Joint, marginal & conditional probability

Click a row header ($A$) or a column header ($B$). The cells you picked light up, and the readout below shows what's left once you renormalize just that slice — the conditional distribution living inside the joint table all along.

A joint distribution $P(A,B)$ assigns a probability to every combination of outcomes at once. Summing a whole row or column marginalizes out the other variable:

• $P(A,B)$ — the joint probability table: one number per (row, column) combination.
• $P(A=a)$ — the marginal probability of $A=a$, found by summing across every value of $B$.

A conditional distribution slices out one row or column and renormalizes it so it sums to 1 again:

• $P(B\mid A=a)$ — the distribution over $B$, restricted to the world where $A=a$ is already known.

1. Marginals are just row and column sums

   No new arithmetic — marginalizing is exactly the sum you'd take to check a row or column adds up.

2. Conditioning is slicing, then renormalizing

   Fix one variable's value, take the sub-table that survives, and divide every entry by whatever that sub-table currently sums to. That division is the entire difference between a joint slice and a conditional distribution.

3. Conditioning on a row and conditioning on a column don't rhyme

   $P(B\mid A{=}a)$ divides by a row sum; $P(A\mid B{=}b)$ divides by a column sum. Nothing forces these two conditionals to look alike, even though they come from the same table.

Pick any cell: the joint value is just that one entry, but the two conditionals built from it divide by completely different denominators — a row sum for $P(B\mid A)$, a column sum for $P(A\mid B)$ — which is exactly why $P(B\mid A) \ne P(A\mid B)$ in general.

1. Read off the joint table

   • Row $a_1$: $P(a_1,b_1)=0.2$, $P(a_1,b_2)=0.1$, $P(a_1,b_3)=0.2$ — sums to $P(a_1)=0.5$
   • Row $a_2$: $P(a_2,b_1)=0.1$, $P(a_2,b_2)=0.3$, $P(a_2,b_3)=0.1$ — sums to $P(a_2)=0.5$
2. Marginalize to get P(b3)

   Sum column $b_3$ down both rows: $P(b_3) = P(a_1,b_3) + P(a_2,b_3) = 0.2 + 0.1 = 0.3$.

3. Condition on b3 to get P(a2 | b3)

   $$P(a_2 \mid b_3) = \frac{P(a_2, b_3)}{P(b_3)} = \frac{0.1}{0.3} = \frac{1}{3} \approx 0.333$$

   Notice this is not the same as $P(a_2)=0.5$ — seeing $B=b_3$ actually shifted the belief about $A$.