Two Checks for a Probability Distribution
In Probability Distribution of a Discrete Random Variable, you learned that a distribution pairs each possible value of a discrete random variable with the probability of that value. A candidate table may look organized and still fail to describe a probability distribution. Before using it to answer questions, check whether its probabilities obey the basic rules of probability.
There are two checks. First, every individual probability must be between 0 and 1, inclusive. Second, the probabilities for all possible values must add to 1. Both checks are necessary: a table that passes just one is not a valid probability distribution.
- Each probability \(P(X=x)\) satisfies \(0\leq P(X=x)\leq1\).
- The probabilities for all possible values of \(X\) sum to 1.
The first condition rules out probabilities below 0 or above 1. A probability of 0 is allowed: it says the model assigns no chance to that value. A probability of 1 is also allowed: it says the model assigns certainty to that value. The sum condition says that the listed values account for the entire chance process, so one of them must occur.
These checks apply to a distribution table for a discrete random variable. As in the earlier tutorial, the table should identify the values of \(X\) and the probability attached to each value. This tutorial focuses on auditing the probabilities; it does not ask you to calculate a missing entry.
A Consistent Way to Audit a Table
Check each row before adding the column. A quick scan can identify a probability that is negative or greater than 1. Then add all the probabilities, including entries equal to 0. Keep the individual-value check separate from the total check: the sum alone cannot tell you whether every entry is allowed.
Verify that each probability is at least 0 and at most 1.
Add the probabilities for all listed values and verify that the sum is 1.
The table is a valid probability distribution only if it passes both checks.
When working with decimals, show the addition so the reader can see how you reached the total. If a total is not 1, a difference of even a small amount means the displayed values do not sum to 1 as written. If entries have been rounded, say so: the underlying, unrounded values may total 1 even when the rounded display does not. Do not silently adjust a value to make the total work.
Worked Example: A Table That Passes Both Checks
A model describes \(X\), the number of service requests received by a small help desk during a chosen hour. Check whether the candidate table is a valid probability distribution.
| \(x\) | \(P(X=x)\) |
|---|---|
| 0 | 0.18 |
| 1 | 0.42 |
| 2 | 0.30 |
| 3 | 0.10 |
State. The question is whether the probabilities assigned to 0, 1, 2, and 3 service requests form a valid distribution.
Plan. Check each probability against the interval from 0 to 1, inclusive. Then add all four probabilities and compare the total with 1.
Do. Every entry—0.18, 0.42, 0.30, and 0.10—is at least 0 and at most 1. The total is \(0.18+0.42+0.30+0.10=1.00\).
Conclude. The table is a valid probability distribution: no probability is outside the allowed range, and the probabilities sum to 1.00.
Passing the Range Check Is Not Enough
A common trap is to see that every entry is a reasonable-looking decimal and conclude that the table must be valid. That checks only the first condition. If the entries total less than 1, some probability is unaccounted for; if they total more than 1, the table assigns more than the full probability of the chance process.
For example, entries of 0.25, 0.35, and 0.45 are individually between 0 and 1. But their total is \(0.25+0.35+0.45=1.05\), not 1. The entries cannot all be probabilities for the complete set of outcomes in one distribution as displayed.
Worked Example: Every Entry Is in Range, but the Total Is Too Large
A quality-control model lists \(X\), the number of blemishes on a randomly selected package, with this candidate distribution. Decide whether the table is valid.
| \(x\) | \(P(X=x)\) |
|---|---|
| 0 | 0.25 |
| 1 | 0.35 |
| 2 | 0.45 |
State. Check whether the three entries can describe the probabilities of the listed values of \(X\).
Plan. First verify the range of each entry. Then calculate their sum; passing the range check by itself is not sufficient.
Do. Each entry is between 0 and 1. However, \(0.25+0.35+0.45=1.05\).
Conclude. The table is not a valid probability distribution as written. Although every individual entry is in range, the total is 1.05 rather than 1. The table assigns more than the full probability of the chance process.
The diagnosis matters. In this example, saying “the probabilities are invalid because they are too large” would be inaccurate: no single entry exceeds 1. The problem is the total. A precise explanation names the check that fails.
A Total of 1 Does Not Rescue an Out-of-Range Entry
The reverse trap is also possible. A set of numbers can add to 1 while still containing an impossible probability. For instance, a negative entry may be offset by a larger positive entry in the sum. The total check cannot replace the check of every individual probability.
Worked Example: The Entries Sum to 1, but One Is Negative
A model assigns the following probabilities to \(X\), the number of alerts a monitoring system produces during a time interval. Is this a valid distribution?
| \(x\) | \(P(X=x)\) |
|---|---|
| 0 | -0.05 |
| 1 | 0.65 |
| 2 | 0.40 |
State. Determine whether the table passes both required checks.
Plan. Check each probability against the range from 0 to 1, then add the entries. Do not infer validity from the total alone.
Do. The entries sum to \(-0.05+0.65+0.40=1.00\). But \(-0.05<0\), so the probability assigned to \(X=0\) is outside the permitted range. The other two entries are in range.
Conclude. The table is not a valid probability distribution. It passes the total check but fails the individual-probability check because one entry is negative.
Similarly, an entry such as 1.08 fails the range check even if the other entries make the total equal 1. The fact that the numbers balance does not turn an out-of-range value into a probability.
Boundary Values Are Allowed
“Between 0 and 1” includes both endpoints. Do not reject a table just because it contains 0 or 1. A zero entry contributes nothing to the total, while an entry of 1 contributes the entire total. For the sum to remain 1, all the other entries in a table with a probability of 1 must be 0.
Worked Example: A Valid Table with Probabilities 0 and 1
A simplified model says that a device produces exactly one alert during a particular interval. Let \(X\) be the number of alerts. The table also lists the other values under consideration. Check whether it satisfies the probability rules.
| \(x\) | \(P(X=x)\) |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 0 |
State. Check the range of each probability and the sum of all three entries.
Plan. Include 0 and 1 when checking the allowed range. Then add all probabilities, including the two zero entries.
Do. Each entry is between 0 and 1, inclusive. The sum is \(0+1+0=1\).
Conclude. The table satisfies both probability checks and is valid. It represents certainty that \(X=1\), with no chance assigned to \(X=0\) or \(X=2\) under this model.
A zero probability does not mean the number is negative or that the table is automatically invalid. It means the model assigns no chance to that listed value. A probability of 1 is not “too large”; it represents certainty. Both boundary cases are permitted by the rules.
Common Mistakes and AP Exam Tips
- Checking only the sum. A total of 1 does not fix a negative entry or an entry greater than 1. Check every probability individually.
- Checking only the entries. Several probabilities can each be between 0 and 1 but still add to something other than 1. Always calculate the total.
- Excluding the endpoints. The conditions are \(0\leq P(X=x)\leq1\), not strict inequalities. Probabilities of 0 and 1 are allowed.
- Rounding or changing entries without saying so. Evaluate the values shown. If the table contains rounded values, explain that rounding may affect the displayed sum; do not quietly alter an entry.
- Giving a verdict without a reason. “Invalid” is clearer when followed by the specific failure, such as “the entries sum to 1.05” or “one probability is negative.”
For a full-credit response, report both checks even when one already shows the table is invalid. For instance: “Each entry is between 0 and 1, but their sum is 1.05, not 1, so this is not a valid probability distribution.” Or: “The entries sum to 1, but one probability is negative, so the table is not valid.” These statements distinguish the two requirements and make the reasoning easy to follow.
Key Takeaway
Before using a discrete probability distribution, audit its probabilities in two parts: every entry must be between 0 and 1, inclusive, and all entries must sum to 1. A candidate table is valid only when it passes both checks.
Check Your Understanding
For each candidate table, state whether it is valid and identify the check or checks that support your conclusion.
- A table has probabilities 0.10, 0.55, and 0.35. Does it pass both checks? Show the sum.
- A table has probabilities 0.20, 0.50, and 0.40. Each entry is in range. Is the table valid? Explain.
- A table has probabilities \(-0.10\), 0.60, and 0.50. The entries sum to 1.00. Is the table valid? Explain.
- Can a valid distribution include a probability of 0? Can it include a probability of 1? State what each means.
- A candidate table contains an entry of 1.04. Which condition does this violate, and why is checking only the total not enough?