
Probability is the science of uncertainty, empowering you to make informed decisions in data science and everyday life. It’s essential because:
Probability measures how likely an event is. Here are the core concepts:
0 (Impossible) ---- Possible ---- 1 (Sure)
The sample space \(\Omega\) lists all outcomes. For an experiment with sample space \(\Omega\) repeated \(n\) times independently, the sample space is \(\Omega^n\) (Cartesian product), with size \(|\Omega|^n\).
Events describe possible outcomes in \(\Omega\).
Common Pitfall: Mutually exclusive (no overlap) differs from independent (no influence). Coin flip and die roll are independent, not mutually exclusive.
Example:
Question: Are “drawing an Ace” and “drawing a King” mutually exclusive? Independent?
Events are sets in \(\Omega\), visualized with Venn diagrams.
Example: \(\Omega = \{1, 2, \ldots, 10\}\), \(A = \{1, 3, 5, 7, 9\}\), \(B = \{1, 2, 3, 4\}\).
Venn Diagram: Visualizing sets \(A\) and \(B\) in \(\\Omega\)

Example: Bag with 2 red, 3 blue balls:
Tip: Check for overlap in “or” problems to avoid double-counting.
Exercise: If \(P(A) = 0.5\), \(P(B) = 0.4\), and \(A \cap B = \emptyset\), find \(P(A \cup B)\).
Chance of \(A\) given \(B\) has occurred:
Bayes’ Theorem: Find \(P(A| B)\) using \(P(B | A)\).
Example: Pregnancy test (Exercise #1)

Example Table:
| \(x\) (Heads) | \(P(X = x)\) | \(x \cdot P(X = x)\) |
|---|---|---|
| 0 | 1/4 | 0 |
| 1 | 1/2 | 0.5 |
| 2 | 1/4 | 0.5 |
| Total | 1 | 1 |
Exercise: Find \(E[X]\) for number of red balls in Exercise #3.
Counting outcomes determines the size of \(\Omega\) and \(A\) in \(P(A) = \frac{|A|}{|\Omega|}\). Key principles include:
Permutations count ways to arrange \(k\) distinct items from \(n\), where order matters.
A, B, C) in 2 positions.
AB, BA, AC, CA, BC, CB.Tip: Use permutations for ordered arrangements (e.g., race rankings).
Combinations count ways to choose \(k\) distinct items from \(n\), where order does not matter.
Apple, Banana, Orange}.
Apple, Banana}, {Apple, Orange}, {Banana, Orange}.MISS (\(2\) S’s, \(1\) M, \(1\) I).
Apple, Banana) with repetition.
Apple, Apple}, {Apple, Banana}, {Banana, Banana}.| Technique | Scenario/Case | Key Questions | Formula | Example |
|---|---|---|---|---|
| Multiplication Principle | Sequential, independent choices. | Are there multiple stages? Can choices repeat? | \(n_1 \cdot n_2 \cdot \ldots \cdot n_k\) | Shirt (3) and pants (2): \(3 \cdot 2 = 6\). |
| Addition Principle | Mutually exclusive choices. | Are options exclusive? No overlap? | \(|A| + |B|\) | Apples (3) or carrots (2): \(3 + 2 = 5\). |
| Permutations | Ordered arrangements of distinct items. | Does order matter? Are items unique? | \(P(n, k) = \frac{n!}{(n-k)!}\) | Arrange 3 letters in 2 spots: \(P(3, 2) = 6\). |
| Combinations | Unordered selections of distinct items. | Does order not matter? Are items unique? | \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) | Choose 2 fruits from 3: \(\binom{3}{2} = 3\). |
| Permutations with Repetition | Ordered arrangements, items can repeat. | Can items repeat? Are choices independent? | \(n^k\) | 3-digit codes (0–9): \(10^3 = 1000\). |
| Indistinguishable Objects | Arrange items with some identical. | Are some items identical? | \(\frac{n!}{n_1! n_2! \ldots}\) | Arrange “MISS”: \(\frac{4!}{2!1!1!} = 12\). |
| Combinations with Repetition | Unordered selections with repetition allowed. | Can items repeat? Order doesn’t matter? | \(\binom{n+k-1}{k}\) | 2 fruits (Apple, Banana): \(\binom{3}{2} = 3\). |
Tip: Ask “Does order matter?” and “Can items repeat?” to start.
Counting determines \(|\Omega|\) and \(|A|\) for \(P(A) = \frac{|A|}{|\Omega|}\).
Why It Matters: Accurate counting ensures correct probabilities.
Exercise: Calculate \(P(4, 2)\) and \(\binom{4}{2}\). How do they differ in a probability context?
If a woman takes an early pregnancy test, she will either test positive, meaning that the test says she is pregnant, or test negative, meaning that the test says she is not pregnant. Suppose that if a woman really is pregnant, there is a 98% chance that she will test positive. Also suppose that if a woman really is not pregnant, there is a 99% chance that she will test negative.

Suppose that 1,000 woman take early pregnancy tests and that 100 of them truly pregnant. What is the probability that a randomly chosen woman from this group will test positive?
Suppose a woman tests positive. What is the probability that she really is pregnant?
This analysis uses a sample of 1,000 women undergoing early pregnancy tests, with the following assumptions:
\[ = 0.98 \times 0.1 + 0.01 \times 0.9 = 0.107 \] - Result: 10.7% chance of testing positive across all 1,000 women.
\[ P(\text{Pregnant | Test Positive}) = \frac{P(\text{Test Positive | Pregnant}) \times P(\text{Pregnant})}{P(\text{Test Positive})} \]
\[ = \frac{0.98 \times 0.1}{0.107} \approx 0.916 \] - Result: 91.6% chance of being pregnant if tested positive.
This diagram illustrates the probability pathways based on pregnancy status and test outcomes:
graph TD;
A[Start: 1,000 Women] -->|10% Pregnant| B[Pregnant]
A -->|90% Not Pregnant| C[Not Pregnant]
B -->|98% Test Positive| D[Test Positive]
B -->|2% Test Negative| E[Test Negative]
C -->|1% Test Positive| F[Test Positive]
C -->|99% Test Negative| G[Test Negative]
style D fill:#4daf4a,color:#ffff
style E fill:#e41a1c,color:#ffff
style F fill:#4daf4a,color:#ffff
style G fill:#4daf4a,color:#ffff
A box contains 3 blue, 2 green, and 4 red marbles. If you draw 2 marbles at random without replacement, what is the probability both are green?
Given a box with 3 blue, 2 green, and 4 red marbles, we are asked to find the probability of drawing two green marbles when drawing two marbles at random without replacement. Here’s the step-by-step calculation:
Using the combination formula \(\binom{n}{k}\) which represents the number of ways to choose \(k\) items from \(n\) items without regard to the order, we calculate:
\[ \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 \]
There are 2 green marbles and we need to draw both:
\[ \binom{2}{2} = \frac{2 \times 1}{2 \times 1} = 1 \]
The probability is the ratio of the number of favorable outcomes (drawing 2 green marbles) to the total number of outcomes (drawing any 2 marbles):
\[ \text{Probability} = \frac{\text{Number of ways to draw 2 green marbles}}{\text{Total number of ways to draw 2 marbles}} = \frac{1}{36} \]
A bag contains 5 white and 7 black balls. Two balls are drawn successively without replacement. What is the probability that one is white, and one is black?
Given a bag containing 5 white and 7 black balls, we are asked to calculate the probability of drawing two balls successively without replacement where one ball is white and the other is black.
Adding these two mutually exclusive probabilities gives: \[ P(\text{One white, one black}) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} \]
This simplifies to: \[ \frac{70}{132} = \frac{35}{66} \approx 0.5303 \]
Find \(E[X]\) for number of heads in three coin tosses.
Table:
| \(k\) (Heads) | \(P(X = k)\) | \(k \cdot P(X = k)\) |
|---|---|---|
| 0 | 1/8 | 0 |
| 1 | 3/8 | 0.375 |
| 2 | 3/8 | 0.75 |
| 3 | 1/8 | 0.375 |
| Total | 1 | 1.5 |
