Find “A or B” by Counting Table Cells
In Finding \(P(A\text{ and }B)\) from the Addition Rule, you used the general addition rule in reverse to find an overlap. Here, a two-way table gives us the counts directly. We will use the table to find the probability that a randomly selected individual is in event A, event B, or both, and then check the result with the addition rule.
The phrase “A or B” means that at least one of the events occurs. It includes individuals in A only, B only, and both A and B. In table terms, the union \(A\cup B\) includes every cell that meets at least one event description. If a cell belongs to both events, it is still counted only once in the union.
A two-way table organizes counts according to two variables. Interior cells represent combinations of categories; row and column totals are marginal counts; and the grand total is the number of individuals represented. As in Using Two-Way Tables to Find Probabilities, when one individual is selected at random from the full group in the table, divide the relevant count by the grand total.
There are two useful ways to count a union from a table. You can add the counts for A only, B only, and both. Or, you can add the marginal counts for A and B and subtract the overlap once. The second method mirrors the general addition rule.
Here, \(n(A\cup B)\) counts individuals in at least one event, and \(n(A\cap B)\) counts individuals in both. The subtraction ensures the overlap is counted only once.
The two approaches should agree when the counts and event definitions are used correctly. That agreement is a helpful check: first count the union in the table, then calculate the probability with the addition rule using the table’s marginal and overlap counts. A disagreement usually signals that a cell was missed, counted twice, or assigned to the wrong event.
A Table-Based Strategy
Before calculating, identify exactly which categories define A and B. Then locate the corresponding row or column totals and the cell where the categories overlap. Do not assume that “or” means one event but not the other: the union includes the overlap too.
Write what A and B mean in the context, and identify the table categories that represent each event.
Add the counts in A only, B only, and both; alternatively, add the two marginal counts and subtract the overlap once.
Divide the union count by the grand total for the full group represented in the table.
Use the table to find \(P(A)\), \(P(B)\), and \(P(A\cap B)\). Confirm that the rule gives the same probability, then state what it means in context.
This strategy assumes that the table describes the group from which the individual is being selected and that the row and column categories are clearly defined. If the question says the person is selected from a particular row or column rather than from everyone represented, the denominator would be that specified group total instead of the grand total. For the examples below, each selection is from the full table population.
Worked Examples
Worked Example: Bus Riders and Refillable Bottles
An invented table summarizes 200 transit-pass holders. Let A be the event that a randomly selected person rides the bus to work, and B the event that the person carries a refillable bottle. Find the probability that a person rides the bus or carries a refillable bottle.
| Carries bottle | Does not carry bottle | Total | |
|---|---|---|---|
| Rides bus | 41 | 37 | 78 |
| Does not ride bus | 51 | 71 | 122 |
| Total | 92 | 108 | 200 |
State: We want \(P(A\cup B)\), the probability that the selected person meets at least one of the two descriptions.
Plan: The union includes bus riders who carry a bottle, bus riders who do not carry one, and non-bus riders who carry one. These are disjoint table cells, so add their counts. Then divide by the grand total. As a check, use the general addition rule with the two marginal totals and the overlap.
Do: Count each qualifying cell once:
There are 200 people in the table, so the probability is:
Check using the addition rule. The bus-rider total is 78, the refillable-bottle total is 92, and the overlap is 41. Thus, \(P(A)=78/200=0.39\), \(P(B)=92/200=0.46\), and \(P(A\cap B)=41/200=0.205\). Substitution gives:
The check agrees with the direct count. The arithmetic can also be checked in counts: \(78+92-41=129\), matching the three-cell total. Conclude: The probability that a randomly selected transit-pass holder rides the bus to work or carries a refillable bottle is \(0.645\), or 64.5%.
The overlap cell is included in both marginal totals, which is why simply adding 78 and 92 would give 170, not the union count. That sum counts the 41 people who meet both descriptions twice. Subtracting the overlap once corrects the double count.
Worked Example: Two Phone Unlock Features
An invented technology survey records whether 160 participants use fingerprint unlock, face unlock, both, or neither on their phones. Let F mean that a participant uses fingerprint unlock, and let G mean that the participant uses face unlock. Find \(P(F\cup G)\).
| Uses face unlock | Does not use face unlock | Total | |
|---|---|---|---|
| Uses fingerprint unlock | 48 | 36 | 84 |
| Does not use fingerprint unlock | 44 | 32 | 76 |
| Total | 92 | 68 | 160 |
The union includes the three cells other than “uses neither.” Count those cells and divide by 160:
Now verify with the general addition rule. The marginal counts are 84 for fingerprint unlock and 92 for face unlock, while 48 participants use both. Therefore, \(P(F)=84/160=0.525\), \(P(G)=92/160=0.575\), and \(P(F\cap G)=48/160=0.300\). Then:
This matches the direct count. The probability that a randomly selected participant uses fingerprint unlock, face unlock, or both is \(0.80\), or 80%.
The cell that represents “both” is not outside the union. A common mistake is to count only the two “only” cells and leave out the overlap. Another is to add the two event totals without correcting for the overlap. Either mistake changes the result, so identify the three included regions before calculating.
Worked Example: Walking or Biking to Work
An invented workplace table classifies 240 employees by their usual way of commuting and whether they attended a safety session. Let A be the event that an employee usually walks to work, and B the event that the employee usually bikes to work. Find \(P(A\cup B)\).
| Usual commute | Attended session | Did not attend | Total |
|---|---|---|---|
| Walks | 28 | 20 | 48 |
| Bikes | 36 | 44 | 80 |
| Drives | 56 | 56 | 112 |
| Total | 120 | 120 | 240 |
An employee’s usual commute is recorded as one category, so no one can be counted as both a walker and a biker in this table. The two event counts do not overlap. Add the walking and biking totals:
The formula check gives the same result. \(P(A)=48/240=0.20\), \(P(B)=80/240=1/3\), and \(P(A\cap B)=0\), because the categories cannot occur together for one employee in this table. Thus:
The probability that a randomly selected employee usually walks or bikes to work is about 0.5333, or 53.33%. This example also shows why the addition rule’s overlap term matters: when the events have no overlap, there is nothing to subtract.
Common Mistakes and AP Exam Tips
- Leaving out the overlap. “A or B” includes people in both A and B. Count the overlap once as part of the union.
- Double-counting the overlap. If you add the A and B marginal totals, subtract the count in both. Otherwise, individuals in the overlap are counted twice.
- Using only a cell when the event is broader. An event defined by one row category may include more than one interior cell. Add all cells that satisfy the event description or use its marginal total.
- Using the wrong denominator. For selection from everyone represented in the table, use the grand total. Use a row or column total only when the question specifies selection from that subgroup.
- Assuming categories overlap or do not overlap without checking. Read the event definitions and table carefully. Some events can occur together; a single-choice category such as the employee’s usual commute cannot place one person in two categories at once.
- Giving a probability without context. A complete answer identifies the randomly selected individual and describes which “or” event the probability represents.
For full credit, show how the relevant counts were selected, state the denominator, and give an interpretation in context. When using the addition rule as a check, identify the overlap instead of merely writing a formula with unexplained numbers. The check should be an independent confirmation of the union count.
Check Your Understanding
Use the table counts and the general addition rule to support your answers.
- A table has 150 people. The count in A only is 32, in B only is 41, in both is 27, and in neither is 50. Find \(P(A\cup B)\) by counting the union, then check with the addition rule.
- In a table of 200 customers, 86 use a digital receipt, 74 use a loyalty app, and 38 use both. Find the probability a randomly selected customer uses a digital receipt or a loyalty app.
- Explain why adding the row total for A and the column total for B may count some individuals twice. What table count corrects this?
- A table describes all 120 members of a club. If 45 members are in A only, 30 in B only, and 15 in both, find the probability that a randomly selected member is in neither event.
- For a random selection from the full table, when should you divide a union count by the grand total rather than by a row total?