From One Table Entry to Several
In Probabilities of Single Values, you learned to find \(P(X=x)\) by reading the table entry for one exact value. Many questions ask about several values at once: “no more than 2,” “at most 2,” or “at least 3.” For a discrete random variable, find these probabilities by adding the entries for all values that satisfy the requested inequality.
The symbols show which side of the cutoff to include. \(P(X \leq x)\) asks for the probability that \(X\) is less than or equal to \(x\), so include every possible value up to and including \(x\). \(P(X \geq x)\) asks for the probability that \(X\) is greater than or equal to \(x\), so include \(x\) and every possible value above it.
The summation notation is a compact way to say “add the probabilities for every listed value that meets the condition.” You do not add the values of \(X\); you add their probabilities. If the distribution table lists values in increasing order, the entries for \(P(X \leq x)\) are at or before the cutoff, and the entries for \(P(X \geq x)\) are at or after it.
Words such as at most, no more than, and up to usually signal \(X \leq x\). Words such as at least and no fewer than usually signal \(X \geq x\). Translate the wording into a symbol before adding. The equality part of \(\leq\) or \(\geq\) is important: the value at the cutoff belongs in the sum.
A Reliable Method
Start by identifying what the random variable counts or measures and what cutoff the question gives. Then mark the table entries that meet the inequality, include the endpoint, add those probabilities, and write a sentence describing the event in context. This approach works whether the possible values are consecutive or have gaps.
Write \(X \leq x\) for “at most \(x\)” or \(X \geq x\) for “at least \(x\).”
Include every possible value satisfying the inequality, including the value \(x\) itself.
Add the selected probabilities and state what the resulting probability means in the situation.
A useful check is to ask whether each included value satisfies the inequality and whether the endpoint was included. The answer must be between 0 and 1. If you add every entry in a complete distribution table, the total is 1, as established in Checking Whether a Probability Distribution Is Valid. A cumulative probability uses only the entries on the requested side of the cutoff, not necessarily the whole table.
Worked Example: At Most Two Customer Callbacks
For an invented model, let \(C\) be the number of customers who call back about an online order during a specified afternoon. The probability distribution is:
| Callbacks, \(c\) | \(P(C=c)\) |
|---|---|
| 0 | 0.12 |
| 1 | 0.23 |
| 2 | 0.28 |
| 3 | 0.19 |
| 4 | 0.12 |
| 5 | 0.06 |
State. Find and interpret the probability that there are at most 2 customer callbacks during the afternoon.
Plan. “At most 2” means \(C \leq 2\). The values that satisfy this inequality are 0, 1, and 2. Because the inequality includes equality, the row for 2 callbacks must be included.
Do. Add the probabilities for \(C=0\), \(C=1\), and \(C=2\):
Conclude. According to this model, the probability of at most 2 customer callbacks during the specified afternoon is 0.63, or 63%.
The table entry for 2 callbacks is part of the event: “at most 2” includes exactly 2. The result describes the combined probability of 0, 1, or 2 callbacks, not the probability of exactly 2.
Include the Cutoff for “At Least”
For \(P(X \geq x)\), begin at the cutoff and move upward through the possible values. Do not begin with the value after the cutoff. For example, \(P(X \geq 2)\) includes the event \(X=2\) as well as any listed values greater than 2.
A table does not have to list every integer for this method to work. If a value is absent because it is not a possible outcome in the distribution, it contributes no probability. Select from the values that are actually listed, using the inequality to decide which rows to include.
Worked Example: At Least Two App Alerts
In an invented model, let \(A\) be the number of urgent app alerts received by a technician during one evening shift. The possible values and probabilities are:
| Urgent alerts, \(a\) | \(P(A=a)\) |
|---|---|
| 0 | 0.08 |
| 1 | 0.17 |
| 2 | 0.30 |
| 4 | 0.25 |
| 5 | 0.20 |
State. Find the probability that the technician receives at least 2 urgent alerts during the shift.
Plan. “At least 2” means \(A \geq 2\). The listed values satisfying this are 2, 4, and 5. Include 2 because it is the cutoff; do not include 0 or 1.
Do. Add the probabilities for those three values:
Conclude. According to the model, the probability that the technician receives at least 2 urgent alerts during one evening shift is 0.75, or 75%.
The table has no row for 3 alerts, so 3 is not a possible value in this model. The sum uses the listed values that meet \(A \geq 2\); it does not require filling in a row for every integer.
Running Totals Make Cumulative Probabilities Easier
When several cumulative probabilities are needed from the same distribution, a running total can organize the work. For each possible value in increasing order, add its probability to the total of all probabilities at smaller values. The result at a value \(x\) is \(P(X \leq x)\). These running totals cannot decrease as \(x\) increases, because each new total keeps the earlier entries and adds another nonnegative probability.
A running-total column is a calculation aid, not a replacement for checking the question’s direction. It gives the probabilities \(P(X \leq x)\) directly. To answer a question about \(P(X \geq x)\), add the entries from \(x\) upward, or use the relevant rows of the original table. In either case, make sure the cutoff row is included.
Worked Example: Cumulative Probabilities for Seedlings
For an invented greenhouse model, let \(S\) be the number of seeds, out of a small planting tray, that produce seedlings by a specified date. The distribution is shown with running totals:
| Seedlings, \(s\) | \(P(S=s)\) | \(P(S \leq s)\), running total |
|---|---|---|
| 0 | 0.04 | 0.04 |
| 1 | 0.11 | 0.15 |
| 2 | 0.26 | 0.41 |
| 3 | 0.31 | 0.72 |
| 4 | 0.18 | 0.90 |
| 5 | 0.10 | 1.00 |
The running totals are formed in order: \(0.04\), then \(0.04+0.11=0.15\), then \(0.15+0.26=0.41\), then \(0.41+0.31=0.72\), then \(0.72+0.18=0.90\), and finally \(0.90+0.10=1.00\). As a check, the individual probabilities also total \(0.04+0.11+0.26+0.31+0.18+0.10=1.00\).
State. Find and interpret the probability that at most 3 seeds produce seedlings.
Plan. “At most 3” means \(S \leq 3\). Include the probabilities for 0, 1, 2, and 3 seedlings. The running-total column at 3 gives the same sum.
Do. Add the entries through \(s=3\):
Conclude. According to the model, the probability that at most 3 seeds produce seedlings by the specified date is 0.72, or 72%.
The same table can answer an “at least” question. For at least 3 seedlings, include 3, 4, and 5. In particular, the probability at \(s=3\) belongs in this sum because “at least 3” includes exactly 3:
Thus, according to this model, the probability that at least 3 seeds produce seedlings by the specified date is 0.59, or 59%. The row for 3 is included in both \(P(S \leq 3)\) and \(P(S \geq 3)\); those two events overlap at \(S=3\), so their probabilities are not expected to add to 1.
Common Mistakes and AP Exam Tips
- Leaving out the endpoint. \(P(X \leq 2)\) includes \(X=2\), and \(P(X \geq 2)\) also includes \(X=2\). A full-credit calculation shows that endpoint’s probability in the sum.
- Adding values instead of probabilities. To find \(P(X \leq 2)\), add the probability entries for qualifying values, not the values \(0+1+2\).
- Reversing the direction. “At most” selects values below the cutoff; “at least” selects values above it. Write the inequality first to prevent choosing the wrong side of the table.
- Using only the endpoint row. \(P(X \leq x)\) and \(P(X \geq x)\) generally involve several values, not just \(P(X=x)\). Use the single-value entry as one part of the sum.
- Assuming values must be consecutive. Use the possible values actually shown. If the table skips a value, do not invent a probability for it.
- Giving a number without context. A complete interpretation names what \(X\) represents and describes the full event, such as “at least 2 alerts during one shift.”
For an AP response, state the inequality, show the selected probability entries and their sum, then interpret the result in context. Words such as “at most” and “at least” should be reflected accurately in both the notation and the concluding sentence.
Key Takeaway
A cumulative probability combines the probabilities of all possible values on one side of a cutoff. The inequality determines which entries to add, and its equality sign means the cutoff itself is included.
Check Your Understanding
Use this invented distribution for \(M\), the number of maintenance requests received by a small community center during one day.
| Requests, \(m\) | \(P(M=m)\) |
|---|---|
| 0 | 0.09 |
| 1 | 0.21 |
| 2 | 0.33 |
| 3 | 0.25 |
| 4 | 0.12 |
- Find \(P(M \leq 1)\). Show which table entries you add.
- Find \(P(M \geq 3)\) and interpret the result in context.
- In \(P(M \leq 2)\), should the row for 2 requests be included? Explain why.
- Find \(P(M \leq 2)\), showing the addition.
- Explain why \(P(M \leq 2)\) and \(P(M \geq 2)\) are not complementary events.