Experiments, Outcomes, and Sample Spaces
Before assigning probabilities, a statistical model must say what results the experiment can produce. The key modeling decision is what counts as one recorded result. A sample space is not just a list of labels: its outcomes must retain the detail needed to answer the questions the model is meant to address.
For example, rolling two dice can mean recording the result of the first die and then the second, or recording only their sum. Those are different ways to describe the experiment. If the question concerns whether the dice match, recording only the sum loses information. In contrast, if the only question is the total, the sum itself can be the recorded outcome.
The sample space depends on both the physical procedure and the recording rule. For a coin toss, the outcome might be just heads or tails. For a sequence of tosses, an outcome is usually the whole ordered sequence, such as HTH. If the experiment records only the number of heads, a different sample space is appropriate. Neither description is automatically wrong; it must match the information the model needs.
A sample space must be exhaustive: every result the experiment can produce is included. Its outcomes must also be distinguishable under the recording rule: results considered different by that rule cannot be represented by the same outcome. These requirements concern the description of outcomes, not their probabilities.
How to Build a Finite Sample Space
A practical way to define a sample space is to specify the experiment’s recording rule first, then write down a typical outcome in that form. Ask whether the order of actions matters, whether the identity of objects matters, and whether the record contains the full result or only a summary. Finally, check that the possible outcomes cover every allowed result.
Specify exactly what the experiment records. “Two dice are rolled” is less precise than “the face on the first die and the face on the second die are recorded in order.”
Use an ordered pair when first and second results are separately recorded, a sequence when the full order of repeated trials matters, or a single value when only a summary is recorded.
List the outcomes or give a precise set description that identifies every allowed outcome and excludes impossible ones.
Confirm that outcomes needed to distinguish relevant cases have not been merged. Also check that no allowed result has been omitted.
At this stage, specifying \(\Omega\) does not assign probabilities to its outcomes. A sample space by itself does not say that its outcomes are equally likely, nor does it provide any other probability distribution. No interest rate or mortality assumption is relevant to the finite experiments below; each example specifies a sample space only.
Worked Example: Recording Two Ordered Die Results
A hypothetical quality-control exercise rolls a blue six-sided die and then a white six-sided die. The record gives the blue result first and the white result second. Define the sample space and identify the event that the results sum to 7. Explain why recording an unordered pair would be a mistake for this experiment.
Method. Represent each outcome by an ordered pair \((b,w)\), where \(b\) is the blue die’s face and \(w\) is the white die’s face. Each die can show any integer from 1 through 6, so the sample space is
For clarity, the set can also be displayed by fixing the blue result in each row:
To find the outcomes whose sum is 7, consider each possible blue result. If \(b=1\), then \(w=6\); if \(b=2\), then \(w=5\); if \(b=3\), then \(w=4\); if \(b=4\), then \(w=3\); if \(b=5\), then \(w=2\); and if \(b=6\), then \(w=1\). Thus the event is
Check. Each listed pair has coordinates from 1 to 6, and each pair sums to 7. The list includes one pair for each possible blue result, so no outcome with sum 7 has been missed.
Common mistake avoided. Treating the pairs as unordered would merge \((1,6)\) with \((6,1)\), even though the first is “blue 1, white 6” and the second is “blue 6, white 1.” The recording rule distinguishes these outcomes, so the sample space must preserve their order. This example specifies no probability distribution; it does not rely on an assumption that the outcomes are equally likely.
Sequences and Identity in Outcomes
Repeated trials are naturally represented by sequences when the record preserves the result at each trial. In a sequence, position matters: the first symbol describes the first trial, the second symbol describes the second trial, and so on. A summary such as a total or count may be useful for a particular question, but it should not silently replace the full sequence when the order could matter.
Identity matters in a similar way. Drawing the ace of hearts and then the three of clubs is a different recorded result from drawing the three of clubs and then the ace of hearts if the draw order is recorded. If the record contains only the set of cards received, the two orders may instead be treated as one result. State the recording rule before choosing between those descriptions.
Worked Example: Three Recorded Coin Tosses
A hypothetical sensor records three tosses of a coin, using H for heads and T for tails, in the order observed. Define the sample space and the event that exactly two tosses are heads. Then explain what information would be lost if the record were only the number of heads.
Method. An outcome is a three-symbol sequence. Each of the three positions can contain H or T. Listing the sequences by their first symbol, then their second symbol, gives
The event “exactly two heads” contains the sequences with two H symbols and one T symbol. Checking the list gives \(HHT\), \(HTH\), and \(THH\). Therefore the event is
Check. Each sequence in the event has three toss results and exactly two H symbols. The remaining five sequences in \(\Omega\) have a different number of heads: \(HHH\) has three, \(HTT\), \(THT\), and \(TTH\) have one each, and \(TTT\) has none.
If the experiment recorded only the number of heads, its sample space would be \(\{0,1,2,3\}\), not the set of eight sequences above. That summary is sufficient to answer the “exactly two heads” question, but it cannot distinguish \(HHT\), \(HTH\), and \(THH\). It therefore cannot answer a question such as whether the first toss was heads. The sample space must preserve order when the question needs order.
Common mistake avoided. Listing only \(0,1,2,3\) while claiming to describe the full ordered sequence experiment discards information. It is a valid sample space only if the recorded outcome is explicitly the number of heads. No probability distribution is assigned here.
Worked Example: Two Cards Drawn in Order
A hypothetical demonstration uses a standard 52-card deck. Two cards are drawn without replacement, and the record gives the first card and then the second card. Define the sample space using card identities, and specify the event that the first card is the ace of hearts and the second card is red.
Method. Let the ranks be \(R=\{A,2,3,4,5,6,7,8,9,10,J,Q,K\}\), and let the suits be \(S=\{\clubsuit,\diamondsuit,\heartsuit,\spadesuit\}\). A card is an ordered pair \((r,s)\) with \(r\in R\) and \(s\in S\). The set of all cards is
Because cards are drawn without replacement, the same card cannot appear twice in one outcome. Because draw order is recorded, an outcome is an ordered pair of distinct cards. The sample space is
The event requires \(c_1=(A,\heartsuit)\), and the second card must have a red suit, \(\diamondsuit\) or \(\heartsuit\). The ace of hearts has already been drawn, so the second card can be any card in either red suit except the ace of hearts itself. In set notation, the event is
There are two red suits, with 13 ranks in each, giving 26 red cards in the deck. The first card is fixed as the ace of hearts, so exactly that one card is excluded from the choices for the second card. The event therefore includes 25 ordered outcomes: the first card stays fixed, and the second card can be any of the other red cards.
Check. Every outcome in the stated event has the ace of hearts first, a red card second, and two distinct cards. The exclusion \(c_2\ne(A,\heartsuit)\) enforces the no-replacement condition.
Common mistake avoided. Writing an outcome as an unordered pair would ignore the stated recording order. Also, allowing the ace of hearts as the second card would violate the no-replacement rule. This is a sample-space description, not a probability calculation; no distribution is assigned.
Why the Recording Rule Matters
The sample space determines what distinctions a model can make. If two physical results are assigned the same outcome, no event defined using that sample space can separate them. For example, after replacing ordered die pairs by their sums, the outcomes \((1,6)\) and \((6,1)\) both become 7. A question about which die showed 1 can no longer be answered from the sum alone.
This is the reason to check the sample space against the intended questions before doing probability calculations. A sample space can be mathematically tidy but still be unsuitable for a particular analysis if it omits relevant detail. Conversely, recording unnecessary detail is not inherently incorrect, though it may make a later calculation less convenient.
Common Mistakes
- Using an incomplete sample space. Omitting a possible recorded result violates exhaustiveness. For three coin tosses, for instance, leaving out \(THT\) means the sequence list is incomplete.
- Discarding order without justification. If first and second results are recorded separately, \((a,b)\) and \((b,a)\) are distinct whenever \(a\ne b\).
- Confusing a summary with the full result. The number of heads is a summary of a coin-toss sequence. It cannot recover the positions of the heads.
- Ignoring restrictions in the experiment. A two-card draw without replacement cannot produce a pair in which the same card appears twice.
- Assuming equal likelihood from a list. Writing down \(\Omega\) describes possible outcomes; it does not establish that all outcomes have the same probability.
Check Your Understanding
For each problem, state the sample space or requested set using the recording rule given. These questions concern possible outcomes only; no probability distribution is assumed.
- A numbered spinner has sectors labeled 1, 2, 3, and 4. It is spun twice, and the first result is recorded before the second. Write the sample space. (Answer: \(\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)\}\))
- A coin is tossed four times, and the complete ordered sequence is recorded. How many possible sequences are there, and give the sequence with tails on the first and fourth tosses and heads on the other two. (Answer: 16 sequences; THHT)
- A die is rolled once, but the record is only whether the result is odd or even. Write the sample space and the event that the recorded result is odd. (Answer: \(\Omega=\{\text{odd},\text{even}\}\); the event is \(\{\text{odd}\}\))
- Two different books are selected from a shelf, one after the other, and their titles are recorded in order. If the titles are \(M\), \(N\), and \(P\), write the sample space. (Answer: \(\{(M,N),(M,P),(N,M),(N,P),(P,M),(P,N)\}\))
- A card is selected from a standard deck and its rank and suit are recorded. Describe the sample space, then describe the event that the card is a black queen. (Answer: \(\Omega=R\times\{\clubsuit,\diamondsuit,\heartsuit,\spadesuit\}\), where \(R=\{A,2,3,4,5,6,7,8,9,10,J,Q,K\}\); event \(=\{(Q,\clubsuit),(Q,\spadesuit)\}\))