Week 3 — Conditional probability

How information changes a probability

The week question

When you learn that something happened, how should the probability of everything else change? Last week you built sample spaces and learned the addition rule for combining events. This week the central question is different: given that one event is known to have occurred, what is the probability of another? That “given” is the whole story. A probability is never a property of an event alone — it is a property of an event together with what you currently know. Conditional probability is the precise tool for revising a probability when new information arrives.

A scheduling note: this is a short week. Because Labor Day falls on Monday, September 7, the class meets only Wednesday and Friday. To keep the development complete, this public note carries the full conditional-probability story as an asynchronous resource — read it as the through-line that the two shorter in-person sessions point at, rather than as a transcript of three lectures.

Why this matters

Almost every interesting probability statement in this course is secretly conditional. When Maya, our commuter student, checks the weather and sees rain, she does not keep using the all-mornings figure for catching the shuttle on time — she uses the rainy-morning figure. That switch is conditioning. The same move drives:

Updating beliefs with evidence. Bayes’ rule (Week 5) is built entirely on the definition you learn this week. You cannot do Bayesian reasoning without conditional probability first.
Diagnostic and screening questions. “Given a positive test, what is the chance the person actually has the condition?” is a conditional probability, and the answer is famously not what intuition guesses.
Sequential and dependent events. Drawing cards without replacement, sampling people one at a time, multi-step processes — each step’s probability depends on what already happened. The multiplication rule, the other star of this week, is what lets you chain those steps together.

Conditioning is also the precise language for independence, which is next week’s topic. Two events are independent exactly when conditioning on one does not change the probability of the other. So this week is the hinge: it gives you the definition that the rest of the course leans on.

Learning goals

By the end of this week you should be able to:

State the definition \(P(A \mid B) = P(A \cap B) / P(B)\) and explain why it requires \(P(B) > 0\).
Use the multiplication rule \(P(A \cap B) = P(A \mid B)\,P(B)\) to find the probability of a joint event from a conditional and a marginal.
Read and build a simple tree diagram for a staged experiment, and recover joint probabilities by multiplying along branches.
Interpret conditioning as updating on information — narrowing the sample space to the outcomes consistent with what you now know.
Verify, in the commuter case, that \(P(\text{on time} \mid \text{rain}) = 0.60\) is consistent with the unconditional \(P(\text{on time}) = 0.81\) and explain why the two numbers differ.

Core vocabulary

Conditional probability \(P(A \mid B)\) — the probability of \(A\) given that \(B\) has occurred; read “\(A\) given \(B\).” Defined only when \(P(B) > 0\).
Conditioning event — the event \(B\) to the right of the bar; the information you are taking as known.
Multiplication rule — \(P(A \cap B) = P(A \mid B)\,P(B)\), the definition rearranged so you can build a joint probability from a conditional one.
Joint event — \(A \cap B\), both happening; its probability is the numerator in the definition.
Reduced sample space — the set of outcomes still possible once you know \(B\); conditioning rescales probabilities so they sum to \(1\) over just this reduced space.
Tree diagram — a branching picture of a staged experiment where each path’s probability is the product of the conditional probabilities along its branches.

Carry the bar notation carefully: \(P(A \mid B)\) and \(P(B \mid A)\) are different questions with generally different answers. Keeping the conditioning event straight is half the battle this week.

Concept development

From “what is the probability” to “the probability given what I know”

Start with a probability you already trust over the whole sample space — the unconditional or marginal probability \(P(A)\). Now you learn that event \(B\) occurred. Outcomes outside \(B\) are no longer possible, so you discard them and look only at the outcomes inside \(B\). Among those, the ones that also give you \(A\) are exactly the outcomes in \(A \cap B\). The conditional probability is the share of \(B\)’s probability that also lands in \(A\):

\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B) > 0. \]

Two things deserve emphasis. First, the denominator \(P(B)\) is what rescales the reduced sample space so its probabilities again sum to \(1\): dividing by \(P(B)\) stretches the leftover probability back up to a full unit. Second, the requirement \(P(B) > 0\) is not a technicality you can wave away — if \(B\) is impossible, the ratio divides by zero and “given \(B\)” has no meaning, because you would be conditioning on something that never happens. Throughout this course, whenever we write \(P(A \mid B)\), take for granted that \(P(B) > 0\).

A useful sanity check: conditioning on the whole sample space \(\Omega\) changes nothing, because \(P(A \mid \Omega) = P(A \cap \Omega)/P(\Omega) = P(A)/1 = P(A)\). The unconditional probability is just the conditional probability given “everything,” which is another way of saying every probability is conditional on some background of assumptions — conditioning simply makes the assumed information explicit.

The multiplication rule: building joint probabilities

The definition has a denominator, which is awkward when what you actually want is the probability that both events happen. Clear the denominator and you get the multiplication rule:

\[ P(A \cap B) = P(A \mid B)\,P(B). \]

This says a joint event happens in two stages: first \(B\) happens (with probability \(P(B)\)), and then, given that \(B\) happened, \(A\) happens (with probability \(P(A \mid B)\)). Multiply the stage probabilities to get the probability of the whole path. By symmetry you can also condition the other way, \(P(A \cap B) = P(B \mid A)\,P(A)\) — both are correct, and choosing the order that matches how the information actually arrives usually makes the arithmetic easiest. This rule is the engine behind every multi-step probability in the course; it is what makes staged experiments computable.

Tree diagrams: conditioning you can see

A tree diagram is the multiplication rule drawn as a picture. Each stage of an experiment becomes a fan of branches; you label every branch with the conditional probability of that outcome given the path that led to it. To get the probability of a complete path, multiply the branch probabilities along it. To get the probability of an event that several paths satisfy, add up those paths.

For the commuter’s morning, the first split is the weather and the second is the shuttle:

Stage 1 — rain or no rain: \(P(\text{rain}) = 0.30\), \(P(\text{no rain}) = 0.70\).
Stage 2 — on time or late, conditioned on the weather:
- given rain: \(P(\text{on time} \mid \text{rain}) = 0.60\), \(P(\text{late} \mid \text{rain}) = 0.40\);
- given no rain: \(P(\text{on time} \mid \text{no rain}) = 0.90\), \(P(\text{late} \mid \text{no rain}) = 0.10\).

The four leaves multiply to the four joint probabilities — for example the “rain and on time” leaf is \(0.30 \times 0.60 = 0.18\). Notice that the second-stage branches are conditional probabilities, and the leaf values are joint probabilities: the tree is literally the multiplication rule, repeated once per path. (All numbers here are synthetic; seed set for the simulated versions below.)

Conditioning as updating on information

The deepest reading of this week is that conditioning is updating. Before checking the weather, Maya’s best probability for catching the shuttle on time is the unconditional \(0.81\). The instant she sees rain, the relevant sample space shrinks to rainy mornings, and her probability drops to \(0.60\). No coin was flipped and no shuttle moved — only her information changed. The probability moved because probability encodes a state of knowledge, not a fixed physical attribute of the morning.

This is the conceptual seed of the whole Bayesian thread of the course. A probability is a summary of what you know; new information conditions on a smaller world; the conditional probability is the updated summary. Week 4 asks when information fails to update a probability (independence), and Week 5 turns the machinery around to ask what the evidence tells you about its cause (Bayes’ rule). All of it is this one definition, viewed from different angles.

Worked examples

Worked example — the commuter’s morning (recurring slice)

Symbolic setup. Let \(R\) be the event “it rains” and \(O\) the event “the shuttle is on time.” We are given the conditional reliability \(P(O \mid R) = 0.60\) and the rain rate \(P(R) = 0.30\), and from Week 1’s marginal we know the unconditional \(P(O) = 0.81\). We want to confirm that the conditional definition is internally consistent: that the joint \(P(O \cap R)\) implied by the multiplication rule, divided by \(P(R)\), returns the \(0.60\) we started with.

By the multiplication rule, the joint probability of rain and an on-time shuttle is

\[ P(O \cap R) = P(O \mid R)\,P(R). \]

The definition of conditional probability run in reverse then recovers the conditional we were given:

\[ P(O \mid R) = \frac{P(O \cap R)}{P(R)}. \]

Numeric. Substitute the locked numbers into the multiplication rule:

\[ P(O \cap R) = 0.60 \times 0.30 = 0.18. \]

Now divide that joint probability by the probability of the conditioning event:

\[ P(O \mid R) = \frac{0.18}{0.30} = 0.60. \]

The loop closes: \(0.18 / 0.30 = 0.60\), exactly the rainy-morning reliability we assumed. The point is the contrast with the unconditional figure. Across all mornings the shuttle is on time with probability \(P(O) = 0.81\), but on the mornings you know it is raining the probability is only \(0.60\). Knowing it rained genuinely changes the probability — from \(0.81\) down to \(0.60\) — because rainy mornings are a worse sub-population for the shuttle, and conditioning restricts attention to exactly that sub-population. Because \(0.60 \neq 0.81\), rain and on-time arrival are not independent, which is the question Week 4 takes up.

Worked example — two cards without replacement (transfer)

Symbolic setup. Move to a fresh context to see the same two rules at work. A standard \(52\)-card deck has \(4\) aces. Draw two cards one after another without replacement and ask for the probability that both are aces. Let \(A_1\) be “first card is an ace” and \(A_2\) be “second card is an ace.” The two draws are dependent: removing the first card changes what is left for the second, so the second draw’s probability must be conditioned on the first. The multiplication rule gives

\[ P(A_1 \cap A_2) = P(A_1)\,P(A_2 \mid A_1). \]

Numeric. On the first draw, \(4\) of \(52\) cards are aces, so

\[ P(A_1) = \frac{4}{52} = \frac{1}{13}. \]

If the first card was an ace, only \(3\) aces remain among the \(51\) remaining cards, so the conditional probability of a second ace is

\[ P(A_2 \mid A_1) = \frac{3}{51} = \frac{1}{17}. \]

Multiply the two stages — exactly the path through a two-level tree:

\[ P(A_1 \cap A_2) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221} \approx 0.0045. \]

The conditioning is doing real work here. The naive “with replacement” guess would be \((4/52)^2 = 1/169 \approx 0.0059\), larger because it ignores that one ace has already been used up. Sampling without replacement makes the second draw depend on the first, and the multiplication rule with its conditional second factor is what records that dependence.

You can confirm the figure by simulation. The chunk below is shown as teaching, not executed here; running it later (with the seed set) will land near \(1/221\).

set.seed(35003)
deck <- rep(c("ace", "other"), times = c(4, 48))  # 4 aces, 48 non-aces
trials <- 100000
both_aces <- replicate(trials, {
  hand <- sample(deck, size = 2, replace = FALSE)   # draw 2 without replacement
  all(hand == "ace")
})
mean(both_aces)          # simulated P(both aces); compare to 1/221 ~ 0.00452
1 / 221                  # exact value for reference

A common mistake

The most common error this week is swapping the conditioning event — treating \(P(A \mid B)\) and \(P(B \mid A)\) as the same number. They are not. In the commuter case, \(P(\text{on time} \mid \text{rain}) = 0.60\) answers “of rainy mornings, how many are on time?” while \(P(\text{rain} \mid \text{on time})\) answers the reverse, “of on-time mornings, how many were rainy?” — a completely different question with a different denominator. The two are linked by Bayes’ rule (Week 5), not equal. A related slip is forgetting the denominator entirely and reporting the joint probability \(P(A \cap B)\) when the question asked for the conditional \(P(A \mid B)\): the joint rain-and-on-time figure is \(0.18\), but the conditional on-time- given-rain figure is \(0.60\), and confusing them is an order-of-magnitude error. When in doubt, write the definition out as a ratio and name the conditioning event in words before you compute.

Low-stakes self-checks (ungraded)

These are ungraded self-checks — work them on paper to test your grip on the definition, then read on. No answers are posted here; that is by design.

Using only the locked commuter numbers, compute \(P(\text{late} \mid \text{rain})\) and confirm it agrees with \(1 - P(\text{on time} \mid \text{rain})\).
Find the joint probability \(P(\text{no rain} \cap \text{on time})\) by multiplying along the relevant tree branch, then check that the four joint probabilities sum to \(1\).
For the two-card example, compute \(P(\text{second is an ace})\) unconditionally and explain why it still equals \(4/52\) even though the draws are dependent.
Explain in one sentence why \(P(A \mid B)\) is undefined when \(P(B) = 0\).

Reading and source pointer

This week’s spine is Grinstead & Snell, Chapter 4 — Conditional Probability, which develops the definition \(P(A \mid B) = P(A \cap B)/P(B)\), the multiplication rule, and the tree-diagram view of staged experiments. For an alternate take on conditioning and trees, the MIT OCW 18.05 materials on conditional probability are a useful supplement. These notes are the course’s own synthesis, grounded in but not copied from the sources; all data are synthetic with seeds set.

Formula-verification status

verified: false. The formulas on this page are drafted but not yet machine-checked, so the math gate is BLOCKED. Treat every expression here — the definition, the multiplication rule, and the worked arithmetic — as provisional pending human sign-off. The numbers are designed to be internally consistent (for instance \(0.18/0.30 = 0.60\) and \(4/52 \times 3/51 = 1/221\)), but consistency is not verification; do not rely on these as confirmed results until the gate is cleared.

Public vs. graded

These notes, the examples, and the practice here are public and ungraded — study material only. No graded prompts, answer keys, rubrics, point values, or due dates appear on this site. Graded checkpoints, quizzes, homework, labs, the midterm, the project, and the final live in Blackboard (the LMS), which is authoritative for due dates, submissions, and grades. If this page and Blackboard ever disagree, follow Blackboard.

Looking ahead

Next week sharpens conditioning into a yes-or-no question: independence. Two events are independent exactly when conditioning on one leaves the other’s probability unchanged — that is, when \(P(A \mid B) = P(A)\). We already have the evidence that rain and on-time arrival are not independent (\(0.60 \neq 0.81\)), and Week 4 will make that observation precise and contrast it with a genuinely independent pair, like two coin flips. Then Week 5 turns the bar around with Bayes’ rule to ask what an observation tells us about its hidden cause.