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: $P(A\cap B) = P(A)P(B)$.
• Distinguish dependent events (without replacement) from independent events (with replacement, or separate trials).

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.

Distinguishability and Cases

If outcomes are generated in stages, check whether order creates different outcomes. If yes, use ordered counting and then convert to probability.

For non-uniform sample spaces, split into clean cases, compute each case probability, and add.

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:

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

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 5: Dependent vs. Independent

Two marbles are drawn from a bag with 4 red and 3 blue marbles.

• With replacement: $P(\text{red then red}) = \frac{4}{7}\cdot\frac{4}{7}$.
• Without replacement: $P(\text{red then red}) = \frac{4}{7}\cdot\frac{3}{6}$.

The denominators change only in the dependent case.

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