Two discrete variables
Two measurements from the same outcome belong in one probability table. Learn to read that table, keep one measurement, condition on the other, and accumulate probabilities below two thresholds.
One outcome, two measurements
Imagine choosing one record at random from a collection of $20$ equally likely records. Each record has two numerical measurements: a tag $X$ that is $0$ or $1$, and a count $Y$ that is $0$, $1$ or $2$. A random variable is a numerical value determined by the random outcome. Here the same selected record determines both variables.
We choose the following example counts: among records tagged $0$, the three counts occur $2$, $3$ and $1$ times; among records tagged $1$, they occur $4$, $2$ and $8$ times. Divide each count by $20$ to get its probability. Rows represent $X$; columns represent $Y$.
| $p_{X,Y}(x,y)$ | $Y=0$ | $Y=1$ | $Y=2$ |
|---|---|---|---|
| $X=0$ | $2/20$ | $3/20$ | $1/20$ |
| $X=1$ | $4/20$ | $2/20$ | $8/20$ |
This is a joint probability mass function, or joint PMF. Its entry $p_{X,Y}(x,y)$ means $P(X=x,Y=y)$: the probability that both equalities hold. For example, $p_{X,Y}(1,2)=8/20=2/5$. The number $2$ is a possible value of $Y$; the fraction $2/5$ is a probability. They have different jobs.
Every entry must be nonnegative, and the entries must total $1$. Here $(2+3+1+4+2+8)/20=1$. A compact way to write that addition is a double sum: first add across the possible $y$ values at each fixed $x$, then add those row totals. The subscripts below the sums name values of the variables, not row numbers.
$$p_{X,Y}(x,y)\ge0,\qquad \sum_x\sum_y p_{X,Y}(x,y)=1$$
Select the records described by an event
- For $P(X=1,Y\le1)$, keep row $X=1$ and columns $Y=0,1$. Add their disjoint possibilities: $(4+2)/20=3/10$.
- For $P(X=0\text{ or }Y=2)$, add the $X=0$ row, with total $6/20$, and the $Y=2$ column, with total $9/20$. Their shared cell $(0,2)$ was counted twice, so subtract $1/20$. The result is $14/20=7/10$.
- For $P(XY\ge1)$, test the product at each pair. All $X=0$ cells fail. In row $X=1$, only $Y=1,2$ qualify, giving $(2+8)/20=1/2$.
If the table had instead been given as unknown multiples $c$ of the six counts, normalization would give $20c=1$, hence $c=1/20$. This is how an unknown probability multiplier is determined: add its weights over the stated possible pairs before solving for it.
A possible value pair can still have zero mass
The positive-mass support consists of the pairs whose entries are positive; every other pair has probability zero. It need not contain every combination of the separate values. For example, a different model puts mass $1/2$ at $(0,0)$ and $1/2$ at $(1,1)$. Both variables can be $0$ or $1$, yet $(0,1)$ and $(1,0)$ never occur. For countably many possible pairs, the same nonnegative-sum definition applies.