A Reliable Plan for Mixed Probability Questions
Probability questions often combine ideas that seem familiar on their own. A question may describe several stages, show counts in a two-way table, or use wording such as “at least one,” “or,” or “neither.” The challenge is to translate the wording into the right event and use a denominator that matches the chance process.
This practice builds on Sample Spaces and Outcomes, Events as Subsets of the Sample Space, The Complement Rule, and Using Two-Way Tables to Find Probabilities. Those tutorials established the tools; here we practice deciding which tool fits a question and explaining why the calculation answers it. As in The Gambler’s Fallacy and Independent Trials, keep the chance model in view: a calculation is justified only when its assumptions match the stated process.
A useful check is to say the event aloud before calculating. “A red tile appears” is not the same event as “both tiles are red.” “Bus or voucher” includes people in both categories, but “bus and voucher” includes only the overlap. Careful wording prevents many avoidable errors.
Worked Example: At Least One Red Tile
Worked Example: At Least One Red Tile
A bag contains five distinct tiles: two red tiles, labeled R1 and R2, and three blue tiles, labeled B1, B2, and B3. Two tiles are drawn at random without replacement. What is the probability that at least one tile is red?
State: Let \(A\) be the event that at least one of the two drawn tiles is red. The tiles are distinct, and the order of the two draws is recorded, so an outcome is an ordered pair of different tiles.
Plan: There are \(5\) choices for the first tile and \(4\) remaining choices for the second, giving \(5\times4=20\) equally likely ordered outcomes. Counting outcomes with at least one red directly is possible, but the complement is simpler: no red means both tiles are blue. We will count those outcomes and subtract their probability from 1.
Do: The sample space consists of all ordered pairs of different tiles, such as (R1, B2) and (B3, B1). For the complement event, both positions must contain blue tiles. There are \(3\) choices for the first blue tile and then \(2\) remaining choices for the second blue tile. Thus, there are \(3\times2=6\) outcomes with no red.
Conclude: The probability that at least one of the two tiles is red is \(7/10\), or \(0.70\). This calculation uses equally likely ordered outcomes and counts the complement, both-blue draws, without treating the second draw as if the first tile had been returned.
The phrase “without replacement” matters because it affects which outcomes are possible after the first draw. In this example, there are 20 possible ordered pairs, not 25: the same tile cannot appear twice. The labeled sample space makes that restriction visible. It also shows why counting outcomes, rather than just color patterns, is appropriate: some color patterns have more specific tile outcomes than others.
A quick independent check is to count event \(A\) directly. If the first tile is red, there are \(2\times4=8\) ordered outcomes; if the first is blue and the second red, there are \(3\times2=6\) more. That gives \(14\) favorable outcomes out of \(20\), agreeing with the complement calculation. A check like this can help catch a mistaken count, although on an exam you need not show two methods unless asked.
Worked Example: “Or” and “Neither” in a Two-Way Table
Worked Example: “Or” and “Neither” in a Two-Way Table
A community center records whether each of 120 visitors arrived by bus and whether each received a meal voucher. The invented counts are shown below. One visitor is selected at random from the 120. What is the probability that the visitor arrived by bus or received a voucher? Also find the probability that the visitor did neither.
| Arrival method | Voucher | No voucher | Total |
|---|---|---|---|
| Walk | 18 | 12 | 30 |
| Bus | 26 | 34 | 60 |
| Car | 14 | 16 | 30 |
| Total | 58 | 62 | 120 |
Define the events: Let \(B\) be the event that the selected visitor arrived by bus, and let \(V\) be the event that the visitor received a voucher. “Bus or voucher” means the visitor is in at least one of those categories, including someone who is in both.
Plan: The bus total is 60, the voucher total is 58, and the overlap—bus and voucher—is 26. Add the two totals, then subtract the overlap once so visitors in both categories are not counted twice. For “neither,” count people who are not in either category, or use the complement of “bus or voucher.”
Do: There are \(60+58-26=92\) visitors who arrived by bus or received a voucher. Divide by 120 because one visitor is selected from the entire group in the table.
For neither event, the visitor must have arrived by a method other than bus and must not have received a voucher. The relevant table cells are walk/no voucher, 12, and car/no voucher, 16. Their total is 28.
Conclude: The probability that the selected visitor arrived by bus or received a voucher is about \(0.7667\). The probability that the visitor did neither is about \(0.2333\). These probabilities sum to 1, as expected for an event and its complement.
The table gives several possible denominators, so the question’s selection statement matters. Because the visitor is selected from all 120 visitors, use the grand total. A row total would be appropriate only for a question that restricts attention to that row; a column total would be appropriate only for a question that restricts attention to that column. Do not change the denominator just because one particular row or column seems relevant.
This example also illustrates why “or” deserves attention. The overlap count of 26 belongs to both the bus total and the voucher total. Adding 60 and 58 without subtracting 26 would count those visitors twice and produce an incorrect favorable count of 118. For “neither,” however, there is no need to apply that union calculation: the two cells outside both event categories can be counted directly.
Worked Example: A Sample Space and Its Complement
Worked Example: A Sample Space and Its Complement
A fair six-sided number cube is rolled once, and a fair coin is flipped once. Assume the two results are independent. What is the probability that the number is even or the coin lands tails?
Define the event: Let \(E\) be the event that the number is even, and let \(T\) be the event that the coin lands tails. We want \(P(E\cup T)\), the probability that at least one of those statements is true.
Plan: The complete outcome records both results, such as (2, H) or (5, T). The number cube has 6 possible results and the coin has 2, so the multiplication principle gives \(6\times2=12\) outcomes. Under the stated fair and independent model, these 12 complete outcomes are equally likely. The complement of “even or tails” is “odd and heads,” which is straightforward to count.
Do: The outcomes in the complement are (1, H), (3, H), and (5, H). There are 3 such outcomes out of 12, so the probability of the complement is \(3/12\). Subtract that probability from 1.
Conclude: The probability that the number is even or the coin lands tails is \(3/4\), or \(0.75\). The complement count is valid because every complete outcome in this sample space has the same probability.
A second count checks the result: there are 6 outcomes with an even number, and 6 outcomes with tails. Three outcomes have both an even number and tails, so the number in the union is \(6+6-3=9\). Thus, \(9/12=3/4\), matching the complement calculation. This check uses the same overlap principle discussed in Choosing Between Counting and Formula Approaches.
The assumption of equally likely complete outcomes is important. If the coin were biased or the number cube did not follow the stated fair model, counting the 12 outcomes equally would not be justified. In an AP response, name the model assumption that supports the count rather than relying on the sample-space list alone.
Common Mistakes and AP Exam Tips
- Using a denominator that does not match the selection. If one person is chosen from the entire table, divide by the grand total. If the problem restricts the selection to a subgroup, identify that subgroup before choosing a denominator.
- Counting a multi-stage outcome incompletely. For a roll followed by a flip, one outcome must include both results. Listing only the number or only the coin result does not describe a complete outcome for the process.
- Assuming every listed outcome is equally likely without support. The equally likely cases method needs a chance model that gives equal probability to each complete outcome. State the fair, independent model when it is part of the question.
- Misreading “or.” In ordinary probability questions, “A or B” includes outcomes in both A and B. When adding table totals, subtract the overlap once; otherwise the overlap is counted twice.
- Using a complement without describing it. Name the event being subtracted from 1. For example, “no red” means both draws are blue, while the complement of “even or tails” is odd and heads.
- Giving a number without a justification. A full-credit explanation connects the count to the event and the model: identify the favorable outcomes or table cells, give the appropriate total, show the probability calculation, and state what the result means in context.
A concise written justification can still be complete. For the tile question, for instance, it is enough to say that the 20 ordered pairs are equally likely, 6 pairs contain two blue tiles, and therefore the probability of at least one red tile is \(1-6/20=0.70\). That sentence explains the sample space, complement, denominator, and result.
Check Your Understanding
For each question, identify the event and give enough work to justify the probability.
- A bag has four distinct green tokens and two distinct yellow tokens. Two are drawn without replacement. What is the probability that both are green? Describe the sample-space denominator you use.
- In a group of 80 people, 30 use a library app, 22 attend a book club, and 10 do both. What is the probability that a randomly selected person uses the app or attends the book club?
- Using the group in question 2, find the probability that a randomly selected person does neither activity. Show how you check your answer against the probability in question 2.
- A fair number cube is rolled and a fair coin is flipped. What is the probability of getting an odd number or heads? State the complement event and count its outcomes.
- In a two-way table, when should you divide a cell count by the grand total rather than a row or column total? Explain using the wording of a selection question.