]
A --> C["Series"]
A --> D["Probability"]
B --> B1["Arithmetic Sequence"]
B --> B2["Geometric Sequence"]
C --> C1["Arithmetic Series Sum"]
C --> C2["Geometric Series Sum"]
C2 --> C3["Infinite Geometric Series (Convergent)"]
D --> D1["Basic Probability"]
D --> D2["Counting Principles"]
D1 --> D1a["P(Event)"]
D1 --> D1b["Independent Events"]
D1 --> D1c["Dependent Events"]
D1 --> D1d["Mutually Exclusive Events"]
D1 --> D1e["Non-Mutually Exclusive Events"]
D2 --> D2a["Permutations (Order Matters)"]
D2a --> D2a1["n! for all items"]
D2a --> D2a2["P(n,k) for some items"]
D2 --> D2b["Combinations (Order Doesn't Matter)"]
D2b --> D2b1["C(n,k)"]
3. Worked Example
Let's say you're dealing with a deck of 52 cards. What's the probability of drawing two aces in a row without replacement?
-
First Event: Drawing an Ace first.
- There are 4 aces in a 52-card deck.
P(Ace on 1st draw) = 4 / 52 = 1 / 13
-
Second Event: Drawing another Ace after the first one (without replacement).
- Now there are only 3 aces left in the deck.
- And there are only 51 cards left in total.
P(Ace on 2nd draw | Ace on 1st draw) = 3 / 51 = 1 / 17
-
Combine probabilities for dependent events:
- Since these are dependent events (drawing the first ace changes the deck for the second draw), you multiply their probabilities.
P(Two Aces in a row) = P(Ace on 1st) * P(Ace on 2nd | Ace on 1st)P(Two Aces in a row) = (1/13) * (1/17)P(Two Aces in a row) = 1 / 221
So, the probability of drawing two aces in a row without replacement is 1/221.
4. Key Takeaways
- Sequences are ordered lists of numbers, with arithmetic sequences adding a constant and geometric sequences multiplying by a constant.
- Series are the sums of sequences, and their formulas depend on whether the sequence is arithmetic or geometric.
- Probability quantifies the likelihood of an event, calculated by dividing favorable outcomes by total possible outcomes.
- Knowing if events are independent, dependent, or mutually exclusive is crucial for combining probabilities correctly.
- Permutations count arrangements where order matters, while combinations count selections where order doesn't.
- An infinite geometric series converges to a finite sum only if its common ratio's absolute value is less than 1.
- Factorials (
n!) are used in permutations and combinations to represent the product of an integer and all the integers below it.
Common Mistakes to Avoid:
- Mixing up arithmetic and geometric sequence/series formulas.
- Forgetting to adjust the total number of outcomes or favorable outcomes when dealing with "without replacement" probability problems.
- Using permutation formulas when the problem requires combinations (or vice-versa), specifically when order doesn't or does matter.
- Incorrectly applying the "or" rule for mutually exclusive vs. non-mutually exclusive events.
5. Now Try It
You're planning a small lottery. You need to pick 3 unique numbers from 1 to 20, and the order of selection doesn't matter for winning. What's the probability that you pick the three winning numbers? Also, if the winning numbers were 2, 4, 6 and you started with 2, what would be the probability of picking 4 next and then 6? (Hint: this part is dependent).
What to do:
1. First, calculate the total number of unique ways to choose 3 numbers from 20 (this is a combination).
2. Then, use this to find the probability of picking the exact three winning numbers.
3. Next, for the second part, treat it as three dependent probability events: picking 2, then 4 (given 2 was picked), then 6 (given 2 and 4 were picked).
What success looks like:
You'll have two probability values: one for picking the 3 winning numbers where order doesn't matter, and another for picking them in a specific order. Your answers should be fractions or decimals.
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