Overview

Probability questions often reduce to counting. Define a clear sample space, count favorable outcomes, and use complements when easier.

When outcomes are not equally likely, use conditional probability, tree diagrams, or symmetry to compute probabilities directly.

Key Ideas

• For equally likely outcomes, use $P(A) =$ favorable outcomes / total outcomes.
• $P(A^c) = 1 - P(A)$ is often simpler to compute.
• Use the rule of product to count outcomes in multi-step experiments.
• Conditional probability: $P(A\mid B) = P(A\cap B)/P(B)$.
• Independence: $P(A\cap B) = P(A)P(B)$.

Core Skills

Define the Sample Space

List the outcomes or count them with a method like the rule of product. Every probability is a ratio of favorable to total outcomes.

Use Complements Early

If the event is "at least" or "not equal", count the opposite event first.

Track Conditional Events

After a condition occurs, update the sample space before counting.

Expected value

For a random variable $X$ with outcomes $x_1, x_2, \ldots, x_n$ occurring with probabilities $p_1, p_2, \ldots, p_n$ is the weighted average:

$E[X] \;=\; \sum_{i=1}^{n} x_i \, p_i.$

The single most important property is the linearity of expectation, for any random variables $X$ and $Y$ (even dependent ones),

$E[X + Y] \;=\; E[X] + E[Y].$

Worked Example

Two fair dice are rolled. What is the probability the sum is $7$?

There are $36$ equally likely outcomes. The favorable pairs are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$, so the probability is $6/36 = 1/6$.

More Examples

Example 1: Conditional Probability

A bag has 3 red and 2 blue marbles. Two marbles are drawn without replacement. What is the probability the second marble is red given the first is red?

After drawing one red, the bag has 2 red and 2 blue left, so the probability is $2/4 = 1/2$.

Example 2: Independence Check

Flip a fair coin twice. Let $A$ be "first flip is heads" and $B$ be "second flip is heads". Then $P(A)=P(B)=1/2$ and $P(A\cap B)=1/4 = P(A)P(B)$, so $A$ and $B$ are independent.

Example 3: Complement

What is the probability that at least one of two fair dice shows a $6$?

Complement: no $6$ on either die. That is $(5/6)^2 = 25/36$. So the probability is $1 - 25/36 = 11/36$.

Example 4: Expected Value

A fair die is rolled once. If it shows $k$, you win $k^2$ dollars. What is the expected winnings?

$\frac{1+4+9+16+25+36}{6} = \frac{91}{6}$

Strategy Checklist

• Decide if outcomes are equally likely.
• Use complements for "at least" or "not" statements.
• Rebuild the sample space after conditioning.
• Check independence before multiplying probabilities.

Common Pitfalls

• Assuming outcomes are equally likely without checking.
• Forgetting to count order when it matters.
• Using $P(A\mid B) = P(A) / P(B)$ by mistake.
• Treating dependent events as independent.

Practice Problems