The Timpson Lyceum · Discrete Math (2025)

Probability: Events, Conditional Probability, and Independence - Lecture 15 (Discrete Math)

Overview

This lecture builds the working machinery of discrete probability on top of last week's sample spaces. An event is formalized as any subset of the sample space, and its probability is the sum of the probabilities of the outcomes it contains. We compute event probabilities across many settings — dice sums, repeated coin flips, poker hands, family compositions, and the birthday problem — leaning heavily on counting and on the complement trick for "at least one" questions. We then prove the inclusion–exclusion identity $P(A) + P(B) = P(A \cap B) + P(A \cup B)$ and read off its consequences. The second half introduces conditional probability and independence, the $n$ -fold repeated trial sample space, and closes with the classic Monty Hall problem.

Key Concepts

Sample space (recap). A sample space is an ordered pair $(S, P)$ where $S$ is the set of all possible outcomes and $P$ is a probability function. Two rules constrain $P$ : every outcome has $P(s) \ge 0$ , and the probabilities of all outcomes sum to $1$ , i.e. $\sum_{s \in S} P(s) = 1$ . For a fair mechanism, every outcome is equally likely; e.g. rolling two dice gives $|S| = 36$ and $P(s) = \tfrac{1}{36}$ for each $s$ .

Event. Given a sample space $(S,P)$ , any subset $A \subseteq S$ is an event. Its probability is the sum of the probabilities of the outcomes inside it: $P(A) = \sum_{a \in A} P(a).$ The full set $S$ and the empty set $\varnothing$ are both valid events; the set of all events is the power set of $S$ .

Equally-likely shortcut. When every outcome has the same probability, $P(A) = \frac{|A|}{|S|}.$ This is why "rolling a 7" with two dice has probability $\tfrac{6}{36}$ — six of the thirty-six outcomes sum to $7$ . This shortcut requires the fairness assumption; it fails for biased coins or uneven spinners.

Counting events with combinations. For five coin flips, $|S| = 2^5 = 32$ . The number of outcomes with exactly $k$ heads is $\binom{5}{k}$ , using $\binom{n}{k} = \frac{n!}{(n-k)!\,k!}.$ So exactly two heads gives $\binom{5}{2} = 10$ outcomes. Note $\binom{n}{k} = \binom{n}{n-k}$ (choosing which to include is the same as choosing which to leave out), and the counts $1,5,10,10,5,1$ form a row of Pascal's triangle — the binomial coefficients.

Complement rule and the "at least one" trick. The complement $A^c$ is everything in $S$ not in $A$ , and $P(A^c) = 1 - P(A).$ For "at least one" questions, count the complement instead. Rolling one die ten times, the chance of at least one 1 is hard to count directly, but its complement (no 1's) has $5^{10}$ outcomes out of $6^{10}$ , so $P(\text{at least one }1) = 1 - \frac{5^{10}}{6^{10}} \approx 0.838.$ Caution: you cannot cancel exponents — $\tfrac{5^{10}}{6^{10}} = \left(\tfrac{5}{6}\right)^{10}$ , not $\tfrac{5}{6}$ .

Inclusion–exclusion identity. For any two events $A, B$ in the same sample space, $P(A) + P(B) = P(A \cap B) + P(A \cup B).$ The proof is a counting argument: for any outcome $s$ , check the four cases (in neither, in $A$ only, in $B$ only, in both) and verify $P(s)$ appears the same number of times on each side. An outcome in both sides shows up twice on each side; an outcome in exactly one set shows up once on each; an outcome in neither shows up zero times. Since both sides accumulate the same terms, they are equal.

Consequences of the identity.

If $A \cap B = \varnothing$ (disjoint events), then $P(A \cup B) = P(A) + P(B)$ .
In general $P(A \cup B) \le P(A) + P(B)$ (subadditivity), with equality exactly when $P(A \cap B) = 0$ .
$P(S) = 1$ and $P(\varnothing) = 0$ (the empty sum is $0$ ).

Conditional probability. The probability of $A$ given $B$ has occurred, written $P(A \mid B)$ , restricts attention to the outcomes inside $B$ . For $P(B) \ne 0$ , $P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$ Geometrically, it is the fraction of $B$ 's region that $A$ covers. Order matters: $P(A \mid B)$ and $P(B \mid A)$ are generally very different.

Independence. Events $A$ and $B$ are independent if and only if $P(A \cap B) = P(A)\,P(B).$ Equivalently, knowing $B$ occurred does not change the odds of $A$ : $P(A \mid B) = P(A)$ . Example: the first flip being heads is independent of the last flip being tails. But conditioning on aggregate information can break independence — "first flip is heads" is not independent of "exactly three of five flips are tails." A common gambler's error is confusing these two kinds of questions (e.g. the roulette "due for red" fallacy).

$n$ -fold repeated trial sample space. Given $(S, P)$ and a positive integer $n$ , let $S^n = S \times S \times \cdots \times S$ ( $n$ copies). Then $(S^n, P)$ is the $n$ -fold repeated trial sample space, where an outcome's probability is the product of its components' probabilities: $P(s_1, s_2, \ldots, s_n) = \prod_{k=1}^{n} P(s_k).$ This lets you reuse one die's sample space for rolling several dice without redefining it. For one fair die, $P(1,1) = \tfrac16 \cdot \tfrac16 = \tfrac{1}{36}$ . For a biased coin with $P(\text{heads}) = p$ (so $P(\text{tails}) = 1 - p$ ), the sequence HHTHH has probability $p^4(1-p)$ .

Monty Hall problem. Three doors hide one car and two goats. You pick a door; the host then opens a different door revealing a goat and offers a switch. You should always switch: your original pick is correct with probability $\tfrac13$ , so the car is behind one of the other doors with probability $\tfrac23$ — and the host's reveal concentrates that $\tfrac23$ onto the single remaining unopened door. Switching wins exactly when your first pick was wrong, which happens $\tfrac23$ of the time.

Worked Example

Two dice. Let $A$ = "rolled an 8" and $B$ = "rolled an even total." Find $P(A \mid B)$ .

There are $5$ ways to total $8$ (namely $2{+}6,\,3{+}5,\,4{+}4,\,5{+}3,\,6{+}2$ ), so $P(A) = \tfrac{5}{36}$ . Half of the $36$ outcomes give an even total, so $|B| = 18$ and $P(B) = \tfrac{18}{36} = \tfrac12$ . Since every "total of 8" outcome is also even, $A \cap B = A$ , giving $P(A \cap B) = \tfrac{5}{36}$ .

Apply the conditional probability definition (symbolic rule first, then substitute):

\begin{aligned} P(A \mid B) &= \frac{P(A \cap B)}{P(B)} \\ &= \frac{5/36}{18/36} \\ &= \frac{5}{36} \cdot \frac{36}{18} \\ &= \frac{5}{18}. \end{aligned}

This matches the intuitive count: knowing the total is even narrows the world to $18$ outcomes, $5$ of which total $8$ .

Summary

An event is any subset $A \subseteq S$ , and $P(A) = \sum_{a \in A} P(a)$ ; for equally likely outcomes this reduces to $P(A) = \tfrac{|A|}{|S|}$ .
Count events with combinations $\binom{n}{k}$ (Pascal's triangle / binomial coefficients), and use the complement $P(A^c) = 1 - P(A)$ to handle "at least one" questions cleanly.
The identity $P(A) + P(B) = P(A \cap B) + P(A \cup B)$ holds for all events; it gives the disjoint-additivity rule, the bound $P(A \cup B) \le P(A) + P(B)$ , and $P(S) = 1$ , $P(\varnothing) = 0$ .
Conditional probability is $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ , restricting the world to $B$ ; note $P(A \mid B) \ne P(B \mid A)$ in general.
Events are independent iff $P(A \cap B) = P(A)P(B)$ , equivalently $P(A \mid B) = P(A)$ ; the $n$ -fold repeated trial space multiplies component probabilities, $P(s_1,\ldots,s_n) = \prod_{k} P(s_k)$ .
In Monty Hall, always switch: staying wins with probability $\tfrac13$ , switching with probability $\tfrac23$ .