Check the Formula by Restricting the Sample Space
In Common Errors with Conditional Probability, you practiced matching the denominator to the condition. Here is a way to check that choice when a chance process has a finite set of equally likely outcomes: list or count only the outcomes that satisfy the condition, then see what fraction of that reduced group also satisfies the target event.
This method makes the reference group visible. If the question asks for \(P(A\mid B)\), first set aside every outcome that is not in \(B\). The remaining outcomes form the reduced sample space. Count how many of those outcomes are also in \(A\). This count-based result should agree with the conditional probability formula.
The method agrees with the formula because the same outcomes are counted in both calculations. Suppose the full sample space has \(N\) equally likely outcomes, \(n(B)\) outcomes satisfy the condition, and \(n(A\cap B)\) satisfy both events. The formula uses probabilities from the full sample space; the reduced-space method uses counts within \(B\).
The full-sample-space total \(N\) cancels in the formula calculation. That is why restricting attention to \(B\) produces the same answer. Use this count check only when the outcomes being counted are equally likely. If outcomes have different probabilities, a simple fraction of outcome counts may not represent probability.
A Reliable Counting Routine
Use the following routine to check a formula result. It is especially useful when you can organize the possible outcomes in a list, table, or other systematic display. For ordered pairs, keep the order: the result on the first die and the result on the second die are distinct parts of an outcome.
Write the requested probability as \(P(A\mid B)\). The event after the bar, \(B\), is the condition.
Identify or count only outcomes that satisfy \(B\). This is the reduced sample space.
Within the reduced sample space, count outcomes that also satisfy \(A\). These are the outcomes in \(A\cap B\).
Divide the favorable count by the reduced-space count. Check that this equals the result from the conditional probability formula.
Counting gives an independent way to inspect your setup: if your formula used the wrong condition probability in the denominator, it will generally disagree with the fraction from the reduced sample space. It also helps catch missed outcomes and accidental double counting.
Worked Example: Two Dice and an Even Sum
Worked Example: Two Dice and an Even Sum
Imagine rolling two fair six-sided dice and recording the ordered pair \((\text{first die},\text{second die})\). Let \(A\) be the event that the first die shows 4, and let \(B\) be the event that the sum is even. Find \(P(A\mid B)\) using the formula, then check the answer by counting outcomes inside the condition.
State: The question asks for \(P(A\mid B)\): the probability that the first die shows 4, given that the sum is even.
Plan: All 36 ordered pairs are equally likely. First use the conditional probability formula, with the even-sum outcomes as the reference group. Then count the ordered pairs with an even sum and check what fraction have a 4 on the first die.
Do with the formula: A sum is even when both die results are even or both are odd. There are \(3\cdot3=9\) even-even pairs and \(3\cdot3=9\) odd-odd pairs, giving 18 even-sum pairs. If the first die is 4 and the sum is even, the second die must also be even: it can be 2, 4, or 6. Thus, 3 outcomes are in \(A\cap B\).
Substitute these probabilities into the conditional probability formula:
Check by counting inside \(B\): The reduced sample space consists of the 18 even-sum ordered pairs. Of those, 3 have a 4 on the first die. The reduced-space fraction is \(3/18=1/6\), matching the formula.
Conclude: Given that the sum is even, the probability that the first die shows 4 is \(1/6\). The answer is a proportion of the even-sum outcomes, not of all 36 dice outcomes.
Worked Example: A Card Draw Given a Face Card
Worked Example: A Card Draw Given a Face Card
A single card is selected at random from a standard 52-card deck. Let \(F\) mean that the card is a face card (a jack, queen, or king), and let \(R\) mean that the card is red. Find \(P(R\mid F)\), first with the formula and then by counting within the condition.
State: The target event is \(R\), and the condition is \(F\), so the requested probability is \(P(R\mid F)\).
Plan: Each individual card is equally likely. There are 12 face cards: 3 ranks in each of 4 suits. The reduced sample space will contain only those 12 cards. Count how many of them are red.
Do with the formula: The red suits are hearts and diamonds. Each has a jack, queen, and king, so 6 cards are both red and face cards. The chance of a face card is \(12/52\), and the chance of a red face card is \(6/52\).
Check by counting inside \(F\): Once the condition is known, the relevant possibilities are the 12 face cards, not all 52 cards. Six of those 12 are red, so the reduced-space fraction is \(6/12=1/2\).
Conclude: Given that the selected card is a face card, the probability it is red is \(1/2\). The formula and the count inside the condition agree.
The reduction does not mean that the deck has physically changed. It means that, for this question, only face cards remain possible. A card that is not a face card is outside the condition and cannot be part of the denominator for \(P(R\mid F)\).
Worked Example: Coin Sequences with at Least Two Heads
Worked Example: Coin Sequences with at Least Two Heads
A fair coin is tossed three times. Let \(B\) mean that the sequence contains at least two heads, and let \(A\) mean that the first toss is heads. Find \(P(A\mid B)\) and verify the result by listing the reduced sample space.
State: We want the probability that the first toss is heads, given that the three-toss sequence contains at least two heads: \(P(A\mid B)\).
Plan: The eight possible three-toss sequences are equally likely. Use the conditional probability formula with \(B\) as the condition, then list just the sequences that meet \(B\) and count those that also meet \(A\).
Do with the formula: The sequences with at least two heads are HHH, HHT, HTH, and THH, so there are 4 outcomes in \(B\). Three of these—HHH, HHT, and HTH—also start with heads. Therefore \(P(B)=4/8\) and \(P(A\cap B)=3/8\).
Check by counting inside \(B\): The reduced sample space is HHH, HHT, HTH, THH. Three of these four sequences start with heads, so the direct count is \(3/4=0.750\).
Conclude: Given that at least two of the three tosses are heads, the probability that the first toss is heads is \(0.750\). The denominator comes from the four sequences that satisfy the condition, not from all eight possible sequences.
What a Mismatch Can Reveal
If the formula and the reduced-space count do not agree, do not average the answers or choose the one that looks more reasonable. Check how the outcomes were defined and counted. The discrepancy usually points to a specific setup error.
- The condition group is wrong. You may have counted outcomes satisfying \(A\) for the denominator instead of outcomes satisfying \(B\). Return to the event after the bar.
- An outcome is missing or counted twice. Write the reduced sample space systematically, or organize it in a table. Check that each possible outcome in \(B\) appears exactly once.
- Order was ignored when it matters. For two dice, \((2,5)\) and \((5,2)\) are different outcomes. Combining them as one sum can distort a count unless the outcomes are handled with their correct multiplicities.
- Outcomes are not equally likely. Counting outcomes as if they had equal weight may give the wrong probability. In that situation, use the probabilities of the outcomes rather than a simple count fraction.
- The condition has no possible outcomes. If no outcome satisfies \(B\), there is no reduced sample space to use, and \(P(A\mid B)\) is undefined.
A count can be a helpful check even if you first solved the problem with probabilities. It verifies that the numerator is genuinely inside the condition group and that the denominator matches the question. In the dice example, for instance, “first die is 4” alone does not define the denominator for \(P(A\mid B)\); only even-sum outcomes belong there.
Common Mistakes and Full-Credit Communication
A common mistake is to divide the number of favorable outcomes by the total number of outcomes in the original experiment. That finds a probability from the full sample space, not a conditional probability. Another is to say “there are 3 favorable outcomes” without naming the group they are favorable within. The count is incomplete until the reference group is clear.
For a clear solution, state the event after “given that,” identify the outcomes in that condition, and show the favorable count divided by the condition count. If also checking the formula, show both \(P(A\cap B)\) and \(P(B)\), then substitute them. Finish with an interpretation that identifies the condition group in context.
This technique is a check, not a replacement for identifying the correct events. The words “given that” still determine the condition. If you swap the condition and target, you will be restricting attention to a different group, even if the overlap count stays the same.
Check Your Understanding
For each question, identify the condition and explain how a reduced sample space can check the conditional probability.
- Two fair dice are rolled. Let \(B\) mean that the first die is odd and \(A\) mean that the sum is 7. How many ordered pairs are in the reduced sample space for \(P(A\mid B)\), and how many satisfy both events?
- A card is selected from a standard deck. Given that the card is a heart, what is the reduced sample space size? How many of those cards are face cards? Write the resulting conditional probability.
- Three fair coin tosses are recorded. Given that the first toss is tails, what is the reduced sample space? What fraction of those sequences contain exactly two heads?
- Why would simply counting the outcomes in an event be unreliable if the possible outcomes were not equally likely?
- A student finds \(P(A\mid B)\) by dividing the number of outcomes in both \(A\) and \(B\) by the total number of outcomes in the original sample space. What probability has the student most likely calculated instead?