Lesson 13.7: Probability

New Concept

Probability quantifies the likelihood of an event. You'll learn to construct models, compute probabilities for different types of events—including unions and complements—and apply counting principles to solve more complex problems involving chance.

What’s next

Now, let's put theory into practice. You'll work through interactive examples and a series of practice cards to master these foundational probability skills.

Property

An experiment is an activity with an observable result. The set of all possible outcomes is called the sample space. An event is any subset of a sample space. The probability of an event p is a number that always satisfies $0 \le p \le 1$. A probability model is a mathematical description of an experiment listing all possible outcomes and their associated probabilities. The sum of the probabilities must equal 1. To construct a model for an event with equally likely outcomes, identify every outcome, determine the total number of outcomes, and express each outcome's probability as a ratio to the total.

Examples

• A fair coin is tossed. The sample space is {Heads, Tails}. The probability model is P(Heads) = $\frac{1}{2}$ and P(Tails) = $\frac{1}{2}$.

• A spinner has four equal sections colored Red, Green, Blue, and Yellow. The probability of landing on any specific color is $\frac{1}{4}$.

Property

The probability of an event $E$ in an experiment with sample space $S$ with equally likely outcomes is given by

$E$ is a subset of $S$, so it is always true that $0 \le P(E) \le 1$.

Examples

• A standard 52-card deck is used. The probability of drawing a Jack is $\frac{4}{52} = \frac{1}{13}$ because there are 4 Jacks in the deck.

• When rolling a single six-sided die, the probability of rolling a number less than 3 (i.e., 1 or 2) is $\frac{2}{6} = \frac{1}{3}$.

Property

The probability of the union of two events $E$ and $F$ (written $E \cup F$) equals the sum of the probability of $E$ and the probability of $F$ minus the probability of $E$ and $F$ occurring together (which is called the intersection of $E$ and $F$ and is written as $E \cap F$).

Examples

• The probability of drawing a King or a Spade from a deck is $P(K) + P(S) - P(K \cap S) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$.

• When rolling a die, the probability of getting an odd number or a number greater than 4 is $P(\text{odd}) + P(>4) - P(\text{odd and } >4) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$.

Property

Events are said to be mutually exclusive events when they have no outcomes in common. The probability of the union of two mutually exclusive events $E$ and $F$ is given by

Examples

• From a deck of cards, drawing a 5 and drawing a Queen are mutually exclusive. The probability of drawing a 5 or a Queen is $P(5) + P(Q) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$.

• When rolling a die, the events 'rolling a 1' and 'rolling a 6' are mutually exclusive. The probability of rolling a 1 or a 6 is $\frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

Property

The complement of an event $E$, denoted $E'$, is the set of outcomes in the sample space that are not in $E$. The probability that the complement of an event will occur is given by

Examples

• If the probability of winning a game is $\frac{1}{10}$, the probability of losing is $1 - \frac{1}{10} = \frac{9}{10}$.

• The probability of rolling a 5 on a die is $\frac{1}{6}$. The probability of not rolling a 5 is $1 - \frac{1}{6} = \frac{5}{6}$.

Property

Many probability problems involve counting principles, permutations, and combinations. To solve these, we find the number of elements in the event and in the sample space. The probability is the ratio of the number of favorable outcomes to the total number of possible outcomes, often calculated using combinations or permutations.

Examples

• A team of 4 is chosen from 12 players. The probability that the 4 best players are chosen is $\frac{C(4,4)}{C(12,4)} = \frac{1}{495}$.

• A drawer has 6 red and 4 blue socks. The probability of randomly picking 2 red socks is $\frac{C(6,2)}{C(10,2)} = \frac{15}{45} = \frac{1}{3}$.