P(A ∩ B) is the chance that both happen; P(A ∣ B) is the chance of A among the outcomes where B happens. Read the question to see which one it asks for.
Two probabilities are easy to mix up: the chance that A and B both happen, and the chance of A among the cases where B happens. You met both in the general multiplication rule, $P(A \cap B) = P(B)\,P(A \mid B)$. Most wrong answers in conditioning problems are right answers to the other question, so this lesson is about reading the question.
🔑 Key idea
$P(A \cap B)$, the joint probability, is a share of everything. $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$, the conditional probability, is a share of B. Words like "of those who", "among", "given" and "if" name the group to divide by; "both" or "and" with no such phrase asks for the joint.
💡 Why it works
Knowing that B happened rules out every outcome outside B, so B becomes the new sample space. To keep the total at 1, every probability inside B is rescaled by $\frac{1}{P(B)}$. The part of A that survives is $A \cap B$, so its new probability is $\frac{P(A \cap B)}{P(B)}$. The function $A \mapsto P(A \mid B)$ satisfies all three axioms, so conditional probabilities obey every rule you know: complements, the addition rule, and so on.
The joint probability is the overlap as a share of the whole square. Conditioning on Late stretches the Late region to fill the square, and the share that Rain covers there is the conditional probability.
In the picture, the overlap has area $\frac{1}{5}$ of the square: that is $P(\text{Rain} \cap \text{Late})$. Inside the Late region (area $\frac{2}{5}$) it covers half: that is $P(\text{Rain} \mid \text{Late}) = \frac{1/5}{2/5} = \frac{1}{2}$. Same overlap, two different denominators.
Worked examples
✏️ Worked example: One table, three questions
Of 200 students, 60 play a sport; 24 of those also have a part-time job; 56 students have a job in all. A student is picked at random. Find (a) the fraction who play a sport and have a job, (b) of the athletes, the fraction with a job, (c) of those with a job, the fraction who play a sport.
(a) No group is named, and "and" asks for both: $P(S \cap J) = \frac{24}{200} = \frac{3}{25}$.
(b) "Of the athletes" divides by the 60 athletes: $P(J \mid S) = \frac{24}{60} = \frac{2}{5}$.
(c) "Of those with a job" divides by the 56 workers: $P(S \mid J) = \frac{24}{56} = \frac{3}{7}$.
Answer: $\frac{3}{25}$, $\frac{2}{5}$ and $\frac{3}{7}$: the same 24 students, three different denominators.
✏️ Worked example: From percents
Of a café's customers, 40% buy coffee and 15% buy both coffee and a pastry. Of the coffee buyers, what fraction buy a pastry?
The 15% is joint: a share of all customers. The question names the coffee buyers as the group.
Answering "of those with B, what fraction have A?" with the joint $P(A \cap B)$, or dividing by the wrong event and giving $P(B \mid A)$ instead of $P(A \mid B)$. Check: name the group the question restricts to, and make sure it is your denominator. A conditional probability is never smaller than the joint one, since $P(B) \le 1$.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.
Grinstead and Snell, Introduction to Probability (2nd revised edition, American Mathematical Society; free from Dartmouth College), chapter 4 (Conditional Probability), section 4.1, discrete conditional probability
MIT OpenCourseWare 18.05, Introduction to Probability and Statistics (Spring 2014), readings on conditional probability, independence and Bayes' theorem