From Counting to Probability
When an experiment has finitely many possible outcomes, counting can turn a probability question into a ratio. The key is to define the outcomes at the right level of detail and to make sure the model assigns the same probability to each one. Then the probability of an event is the fraction of outcomes in the sample space that satisfy it.
The counting method depends on what counts as a different outcome. If positions or order matter, use ordered selections. If only the selected group matters, use unordered selections. A correct count with the wrong interpretation of “outcome” gives the wrong probability.
The assumption that all outcomes in the chosen sample space are equally likely is essential. The ratio is not justified merely because the experiment has a finite number of possible outcomes. First specify what one outcome represents; then establish that the outcomes in that representation are equally likely.
Choose the Counting Method
For \(n\) distinct objects, selecting \(r\) of them in order gives a permutation. There are \(n\) choices for the first position, \(n-1\) for the next, and so on. Selecting \(r\) objects without regard to order gives a combination. The formulas below apply when the objects are distinct and selections are made without replacement.
For a selection from distinct objects divided into categories, count the possible selections within each category and multiply. For an event that can occur in several mutually exclusive ways, count each case and add. These principles prevent both omissions and double counting.
| What defines one outcome? | Typical count | Example interpretation |
|---|---|---|
| The order of selected objects matters | \(P(n,r)\) | Three distinct tags assigned to first, second, and third positions |
| Only the selected group matters | \(\binom{n}{r}\) | A committee selected without assigned roles |
| A selection must contain specified category counts | Multiply category combinations | Choose two from one group and one from another |
Worked Example: Ordered Tags in Three Positions
A machine randomly selects three distinct tags from eight tags numbered 1 through 8 and places them in first, second, and third positions. Assume every ordered selection of three distinct tags is equally likely. Find the probability that the first tag is odd and the third tag is greater than 5.
Method. The positions distinguish outcomes, so count ordered selections. The total number is \(P(8,3)=8\cdot7\cdot6=336\).
For a favorable selection, the first tag must be odd: 1, 3, 5, or 7. The third tag must be 6, 7, or 8, and it cannot equal the first tag because tags are selected without replacement. The second position can then be filled by any of the six remaining tags.
If the first tag is 1, 3, or 5, there are three choices for it and three choices for the third tag (6, 7, or 8). This gives \(3\cdot3\cdot6=54\) ordered selections. If the first tag is 7, the third tag can only be 6 or 8, giving \(1\cdot2\cdot6=12\) more.
Thus there are \(54+12=66\) favorable outcomes, and \(P(A)=66/336=11/56\), or 0.1964 to four decimal places.
Check and common mistake. The favorable count, 66, is less than the total, 336, and the probability is between 0 and 1. A common error is to allow tag 7 in both the first and third positions; the selections are without replacement, so those positions cannot contain the same tag.
Counting by Categories and Cases
When a selection is unordered but the objects fall into categories, combinations count the choices from each category. If the event allows several possible category totals, calculate each total separately and add. The cases must be disjoint—for example, “exactly two priority files,” “exactly three priority files,” and “exactly four priority files” cannot overlap.
Worked Example: Selecting Priority Files
A review team selects four files from a batch containing four priority files and three standard files. Assume every four-file subset is equally likely. Find the probability that the selection contains at least two priority files.
Method. The order of selection does not matter. The total number of four-file subsets is \(\binom{7}{4}=35\).
Count favorable subsets by the number of priority files. With exactly two priority files, choose two of the four priority files and two of the three standard files: \(\binom{4}{2}\binom{3}{2}=6\cdot3=18\).
With exactly three priority files, choose three priority files and one standard file: \(\binom{4}{3}\binom{3}{1}=4\cdot3=12\). With exactly four priority files, choose all four priority files and no standard files: \(\binom{4}{4}\binom{3}{0}=1\cdot1=1\).
“At least two” includes exactly two, exactly three, and exactly four priority files. The event does not require any standard files, so the four-priority selection qualifies. The favorable count is \(18+12+1=31\), and the probability is \(31/35=0.8857\) to four decimal places.
Check and common mistake. There are three disjoint cases, and their counts sum to 31, which is no greater than the total of 35. Omitting the four-priority case would incorrectly treat “at least two priority files” as though it also required selecting a standard file.
Using a Complement to Count Favorable Outcomes
Sometimes it is simpler to count all outcomes meeting a main condition and subtract the ones that fail an additional requirement. This works when the unwanted outcomes are exactly the complement of the desired outcomes within that main condition. The subtraction must preserve every part of the main condition.
Worked Example: Two Aces and at Least One Heart
A standard deck has 52 distinct cards. A five-card hand is dealt, and every five-card subset is equally likely. Find the probability that the hand contains exactly two aces and at least one heart.
Method. A hand is unordered, so the total number of hands is \(\binom{52}{5}=2{,}598{,}960\). First count all hands with exactly two aces: \(\binom{4}{2}\binom{48}{3}=6\cdot17{,}296=103{,}776\).
Among those hands, subtract the ones with no hearts. There are 39 non-heart cards, including three non-heart aces. Of the 39 non-heart cards, 36 are not aces. A no-heart hand with exactly two aces therefore has \(\binom{3}{2}\binom{36}{3}=3\cdot7{,}140=21{,}420\) possibilities.
The favorable count is \(103{,}776-21{,}420=82{,}356\). Therefore, $$ P(A)=\frac{82{,}356}{2{,}598{,}960} \approx 0.0316881. $$ The probability is 0.0317 to four decimal places.
Check and common mistake. The favorable count is positive and less than the count of all hands with exactly two aces. When subtracting the no-heart hands, it would be incorrect to choose the two aces from all four aces: a no-heart hand can use only the three non-heart aces.
Why the Favorable-Count Ratio Holds
The ratio follows directly from equal likelihood. Suppose the sample space has \(N\) outcomes, each with probability \(c\). Because the outcomes are disjoint and together exhaust the sample space, their probabilities sum to 1. Thus \(Nc=1\), so each outcome has probability \(c=1/N\). If event \(A\) contains \(K\) outcomes, add the probability \(1/N\) for each of its outcomes:
This is also why the outcome definition matters: the calculation adds probabilities of individual outcomes in the chosen sample space. In a selection of distinct objects, each unordered group corresponds to the same number, \(r!\), of possible orders. Consequently, if order is irrelevant, counting all \(P(n,r)\) arrangements would count each group \(r!\) times. Dividing by \(r!\) gives \(\binom{n}{r}=P(n,r)/r!\), the number of distinct groups.
Common Counting Checks
- Ask whether order matters. Assigned positions or roles usually distinguish outcomes; an unassigned group usually does not.
- Check the selection rule. The permutation and combination formulas above assume distinct objects selected without replacement.
- Translate “at least” carefully. “At least two” includes two and every larger feasible count, including selection of all objects in a category when allowed.
- Keep restrictions in complement counts. When subtracting unwanted outcomes, continue to enforce the other conditions of the event.
- Check the result. A favorable count cannot exceed the total, and a probability calculated from the ratio must lie between 0 and 1.
Check Your Understanding
For each problem, identify whether order matters, count the total outcomes, and then count the favorable outcomes.
- A container has three red and five blue distinct tokens. Three tokens are selected without replacement, and every three-token subset is equally likely. What is the probability of selecting exactly two red tokens? (Answer: \(15/56\), or 0.2679 to four decimal places.)
- Seven distinct numbered markers, 1 through 7, are placed without replacement in two ordered positions, with every ordered pair equally likely. What is the probability that the first marker is in \(\{1,2,3\}\) and the second is in \(\{5,6,7\}\)? (Answer: \(3/14\), or 0.2143 to four decimal places.)
- A committee of three is selected uniformly from four engineers and six actuaries. Two of the actuaries are specifically designated. Find the probability that the committee has exactly two engineers and includes at least one designated actuary. (Answer: \(1/10\), or 0.1000 to four decimal places.)
- A four-card hand is selected uniformly from a standard 52-card deck. What is the probability that it contains at least one of two specified cards? (Answer: \(1-\binom{50}{4}/\binom{52}{4}=1617/10829\), or 0.1493 to four decimal places.)
- A batch has six priority files and four standard files. A uniform five-file subset is selected. What is the probability that it contains at least three priority files? (Answer: \(\bigl[\binom{6}{3}\binom{4}{2}+\binom{6}{4}\binom{4}{1}+\binom{6}{5}\binom{4}{0}\bigr]/\binom{10}{5}=31/42\), or 0.7381 to four decimal places.)