\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]

Properties of Probability:

• \( 0 \leq P(E) \leq 1 \)

• \( P(\text{impossible event}) = 0 \)

• \( P(\text{certain event}) = 1 \)

• \( P(E) + P(\text{not } E) = 1 \)

Rolling a fair die, find P(rolling a 3)

Favorable outcomes: 1 (only one 3)

Total outcomes: 6

Answer: \( P(3) = \frac{1}{6} \)

If an event can occur in \( m \) ways and a second independent event can occur in \( n \) ways, then the two events together can occur in \( m \times n \) ways

\[ \text{Total outcomes} = n_1 \times n_2 \times n_3 \times \cdots \times n_k \]

If an event can occur in \( m \) ways OR another mutually exclusive event can occur in \( n \) ways, then one or the other can occur in \( m + n \) ways

A restaurant offers 4 appetizers, 5 main courses, and 3 desserts. How many different meals?

Total meals = \( 4 \times 5 \times 3 = 60 \)

Answer: 60 different meals

Number of ways to arrange \( r \) objects from \( n \) objects

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

Special Case - All objects:

\[ P(n,n) = n! \]

Number of ways to choose \( r \) objects from \( n \) objects

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

Key Relationship:

\[ P(n,r) = r! \times C(n,r) \]

\[ P(E) = \frac{\text{Number of favorable arrangements/selections}}{\text{Total number of arrangements/selections}} \]

Example 1: A committee of 3 is chosen from 10 people. What's the probability a specific person is on the committee?

Answer: \( \frac{3}{10} \) or 0.3 or 30%

Example 2: Drawing 5 cards from deck, what's probability of all hearts?

Answer: ≈ 0.0495% or about 1 in 2020

A table that displays data for two categorical variables

Example Table: Student Survey

Example Probabilities:

P(Male) = \( \frac{50}{100} = 0.5 \)

P(Plays Sports) = \( \frac{55}{100} = 0.55 \)

P(Male AND Plays Sports) = \( \frac{30}{100} = 0.3 \)

Two events are independent if the occurrence of one event does NOT affect the probability of the other event

\[ P(A \cap B) = P(A) \times P(B) \]

or equivalently

\[ P(A|B) = P(A) \]

Check if: \( P(A \cap B) = P(A) \times P(B) \)

• If TRUE → Events are independent

• If FALSE → Events are dependent

Independent:

• Flipping a coin twice

• Rolling two dice

Dependent:

• Drawing cards without replacement

• Choosing 2 students from a class

\[ P(A \text{ AND } B) = P(A) \times P(B) \]

Example:

Flip a coin and roll a die. P(Heads AND 6)?

\( P(\text{H AND 6}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \)

\[ P(A \text{ AND } B) = P(A) \times P(B|A) \]

Example:

Draw 2 cards without replacement. P(both Aces)?

\( P(\text{1st Ace}) = \frac{4}{52} \)

\( P(\text{2nd Ace}|\text{1st Ace}) = \frac{3}{51} \)

\( P(\text{both Aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{221} \)

The probability of event B occurring given that event A has already occurred

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

Read as: "Probability of B given A"

A bag has 3 red and 2 blue marbles. If you draw a red, what's P(draw another red)?

Answer: \( \frac{1}{2} \) or 50%

For independent events:

\[ P(B|A) = P(B) \]

This means knowing that A occurred doesn't change the probability of B

Using the survey table from Section 5:

Example: P(Plays Sports | Male)?

Look only at Male row: 30 play sports out of 50 males

\( P(\text{Sports}|\text{Male}) = \frac{30}{50} = \frac{3}{5} = 0.6 \)

\[ P(A \text{ OR } B) = P(A) + P(B) - P(A \cap B) \]

We subtract \( P(A \cap B) \) to avoid counting the overlap twice

If A and B cannot occur together, \( P(A \cap B) = 0 \)

\[ P(A \text{ OR } B) = P(A) + P(B) \]

Example 1: Drawing a card, P(King OR Heart)?

Answer: \( \frac{4}{13} \) ≈ 30.8%

Example 2: Rolling a die, P(2 OR 5)?

Answer: \( \frac{1}{3} \) ≈ 33.3%