Topic 4.5 Conditional Probability: calculate and interpret conditional probabilities using the definition and the multiplication rule, including from two-way tables and tree diagrams.

What this topic is asking

The College Board (Topic 4.5) wants you to calculate and interpret conditional probabilities using the definition and the multiplication rule, and to read them off two-way tables and tree diagrams.

The definition

The intuition is the same as conditional distributions from Topic 2.3: you condition on $B$ by treating $B$ as the new whole, and ask what fraction of that world also has $A$. The denominator is the probability of the condition (what you are given), and the numerator is the probability that both occur. Getting the denominator right, the probability of the thing after the bar, is the crux.

The multiplication rule

The multiplication rule and the conditional definition are two views of one relationship. When you know two of the three quantities ($P(A \text{ and } B)$, $P(B)$, $P(A \mid B)$), you can find the third. In a tree diagram, each branch carries a conditional probability, and multiplying the branch probabilities along a path gives the joint probability of that sequence of events, the systematic way to handle multi-stage chance processes.

Reading conditionals from tables and trees

Two representations dominate exam questions. From a two-way table, a conditional probability is a cell count divided by the row or column total of the condition, exactly the conditional relative frequencies of Topic 2.3. "Given the person is female, what is the probability they prefer tea?" means divide the female-tea cell by the female total, not by the grand total. From a tree diagram, you build the process in stages: the first set of branches shows the initial event with its probabilities, and each later branch shows a conditional probability given the branch you are on. Multiplying along a path gives that path's joint probability, and adding the joint probabilities of all paths that produce an outcome gives its total probability. The medical-test example shows both moves: multiply along branches to get each path's probability, sum the "positive" paths to get $P(\text{positive})$, then divide to get the conditional $P(\text{condition} \mid \text{positive})$. Fluency in moving between tables, trees, and the formula is what the topic is really training.

Order matters: a famous trap

The most consequential idea here is that $P(A \mid B)$ and $P(B \mid A)$ are not the same, and confusing them produces badly wrong conclusions. In the medical-test example, $P(\text{positive} \mid \text{condition}) = 0.90$ is high, yet $P(\text{condition} \mid \text{positive}) \approx 0.37$ is low, because the condition is rare, so most positives come from the large healthy group's false positives. Reversing the conditional, assuming a positive test means a $90\%$ chance of disease, is a serious error that this topic is designed to inoculate against. The general lesson is to read carefully which event is given (the condition, after the bar) and which is being asked about, and to remember that a small base rate can make a "good" test's positive result surprisingly unreliable. This sensitivity to what is being conditioned on, and on the base rate, is one of the most practically important ideas in the whole probability unit.

Try this

Q1. $P(A \text{ and } B) = 0.12$ and $P(B) = 0.3$. Find $P(A \mid B)$. [2 points]

• Cue. $P(A \mid B) = \dfrac{0.12}{0.3} = 0.4$.

Q2. A test is positive $95\%$ of the time when a rare disease is present. Why might $P(\text{disease} \mid \text{positive})$ still be low? [1 point]

• Cue. Because the disease is rare, most positives are false positives from the large healthy group, so the reversed conditional can be small even when $P(\text{positive} \mid \text{disease})$ is high.