Small Words Change the Event
In Cumulative Probabilities for Discrete Variables, you learned to add probabilities for values at or below or above a cutoff. This tutorial focuses on a distinction that can change which table entries belong in the sum: more than 2 does not mean the same thing as at least 2. The first excludes 2; the second includes it.
For a discrete random variable, translate the wording into an inequality before selecting table entries. The symbol \(>\) means strictly greater than, so the cutoff is excluded. The symbol \(\geq\) means greater than or equal to, so the cutoff is included. Similarly, \(<\) excludes its cutoff, while \(\leq\) includes it.
Common wording has predictable translations. “More than \(a\)” means \(X>a\); “at least \(a\)” means \(X\geq a\). “Less than \(b\)” means \(X<b\); “at most \(b\)” means \(X\leq b\). Once the event is written, use the distribution table to select every possible value that satisfies it and add those probabilities.
A range has two endpoints, and each endpoint needs its own decision about inclusion. For example, \(a<X\leq b\) excludes \(a\) and includes \(b\). The event \(a\leq X<b\) does the opposite. Do not assume that “between” always tells you whether the endpoints count. If a question uses “between” without clarifying inclusivity, use any definitions supplied in the context; otherwise, recognize that the wording may need clarification.
Translate First, Then Select Values
A dependable approach is to write the event in symbols, then check the possible values one by one. This is especially useful when a table has gaps: the inequality describes which values qualify, but only possible values from the distribution can contribute probability. You do not need to invent entries for values that cannot occur.
Decide whether each boundary is strict or includes equality.
For a range, write both comparisons, such as \(2<X\leq 5\).
Check each value in the distribution against every part of the inequality.
Add the selected probabilities and describe the event in context.
If two events differ only in whether they include a cutoff, their probabilities differ by the probability at that cutoff, provided the cutoff is a possible value. For instance, \(X>2\) leaves out \(X=2\), while \(X\geq 2\) includes it. That gives a useful check on your selection and arithmetic.
Worked Example: More Than Two Versus At Least Two
For an invented model, let \(D\) be the number of delayed packages arriving at a neighborhood pickup point during one morning. The distribution is:
| Delayed packages, \(d\) | \(P(D=d)\) |
|---|---|
| 0 | 0.11 |
| 1 | 0.21 |
| 2 | 0.29 |
| 3 | 0.20 |
| 4 | 0.13 |
| 5 | 0.06 |
State. Find and interpret (a) the probability of more than 2 delayed packages and (b) the probability of at least 2 delayed packages during the morning.
Plan. “More than 2” means \(D>2\), so the possible values 3, 4, and 5 qualify. “At least 2” means \(D\geq 2\), so 2, 3, 4, and 5 qualify. The difference is whether exactly 2 packages are included.
Do. Add the probabilities for each event:
Conclude. According to the model, the probability of more than 2 delayed packages during the morning is 0.39. The probability of at least 2 is 0.68.
The second probability is 0.29 greater than the first, exactly the probability of \(D=2\). This verifies that “at least 2” includes the cutoff, while “more than 2” does not.
Ranges Require Two Endpoint Decisions
A question may specify a lower and an upper boundary. Translate the complete phrase rather than focusing only on one side. “More than 2 but no more than 7” becomes \(2<X\leq 7\): values must be above 2 and at or below 7. Both parts must be true for a value to be included.
A value exactly at an excluded endpoint does not qualify, even if it satisfies the other comparison. Likewise, a value at an included endpoint qualifies if it also meets the other boundary. This both-boundaries check prevents accidentally including a value that falls outside the range.
Worked Example: A Range with Gaps in the Possible Values
In an invented model, let \(R\) be the number of equipment repairs requested by a small theater during a week. Only the values shown are possible in this model:
| Repair requests, \(r\) | \(P(R=r)\) |
|---|---|
| 0 | 0.12 |
| 1 | 0.16 |
| 2 | 0.22 |
| 4 | 0.31 |
| 7 | 0.19 |
State. Find and interpret the probability of more than 2 but no more than 7 repair requests during the week.
Plan. “More than 2” excludes 2 and means \(R>2\). “No more than 7” includes 7 and means \(R\leq 7\). Together, the event is \(2<R\leq 7\). Among the listed possible values, only 4 and 7 satisfy both parts.
Do. Add the probabilities for 4 and 7 requests:
Conclude. According to the model, the probability of more than 2 but no more than 7 repair requests during the week is 0.50, or 50%.
The table skips 3, 5, and 6 because those are not possible values in this distribution; they contribute no entries to the sum. The value 2 is excluded by the strict lower boundary, and 7 is included by the upper boundary with equality.
As a comparison, “at least 2 but less than 7” would translate to \(2\leq R<7\). It includes 2 and 4, but excludes 7. Its probability is \(0.22+0.31=0.53\). The words “at least” and “less than” make the endpoint choices different from the first question.
Integer Values and Noninteger Cutoffs
Counts are integer-valued, so a noninteger cutoff can sometimes be translated into the nearest qualifying integer values. For example, if \(X\) counts items, the event \(X>1.6\) includes integer values 2 and higher; it cannot include a value between 1.6 and 2 if the model allows only whole-number counts.
This observation can simplify selecting table entries, but it does not change the original meaning of the inequality. Write the inequality as stated first, then use the variable’s possible values to decide which rows qualify. Do not round a cutoff before checking whether doing so would change the event.
Worked Example: Noninteger Boundaries for a Count
For an invented model, let \(V\) be the number of video clips a student reviews during a scheduled study session. The distribution is:
| Clips reviewed, \(v\) | \(P(V=v)\) |
|---|---|
| 0 | 0.08 |
| 1 | 0.17 |
| 2 | 0.25 |
| 3 | 0.28 |
| 4 | 0.15 |
| 5 | 0.07 |
State. Find the probability that the student reviews more than 1.6 but at most 4.4 clips during the session.
Plan. “More than 1.6” means \(V>1.6\), and “at most 4.4” means \(V\leq 4.4\). Since \(V\) counts whole clips, the possible values satisfying both inequalities are 2, 3, and 4. Values 1 and 5 do not qualify.
Do. Add the probabilities for 2, 3, and 4 clips:
Conclude. According to the model, the probability that the student reviews more than 1.6 but at most 4.4 clips during the session is 0.68.
Although a count cannot equal either decimal cutoff, the inequality signs still determine the event. Checking the possible integer values gives the same selection as recognizing that the qualifying counts are 2 through 4, inclusive.
Common Mistakes and AP Exam Tips
- Treating “more than” as “at least.” More than 2 is \(X>2\), not \(X\geq 2\). A full-credit answer excludes the row for 2.
- Forgetting that “at least” includes equality. At least 2 is \(X\geq 2\), so include \(P(X=2)\) if 2 is possible.
- Checking only one endpoint in a range. For \(2<X\leq 7\), a value must be greater than 2 and no greater than 7. State both comparisons and check both.
- Assuming “between” specifies endpoint rules. “Between 2 and 7” may be used differently in different contexts. Look for wording that says whether endpoints are included.
- Rounding a noninteger cutoff too early. Keep the stated inequality, then check the actual possible values of the variable. This avoids changing a strict comparison into an inclusive one.
- Reporting only a sum without identifying the event. Show the inequality, the qualifying values or probability entries, and a sentence interpreting the result in context.
For a clear AP response, translate the wording before calculating. In a range, show both endpoint symbols; then identify the table entries that satisfy the complete inequality. Add those probabilities and interpret the result using the variable and situation.
Key Takeaway
The words at a cutoff determine whether the endpoint belongs to the event. Apply the same careful translation at both ends of a range, then add probabilities only for possible values that meet every part of the inequality.
Check Your Understanding
Use this invented distribution for \(N\), the number of new messages received by a volunteer coordinator during one afternoon.
| Messages, \(n\) | \(P(N=n)\) |
|---|---|
| 0 | 0.10 |
| 1 | 0.18 |
| 2 | 0.27 |
| 3 | 0.25 |
| 4 | 0.14 |
| 5 | 0.06 |
- Translate “more than 3 messages” into an inequality and find its probability.
- Translate “at least 3 messages” into an inequality and find its probability. How does this probability differ from your answer to question 1?
- Find the probability of \(1\leq N<4\). Which endpoints are included?
- For a count \(N\), which possible values satisfy \(1.4<N\leq 4.2\)?
- Explain why \(N>2\) and \(N\geq 2\) describe different events when 2 is a possible value.