Unit 11 · Chapter 11.7

11.7Probability

Calculate classical probability, apply the addition rule and complement, work with independent and mutually exclusive events, and use conditional probability.

Probability is the mathematical foundation of statistics, data science, machine learning, and risk analysis. Every decision under uncertainty — from medical testing to financial modeling — relies on probabilistic reasoning.

Essential Question

How do we quantify uncertainty? How can counting techniques help us calculate the likelihood of complex events?

Overview

Probability measures how likely an event is to occur, expressed as a number between 0 (impossible) and 1 (certain). The sample space S is the set of all possible outcomes; an event E is any subset of S.

  • Classical probability: P(E) = |E| / |S| (equally likely outcomes)
  • Complement rule: P(E′) = 1 − P(E)
  • Addition rule: P(A∪B) = P(A) + P(B) − P(A∩B)
  • Mutually exclusive: P(A∩B) = 0, so P(A∪B) = P(A) + P(B)
  • Independent events: P(A∩B) = P(A) · P(B)
  • Conditional probability: P(A|B) = P(A∩B) / P(B)
SP(A only)P(A∩B)P(B only)P(neither)AB

P(A∪B) = P(A) + P(B) − P(A∩B)

Venn diagram: the intersection A∩B (blue) is subtracted to avoid double-counting.

Worked Examples

Example 1

Roll a fair six-sided die. Find P(even).

Identify the sample space: S = {1, 2, 3, 4, 5, 6}, so |S| = 6.

Identify the event: E = {2, 4, 6} (even numbers), so |E| = 3.

Apply classical probability: P(E) = |E| / |S| = 3/6.

Simplify: P(even) = 1/2 = 0.5.

Answer:P(even) = 1/2
Example 2

Draw one card from a standard 52-card deck. Find P(heart or face card).

Count hearts: |A| = 13.

Count face cards (J, Q, K in each suit): |B| = 12.

Count cards that are both heart AND face card: |A∩B| = 3 (J♥, Q♥, K♥).

Apply addition rule: P(A∪B) = 13/52 + 12/52 − 3/52.

P(heart or face card) = 22/52 = 11/26 ≈ 0.423.

Answer:P(heart or face card) = 11/26 ≈ 0.423
Example 3

Two fair coins are flipped. Find P(at least one head).

Sample space: S = {HH, HT, TH, TT}, so |S| = 4.

Use the complement: P(at least one head) = 1 − P(no heads).

P(no heads) = P(TT) = 1/4.

P(at least one head) = 1 − 1/4 = 3/4.

Answer:P(at least one head) = 3/4
Example 4

A bag has 3 red, 5 blue, and 2 green marbles (10 total). Two marbles are drawn without replacement. Find P(both red).

Total ways to choose 2 from 10: C(10,2) = 45.

Ways to choose 2 red from 3: C(3,2) = 3.

P(both red) = C(3,2) / C(10,2) = 3/45.

Simplify: P(both red) = 1/15 ≈ 0.0667.

Answer:P(both red) = 1/15 ≈ 0.067
Example 5

P(A) = 0.4, P(B) = 0.3, P(A∩B) = 0.1. Find P(A|B).

Recall the conditional probability formula: P(A|B) = P(A∩B) / P(B).

Substitute the known values: P(A|B) = 0.1 / 0.3.

P(A|B) = 1/3 ≈ 0.333.

Interpretation: given that B has occurred, there is a 33.3% chance A also occurs.

Answer:P(A|B) = 1/3 ≈ 0.333

Guided Practice

Guided Problem 1

Roll a fair six-sided die. Find P(prime number).

Hint: The prime numbers on a die are 2, 3, and 5. How many outcomes are in the sample space?

Guided Problem 2

Draw one card from a 52-card deck. Find P(red or king).

Hint: Use the addition rule. Count red cards (26), kings (4), and red kings (2). Don't forget to subtract the overlap!

Guided Problem 3

P(A) = 0.5, P(B) = 0.4, and A and B are mutually exclusive. Find P(A∪B).

Hint: Mutually exclusive means P(A∩B) = 0. The addition rule simplifies to P(A∪B) = P(A) + P(B).

Guided Problem 4

A jar has 4 red and 6 blue marbles (10 total). Two are drawn without replacement. Find P(one red and one blue).

Hint: Count favorable outcomes: C(4,1)·C(6,1). Divide by total ways to choose 2 from 10: C(10,2) = 45.

Guided Problem 5

P(A) = 0.6, P(B) = 0.5, P(A∩B) = 0.3. Find P(A|B) and P(B|A).

Hint: Use P(A|B) = P(A∩B)/P(B) and P(B|A) = P(A∩B)/P(A). Substitute carefully.

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Common Mistakes

P(A∪B) = P(A) + P(B) always

Only when A and B are mutually exclusive. Otherwise subtract P(A∩B) to avoid double-counting.

Confusing P(A|B) with P(B|A)

These are generally different. P(A|B) = P(A∩B)/P(B) and P(B|A) = P(A∩B)/P(A).

Assuming independent events are mutually exclusive

Independent means P(A∩B) = P(A)·P(B) ≠ 0. Mutually exclusive means P(A∩B) = 0. They are opposite concepts (for non-trivial events).

Forgetting to reduce the denominator when drawing without replacement

After drawing one marble, there are n−1 items left. Use combinations: C(n,k) for unordered draws.

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Math Tips

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Always check: does your probability answer fall between 0 and 1? If not, recheck your work.

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The complement rule P(E′) = 1 − P(E) is often the fastest path when 'at least one' appears.

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For independent events, P(A∩B) = P(A)·P(B) — this is both the definition and a test for independence.

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Draw a Venn diagram or tree diagram to visualize overlapping events before calculating.

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When using combinations for probability, set up the fraction: (favorable combinations) / (total combinations).

Sample Space (S)

The set of all possible outcomes of a random experiment.

Event (E)

Any subset of the sample space; a collection of outcomes.

Classical Probability

P(E) = |E|/|S|, valid when all outcomes are equally likely.

Complement (E′)

All outcomes NOT in E; P(E′) = 1 − P(E).

Mutually Exclusive

Events that cannot occur simultaneously; P(A∩B) = 0.

Independent Events

Events where the occurrence of one does not affect the other; P(A∩B) = P(A)·P(B).

Conditional Probability

P(A|B) = P(A∩B)/P(B); the probability of A given B has occurred.

Addition Rule

P(A∪B) = P(A) + P(B) − P(A∩B).

Interactive Practice — 1 Questions

1

A fair die is rolled. Which formula gives P(A∪B) when A and B are NOT mutually exclusive?