Why the Two Draws Must Be Treated Together
When two objects are drawn from a finite population without replacement, the first draw changes which objects remain for the second. A joint probability table records the probabilities of all pairs of outcomes and makes that change visible. It also provides a direct way to obtain each draw's marginal distribution and to calculate probabilities involving both draws.
The order of the draws matters when the outcomes are named \(X\) for the first draw and \(Y\) for the second. For example, the ordered outcome \((1,3)\) is different from \((3,1)\), even though both contain the same two values. The calculations below use two ordered draws, with every remaining population member equally likely to be drawn at each step.
Members of the population are distinct even when they share the same recorded value. Thus, if three different objects all have value 0, drawing a 0 does not identify which object was selected. This distinction is essential when counting possible pairs.
Building the Joint Table
If the recorded values for each member are the same kind of quantity, let \(n_x\) be the number of population members with value \(x\). There are \(N(N-1)\) possible ordered pairs of distinct members, all equally likely. For any values \(x\) and \(y\), the number of pairs giving \(X=x\) and \(Y=y\) is \(n_xn_y\) when \(x\ne y\). When \(x=y\), it is \(n_x(n_x-1)\), because the same member cannot be selected twice.
Here \(\mathbf{1}_{\{x=y\}}\) is 1 when \(x=y\) and 0 otherwise. The formula applies to uniformly selected ordered draws without replacement. It does not apply unchanged to independent draws, sampling with replacement, or a nonuniform selection procedure.
When both draws record the same population characteristic, each marginal gives the population frequency distribution: \(P(X=x)=P(Y=x)=n_x/N\). The joint distribution still need not factor into the product of those marginals. The restriction that the same member cannot appear twice can create dependence between the recorded outcomes.
Worked Tables and Marginals
Worked Example: Indicators for Two Colors
A hypothetical bag contains 2 red tokens, 2 blue tokens, and 1 green token. Two tokens are drawn in order without replacement. Let \(X=1\) if the first token is red and \(X=0\) otherwise; let \(Y=1\) if the second token is blue and \(Y=0\) otherwise. Find the joint table, both marginal distributions, and \(P(X=1,Y=1)\). The model is a uniform ordered sample from a finite population of \(N=5\) tokens; no interest or mortality assumption is involved.
There are \(5\cdot4=20\) equally likely ordered pairs of distinct tokens. For \(X=1,Y=1\), there are 2 choices for the first red token and 2 choices for the second blue token, giving \(2\cdot2=4\) pairs and probability \(4/20=1/5\).
For \(X=1,Y=0\), there are 2 choices for the first red token. After that choice, the second token must be nonblue; the remaining population has 1 red and 1 green token that satisfy this requirement. There are \(2\cdot2=4\) pairs, so the probability is \(4/20=1/5\).
For \(X=0,Y=1\), the first token must be blue or green. If it is blue, there are \(2\cdot1=2\) ordered pairs with a blue token second; if it is green, there are \(1\cdot2=2\) such pairs. The total is \(4\), giving probability \(4/20=1/5\). The final cell is \(P(X=0,Y=0)=1-3/5=2/5\), since the other three cells sum to \(3/5\).
| \(X\) / \(Y\) | \(Y=0\) | \(Y=1\) | Row total |
|---|---|---|---|
| \(X=0\) | \(2/5\) | \(1/5\) | \(3/5\) |
| \(X=1\) | \(1/5\) | \(1/5\) | \(2/5\) |
| Column total | \(3/5\) | \(2/5\) | 1 |
The row totals give \(P(X=0)=3/5\) and \(P(X=1)=2/5\). The column totals give \(P(Y=0)=3/5\) and \(P(Y=1)=2/5\). In particular, \(P(X=1,Y=1)=1/5\). The marginals also check directly: 2 of the 5 tokens are red, and 2 of the 5 are blue.
Method and check: Count ordered pairs in each cell, then divide by 20. All four cells are nonnegative and sum to 1. A common mistake would be to multiply the red proportion \(2/5\) by the blue proportion \(2/5\) and report \(4/25\); that treats the draws as independent. The correct joint probability is \(4/20=1/5\), because the first red draw leaves all 2 blue tokens among 4 remaining tokens.
Worked Example: A Joint Table for Numerical Outcomes
A hypothetical set of 6 inspection cards has recorded values 0 on 3 cards, 1 on 2 cards, and 2 on 1 card. Two cards are drawn in order without replacement. Let \(X\) and \(Y\) be the recorded values on the first and second cards. Construct the joint table, find the marginals, and calculate \(P(X+Y\ge2)\). The distribution is the finite-population distribution with frequencies \(n_0=3,n_1=2,n_2=1\) and \(N=6\); no interest or mortality assumption applies.
There are \(6\cdot5=30\) equally likely ordered pairs. The count in each cell is \(n_x(n_y-1)\) if \(x=y\), and \(n_xn_y\) otherwise. For example, the \((0,0)\) cell has \(3\cdot2=6\) pairs, while the \((2,2)\) cell has \(1\cdot0=0\) pairs.
| \(X\) / \(Y\) | \(Y=0\) | \(Y=1\) | \(Y=2\) | Row total |
|---|---|---|---|---|
| \(X=0\) | \(6/30\) | \(6/30\) | \(3/30\) | \(15/30\) |
| \(X=1\) | \(6/30\) | \(2/30\) | \(2/30\) | \(10/30\) |
| \(X=2\) | \(3/30\) | \(2/30\) | \(0/30\) | \(5/30\) |
| Column total | \(15/30\) | \(10/30\) | \(5/30\) | 1 |
The row totals give \(P(X=0)=15/30=1/2\), \(P(X=1)=10/30=1/3\), and \(P(X=2)=5/30=1/6\). The column totals give the same marginal probabilities for \(Y\).
The event \(X+Y\ge2\) contains the cells \((0,2),(1,1),(1,2),(2,0),(2,1),(2,2)\). Their counts are \(3,2,2,3,2,0\), respectively, so \(P(X+Y\ge2)=(3+2+2+3+2+0)/30=12/30=2/5\).
Method and check: Sum the joint cells meeting the event, rather than multiplying marginal probabilities. The table's numerators sum to \(15+10+5=30\), and the \((2,2)\) cell is zero because there is only one card with value 2. Treating the draws as independent would incorrectly assign that cell probability \((1/6)^2=1/36\).
Why the Counting Formula Holds
The denominator in the joint probability formula counts every possible ordered pair: there are \(N\) choices for the first member and \(N-1\) remaining choices for the second. Each such pair has the same probability under uniform sequential sampling. For distinct recorded values \(x\) and \(y\), any of the \(n_x\) members with value \(x\) can be paired with any of the \(n_y\) members with value \(y\), giving \(n_xn_y\) favorable pairs.
If the two recorded values are equal, the first member must still be different from the second. There are \(n_x\) choices for the first and \(n_x-1\) remaining members with that same value for the second. Thus the favorable-pair count is \(n_x(n_x-1)\). Dividing each count by the total \(N(N-1)\) proves the stated formula.
Record the frequency of each value in the finite population, including zero frequencies if useful for displaying the table.
Use \(n_xn_y\) for different values and \(n_x(n_x-1)\) for equal values, or count the members satisfying two different outcome rules directly.
For two draws without replacement, the total is \(N(N-1)\). The resulting cell probabilities form the joint table.
Sum across rows or columns for marginal probabilities; add every cell that belongs to the requested pair event.
Pair Events and Structural Zeros
A pair event can impose a relation between the two recorded values, such as \(X+Y\ge c\), \(X\ne Y\), or \(X\) belonging to one category while \(Y\) belongs to another. Once the joint table is available, identify the cells satisfying the relation and add their probabilities. The outcome order must match the definitions: a condition on the first draw is a condition on \(X\), and one on the second is a condition on \(Y\).
A structural zero is a cell that cannot occur under the model. For example, if only one population member has a particular value \(a\), then \(P(X=a,Y=a)=0\). Such a zero is not missing data and should not be replaced with a positive product of marginal probabilities.
Worked Example: A Sum Event with Unequal Frequencies
A hypothetical collection has 6 distinct tags with recorded values \(1,1,2,3,3,4\). Two tags are drawn in order without replacement. Let \(X\) and \(Y\) be the first and second values. Find the marginals and \(P(X+Y=5)\). This is a uniform ordered sample from frequencies \(n_1=2,n_2=1,n_3=2,n_4=1\), with \(N=6\); no interest or mortality assumption is involved.
There are \(6\cdot5=30\) ordered pairs. The row counts for \(X=1,2,3,4\) are \(2\cdot5=10\), \(1\cdot5=5\), \(2\cdot5=10\), and \(1\cdot5=5\). Dividing by 30 gives \(P(X=1)=1/3\), \(P(X=2)=1/6\), \(P(X=3)=1/3\), and \(P(X=4)=1/6\). The column totals give these same probabilities for \(Y\), since the same recorded values are available on either draw.
For \(X+Y=5\), the possible ordered value pairs are \((1,4),(2,3),(3,2),(4,1)\). Their ordered-pair counts are \(2\cdot1=2\), \(1\cdot2=2\), \(2\cdot1=2\), and \(1\cdot2=2\). Thus there are \(2+2+2+2=8\) favorable pairs, and \(P(X+Y=5)=8/30=4/15\).
Method and check: Count each ordered pair of values separately and divide by 30. The four cell counts sum to 8, which is less than the 30 total pairs, and the reduced probability \(4/15\) is between 0 and 1. Reversing the draw order does not change the event here, but both orders must still be counted because the sampling experiment is ordered.
Common Errors to Avoid
- Using \(N^2\) as the denominator. There are \(N(N-1)\) ordered pairs without replacement, not \(N^2\). The latter denominator corresponds to independent draws with replacement from the same finite set.
- Allowing a member to be drawn twice. For equal recorded values, use \(n_x(n_x-1)\), not \(n_x^2\). Multiple population members can share a value, but the same member cannot appear twice.
- Confusing values with population members. Two different members can have the same recorded value, so a diagonal cell may be possible when \(n_x\ge2\), and impossible when \(n_x=1\).
- Assuming the two draws are independent. Without replacement, the first selection changes the remaining population. Calculate joint cells from ordered-pair counts rather than multiplying marginals.
- Summing the wrong cells. For an event involving both outcomes, check the defining condition for each ordered pair \((x,y)\) before adding its probability.
Check Your Understanding
For each problem, treat the two draws as ordered and uniform, with no replacement. Give exact probabilities.
- A box contains 3 red, 2 blue, and 2 white counters. Let \(X=1\) if the first counter is red and \(Y=1\) if the second is white, with 0 otherwise. Find \(P(X=1,Y=1)\), \(P(Y=1)\), and \(P(X+Y=2)\). (Answer: \(1/7,\ 2/7,\ 1/7\), respectively.)
- A population of 6 cards has values \(0,0,1,1,1,3\). Let \(X,Y\) be the values drawn. Find \(P(X=1,Y=1)\), \(P(X+Y=2)\), and \(P(X=3)\). (Answer: \(1/5,\ 1/5,\ 1/6\), respectively.)
- A set of 8 files contains 4 tagged A, 3 tagged B, and 1 tagged C. Let \(X=1\) if the first file is A and \(Y=1\) if the second file is C, with 0 otherwise. Find \(P(X=1,Y=1)\), both marginal probabilities \(P(X=1)\) and \(P(Y=1)\), and \(P(X=1\text{ or }Y=1)\). (Answer: \(1/14,\ 1/2,\ 1/8,\ 31/56\), respectively.)
- A population has values \(1,1,2,2,2,4\). For the two recorded values \(X,Y\), find \(P(X+Y=5)\), \(P(X=2)\), and \(P(X=Y)\). (Answer: \(2/15,\ 1/2,\ 4/15\), respectively.)
- A collection has 10 distinct items: 4 tagged, 3 cleared, and 3 pending. Let \(X=1\) if the first item is tagged and \(Y=1\) if the second item is cleared, with 0 otherwise. Find the joint probability \(P(X=1,Y=1)\), the marginal \(P(Y=1)\), and the probability that at least one of \(X,Y\) equals 1. (Answer: \(2/15,\ 3/10,\ 17/30\), respectively.)