Turning a Word Problem into \(n\), \(p\), and \(X\)
In Recognizing a Binomial Setting and the BINS Checklist for Binomial Conditions, you learned to identify a trial, label its two outcomes success and failure, and check the conditions for a binomial setting. After those checks, the next task is to name the model’s ingredients clearly. For a binomial setting, those ingredients are the fixed number of trials \(n\), the probability of success on each trial \(p\), and the random variable \(X\), which counts the successes.
This translation matters because a problem may use “success” in a way that does not mean a desirable result. Success is simply the outcome being counted. If the question asks how many products are defective, a defective product can be labeled a success. If the question asks how many tests are negative, a negative test can be labeled a success. The labels help organize the model; they do not judge the outcome.
Keep the roles separate. The number \(n\) describes how many opportunities there are for success. The number \(p\) describes the chance of success on one opportunity. The variable \(X\) describes the total number of successes across the trials. In particular, \(p\) is not the probability of getting a particular total number of successes, and \(X\) is not the name of one trial.
A Reliable Translation Routine
Read the problem once for the story, then read it again looking for the trial, the counted outcome, the number of repetitions, and the per-trial chance. A small setup table can keep the three pieces distinct.
| Question to ask | What to record |
|---|---|
| What is one trial, and which outcome will be called success? | Write a short definition of success. |
| How many trials are planned in advance? | Write that fixed count as \(n\). |
| What is the chance of success on one trial? | Write that probability as \(p\), as a decimal or fraction. |
| What count does the question describe? | Define \(X\) in words, including the event counted and the trial set. |
The order is useful: decide what counts as success before copying a probability. The problem might give the chance of the opposite outcome. If so, either define success as the outcome whose probability is given or use the complement to find the probability of the event you want to count. Earlier in this course, Using the Complement with Random Variables explained how complementary outcomes relate; here, the key is to make the definition and the chosen \(p\) agree.
Finally, use the BINS reasoning from earlier tutorials to confirm that a binomial model is appropriate. This tutorial focuses on naming its ingredients, not calculating probabilities for possible values of \(X\). The next tutorial introduces the binomial probability formula.
Worked Example: Seeds That Germinate
Worked Example: Seeds That Germinate
A greenhouse worker plants 18 seeds of the same variety. Assume the seeds’ germination outcomes are independent, and each seed has probability 0.84 of germinating. Let \(X\) be the number of seeds that germinate. Identify \(n\), \(p\), and \(X\).
State. One trial is planting and observing one seed. Define success as that seed germinating. The question asks for the count of germinating seeds.
Plan. Use the stated number of seeds for \(n\), the per-seed chance of germination for \(p\), and define the random variable as the count of successes among those seeds. The assumptions establish independent trials with a common success probability; each trial has the two relevant outcomes, germinate or not germinate.
Do. There are 18 seeds, so \(n=18\). The probability of success on one trial is \(p=0.84\). Define \(X\) as the number of the 18 planted seeds that germinate. Thus \(X\) counts successes, and it can take any whole-number value from 0 to 18.
Conclude. In this setting, \(n=18\), \(p=0.84\), and \(X\) is the number of the 18 seeds that germinate. The probability \(0.84\) belongs to one seed at a time; it does not say that exactly 84% of these 18 seeds will germinate.
That last distinction is important. A probability describes the chance of success on an individual trial under the model. The realized count \(X\) is the number of successes in the whole group. The count need not equal the number of trials multiplied by the per-trial probability.
Worked Example: Counting Defective Items
Worked Example: Counting Defective Items
A factory checks 12 independently produced sensors. The quality-control model assigns each sensor a probability of 0.025 of being defective. Let \(D\) be the number of defective sensors among the 12. Identify \(n\), \(p\), and the random variable. Here, “success” will mean “defective.”
State. One trial is checking one sensor. The event being counted is that the sensor is defective, so label a defective sensor as a success.
Plan. The fixed number of sensors gives \(n\). Because success is defined as defective, use the probability of a defect—not the probability of a nondefect—for \(p\). Define \(D\) as the count of defective sensors in the batch.
Do. Exactly 12 sensors are checked, so \(n=12\). A single sensor has probability \(p=0.025\) of being defective. Define \(D\) as the number of defective sensors among the 12 checked. The assumptions in the model supply independence and the same defect probability for every sensor.
Conclude. For this count, \(n=12\), \(p=0.025\), and \(D\) is the number of defective sensors among the 12 checked. Calling a defect a success is appropriate because defects are the outcomes the random variable counts.
A different question about the same batch could use a different success definition. If the question asked for the number of sensors that are not defective, success could mean “not defective,” and the corresponding per-trial probability would be \(1-0.025=0.975\). The count would then be defined as the number of nondefective sensors among the 12. Both descriptions are valid when the variable, success definition, and probability all match the question.
Worked Example: When the Given Chance Is for the Opposite Outcome
Worked Example: When the Given Chance Is for the Opposite Outcome
A basketball player takes 10 free throws. Assume the shots are independent and the probability of making each shot is 0.72. Let \(M\) be the number of missed shots. Identify \(n\), \(p\), and \(M\).
State. One trial is one free throw. Since \(M\) counts misses, define success for this model as missing a shot, even though making a shot is usually the desired outcome.
Plan. There are 10 planned trials, so use \(n=10\). The problem gives the probability of making a shot, but the success event for \(M\) is missing. Use the complementary probability for \(p\), then define the variable as a count of misses.
Do. The fixed number of shots is \(n=10\). The probability of missing one shot is $$ p=1-0.72=0.28. $$ Define \(M\) as the number of missed shots in the 10 free throws. Under the stated assumptions, each shot has the same miss probability and the results are independent.
Conclude. For the random variable \(M\), \(n=10\), \(p=0.28\), and \(M\) counts the missed shots among the 10 attempts. Using \(p=0.72\) for this particular definition would mismatch the probability with the success event.
The same process can be modeled with a different random variable. If \(K\) counts made shots, then success would mean making a shot, \(n=10\), and \(p=0.72\). The problem’s target count determines which outcome is called success. This choice changes the definition of the variable and the matching value of \(p\), but it does not change the number of trials.
Worked Example: A Percentage as a Per-Trial Probability
Worked Example: A Percentage as a Per-Trial Probability
A community center sends the same reminder message to 15 families. A planning model assumes each family independently has a 40% chance of opening the message. Let \(O\) be the number of families who open it. Identify \(n\), \(p\), and \(O\).
State. One trial is sending the reminder to one family and recording whether it is opened. Define success as the family opening the message.
Plan. Translate the percentage into a probability, identify the fixed number of families, and define the random variable as the count of families with the success outcome. Keep the probability attached to one family rather than to all 15 together.
Do. The number of trials is \(n=15\). The stated per-family probability is \(p=40/100=0.40\). Define \(O\) as the number of the 15 families who open the reminder message. The problem states independence and a common probability for the families, so those assumptions support a binomial model.
Conclude. Here \(n=15\), \(p=0.40\), and \(O\) is the number of the 15 families who open the message. The value \(0.40\) describes the chance for one family; it is not the number of families expected to open the message.
Notice how the words “each family” identify the per-trial probability. If a prompt instead gives a total count or a percentage from a particular observed group, do not automatically treat that number as the model’s \(p\). First determine whether the context states a probability for each trial and whether it is reasonable to use the same probability for all trials.
Common Mistakes and AP Exam Tip
A complete setup should make it possible for another reader to identify the trial, the counted outcome, and the roles of \(n\) and \(p\) without guessing. State the random variable in context rather than writing only a letter. Then check that the probability you wrote belongs to the success event in your definition.
- Calling \(p\) the probability of the whole count. \(p\) is the chance of success on one trial. The probability of a particular total for \(X\) is a different quantity.
- Using “success” to mean “a good result.” Success is the outcome the variable counts. A defect or a missed shot can be success if that is what the question asks you to count.
- Copying the wrong probability. If the variable counts misses but the prompt gives the chance of making a shot, use the complementary probability for the miss event.
- Writing \(X\) without defining it. “Let \(X\) be the number of successes” is incomplete unless success and the group of trials are clear. Name both in context.
- Confusing the trial count and the success count. \(n\) is fixed by how many trials are performed; \(X\) is the random count of successes within them. They need not be equal.
- Ignoring whether the trials support a binomial model. Naming \(n\), \(p\), and \(X\) does not replace checking BINS. Cite the relevant evidence for the conditions, as in the earlier tutorials.
A concise, well-communicated setup might read: “Let one trial be checking one sensor, with a defective sensor defined as a success. The 12 sensors are the fixed trials, so \(n=12\), and the probability of success on each trial is \(p=0.025\). Let \(D\) be the number of defective sensors among the 12.” This connects every symbol to the context and makes the chosen success outcome explicit.
Key Takeaway
Translate the wording before doing any binomial calculations. Identify one trial, decide which outcome the random variable will count, find the probability of that exact outcome on a single trial, and record the fixed number of trials. Then verify the BINS conditions using the reasoning from the earlier tutorials.
Check Your Understanding
For each situation, identify the trial, define success to match the variable, and state the appropriate \(n\), \(p\), and random variable.
- A clinic processes 20 independent samples. Each sample has probability 0.06 of needing a repeat test. Let \(R\) count samples needing a repeat. What are \(n\), \(p\), and \(R\)?
- A delivery service makes 8 independent deliveries, and each delivery has probability 0.91 of arriving on time. Let \(L\) count late deliveries. What probability should be used for \(p\), and why?
- A student answers 12 independent true-or-false questions by guessing. Let \(C\) count correct answers. Identify the trial and state \(n\), \(p\), and \(C\).
- A store sends coupons to 25 customers. The model gives each customer a 30% chance of using a coupon. Let \(U\) count customers who do not use a coupon. State the matching success event and value of \(p\).
- In your own words, explain the difference between \(p\) and \(X\) in one of the situations above.