AP Precalculus: Probability

Master probability formulas, counting techniques, and probability rules

🎲 Basic Probability πŸ”’ Counting πŸ”— Conditional βž• Addition Rule

πŸ“š Understanding Probability

Probability measures the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). AP Precalculus covers fundamental probability concepts including counting techniques, conditional probability, and probability rules for combining events.

1 Basic Probability

The probability of an event A is the ratio of favorable outcomes to total possible outcomes in the sample space.

Basic Probability Formula \(P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\)
Probability Range
\(0 \leq P(A) \leq 1\)
Certain Event
\(P(A) = 1\)
Impossible Event
\(P(A) = 0\)
Complement
\(P(A') = 1 - P(A)\)
πŸ“Œ Example

Problem: A bag contains 3 red, 5 blue, and 2 green marbles. Find P(blue).

Solution: \(P(\text{blue}) = \frac{5}{3+5+2} = \frac{5}{10} = 0.5\)

2 Permutations & Combinations

Permutations count arrangements where order matters. Combinations count selections where order doesn't matter.

Permutations (Order Matters)
\(P(n, r) = \frac{n!}{(n-r)!}\)
Arranging r items from n items
Combinations (Order Doesn't Matter)
\(C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}\)
Choosing r items from n items
Factorial Definition \(n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1\), and \(0! = 1\)
πŸ“Œ Example: Permutation

How many ways can 3 people finish 1st, 2nd, 3rd from 8 runners?

\(P(8, 3) = \frac{8!}{5!} = 8 \times 7 \times 6 = 336\) ways

πŸ“Œ Example: Combination

How many ways can a committee of 3 be chosen from 8 people?

\(C(8, 3) = \frac{8!}{3! \cdot 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56\) ways

πŸ’‘ When to Use Which

Permutation: rankings, arrangements, passwords. Combination: committees, teams, selections.

3 Probability with Counting Techniques

Use permutations or combinations to count favorable outcomes and total outcomes, then apply the basic probability formula.

Probability with Counting \(P(\text{event}) = \frac{\text{Number of ways event can occur}}{\text{Total number of possible outcomes}}\)
πŸ“Œ Example

Problem: From a deck of 52 cards, find P(drawing 2 aces when selecting 2 cards).

Favorable: \(C(4, 2) = 6\) ways to choose 2 aces from 4

Total: \(C(52, 2) = 1326\) ways to choose any 2 cards

Probability: \(P = \frac{6}{1326} = \frac{1}{221} \approx 0.0045\)

4 Two-Way Frequency Tables

A two-way frequency table displays data for two categorical variables. Use it to find joint, marginal, and conditional probabilities.

Likes Math Doesn't Like Math Total
Boys 30 20 50
Girls 25 25 50
Total 55 45 100

Joint Probability

\(P(\text{Boy AND Likes Math}) = \frac{30}{100} = 0.30\)

Marginal Probability

\(P(\text{Likes Math}) = \frac{55}{100} = 0.55\)

Conditional Probability

\(P(\text{Likes Math | Boy}) = \frac{30}{50} = 0.60\)

5 Independent Events

Two events are independent if the occurrence of one does not affect the probability of the other.

Test for Independence
\(P(A \cap B) = P(A) \cdot P(B)\)
If true, events are independent
Alternative Test
\(P(A|B) = P(A)\)
Knowing B doesn't change P(A)
Probability of Both Independent Events \(P(A \text{ and } B) = P(A) \times P(B)\)
πŸ“Œ Example

Problem: A coin is flipped twice. Find P(heads on both).

Solution: Flips are independent. \(P(H_1 \cap H_2) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)

⚠️ Independent β‰  Mutually Exclusive

Independent events can both occur. Mutually exclusive events cannot both occur (if one happens, the other can't).

6 Conditional Probability

Conditional probability is the probability of event A occurring given that event B has already occurred.

Conditional Probability Formula \(P(A|B) = \frac{P(A \cap B)}{P(B)}\), provided \(P(B) > 0\)

Read as "probability of A given B"

Rearranged: Find P(A ∩ B)

\(P(A \cap B) = P(A|B) \cdot P(B)\)

Also Written As

\(P(A \cap B) = P(B|A) \cdot P(A)\)

πŸ“Œ Example

Problem: \(P(\text{rain}) = 0.3\), \(P(\text{traffic jam | rain}) = 0.6\). Find P(rain and traffic jam).

Solution: \(P(\text{rain} \cap \text{jam}) = P(\text{jam | rain}) \times P(\text{rain}) = 0.6 \times 0.3 = 0.18\)

7 Addition Rule

The addition rule finds the probability that at least one of two events occurs (A or B or both).

General Addition Rule
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
Subtracts the overlap to avoid double-counting
Mutually Exclusive Events
\(P(A \cup B) = P(A) + P(B)\)
No overlap when events can't both occur
πŸ“Œ Example

Problem: \(P(\text{rain}) = 0.4\), \(P(\text{cold}) = 0.3\), \(P(\text{rain and cold}) = 0.1\). Find P(rain or cold).

Solution: \(P(\text{rain} \cup \text{cold}) = 0.4 + 0.3 - 0.1 = 0.6\)

πŸ’‘ "Or" vs "And"

Addition rule: P(A or B). Multiplication: P(A and B). Watch for these keywords in problems!

8 Multiplication Rule

The multiplication rule finds the probability that both events A and B occur.

General Multiplication Rule
\(P(A \cap B) = P(A) \cdot P(B|A)\)
For any two events
Independent Events
\(P(A \cap B) = P(A) \cdot P(B)\)
When events don't affect each other
πŸ“Œ Example (Dependent)

Problem: Draw 2 cards without replacement. Find P(both red).

Solution: \(P = \frac{26}{52} \times \frac{25}{51} = \frac{650}{2652} = \frac{25}{102} \approx 0.245\)

πŸ“‹ Quick Reference

Basic Probability

\(P(A) = \frac{\text{favorable}}{\text{total}}\)

Complement

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

Permutation

\(P(n,r) = \frac{n!}{(n-r)!}\)

Combination

\(C(n,r) = \frac{n!}{r!(n-r)!}\)

Conditional

\(P(A|B) = \frac{P(A \cap B)}{P(B)}\)

Addition Rule

\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

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