Joint Probability vs Marginal Probability

A joint probability is $P(A \cap B)$, the probability of the intersection ($\cap$ is read "and"). It is symmetric, since $P(A \cap B) = P(B \cap A)$, which is one clean difference from a conditional probability, where swapping the order changes the answer. Read off a two-way table it is always one interior cell divided by the grand total.

Take a survey of 200 students, classified by year and by how they get to school.

So $P(\text{senior and drives}) = 45/200 = 0.225$. The four interior cells give the four joint probabilities, 0.225, 0.175, 0.15 and 0.45, and those add to 1 because every student sits in exactly one cell.

"45 of the 80 seniors drive, so the probability of senior and drives is $45/80 = 0.5625$." That value is real, but it is the conditional probability $P(\text{drives} \mid \text{senior})$, which has already discarded the 120 non-seniors. A joint probability describes a student drawn from all 200, so the denominator is 200 and the answer is 0.225. Any time your denominator is a row or column total, you have computed a conditional.

Multiplying is only safe under independence. Here $P(\text{senior}) = 0.40$ and $P(\text{drives}) = 0.375$, whose product is 0.15, well short of the true joint value of 0.225. That gap is itself the evidence that year and driving are associated. When events are not independent, the general multiplication rule gets it right: $0.40 \times 0.5625 = 0.225$.

Adding the joint probabilities across the senior row returns $0.225 + 0.175 = 0.40$, which is the marginal probability of being a senior.

A marginal probability answers a question about one variable while ignoring the other. The numerator is a row total or a column total and the denominator is always the grand total, so nobody is discarded: every individual is still counted, just sorted on one variable instead of two. The name comes from where those totals sit, out in the margins of the two-way table.

Use the same survey of 200 students that the joint probability entry tabulates: 80 are seniors and 120 are not, while 75 drive to school and 125 do not. So $P(\text{senior}) = 80/200 = 0.40$ and $P(\text{drives}) = 75/200 = 0.375$. Each of those is also the sum of the joint probabilities along its own margin, since the senior row splits into $45/200 = 0.225$ and $35/200 = 0.175$, which add to 0.40.

"The marginal probabilities are 0.40, 0.60, 0.375 and 0.625, and they add to 2, so I have made a mistake." Nothing is wrong. Marginal probabilities add to 1 within a single variable, not across the table: $0.40 + 0.60 = 1$ for class and $0.375 + 0.625 = 1$ for driving. Adding all four counts every student twice, once by year and once by transport. The set that does add to 1 across the whole table is the four joint probabilities.

A marginal probability tells you nothing about the relationship between the two variables. $P(\text{drives}) = 0.375$ is a fact about the whole school and it would read the same whether or not seniors drive more than anyone else. Seeing the relationship takes conditionals: $P(\text{drives} \mid \text{senior}) = 45/80 = 0.5625$ against $P(\text{drives} \mid \text{not senior}) = 30/120 = 0.25$.

Two-way tables and the probabilities read from them open Unit 2, at topic 2.1.