Joint 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.