Count the Outcomes in Both Events
In Drawing a Venn Diagram of Two Events, the shared part of the circles represented the outcomes in both events. In Probability of “Not” and “Neither” Events, the focus was on outcomes outside the circles. Now we will calculate the probability of the shared region directly by counting its outcomes.
The intersection \(A\cap B\) is the event that both \(A\) and \(B\) occur. If a chance process has a finite sample space in which every outcome is equally likely, count the outcomes that belong to both events and divide by the total number of outcomes. The method works because each individual outcome has the same chance of occurring.
The word both is the key. An outcome belongs in the numerator only if it satisfies the definition of \(A\) and the definition of \(B\). Count each shared outcome once, even though it belongs to both event sets.
This calculation uses the equally likely cases idea from Equally Likely Outcomes and Counting Probability. The denominator must count the complete outcomes of the chance process—not just outcomes in \(A\), outcomes in \(B\), or categories that do not have equal chances. As emphasized in Sample Spaces and Outcomes, first be clear about what one complete outcome records.
A Direct Counting Method
A reliable method is to write down the sample space, identify the outcomes that meet each event definition, and then select only the outcomes that meet both. For a small sample space, listing is often the clearest approach. For a larger one, a carefully organized table can prevent missed or repeated outcomes.
Describe what one outcome records, and identify the complete sample space \(S\).
Make sure the individual outcomes being counted are equally likely under the chance model.
Apply both event definitions and count the outcomes that satisfy both.
Divide the overlap count by the total sample-space count, then state what the probability means in context.
The equal-likelihood check matters especially for multi-stage processes. For example, when two fair dice are rolled, the 36 ordered pairs are equally likely, but the possible sums are not. There is only one way to get a sum of 2, while there are six ways to get a sum of 7. Counting sums as if each sum had the same probability would use the wrong sample space.
Worked Example: Both Events on One Fair Die
Worked Example: Both Events on One Fair Die
A fair six-sided die is rolled once. Let \(A\) be the event that the result is even, and let \(B\) be the event that the result is greater than 3. Find \(P(A\cap B)\).
Define the sample space: One outcome is the number showing on the die. The six possible outcomes are \(S=\{1,2,3,4,5,6\}\). Because the die is fair, these six outcomes are equally likely.
Identify each event: The even results are \(A=\{2,4,6\}\). The results greater than 3 are \(B=\{4,5,6\}\). An outcome in the intersection must appear in both sets.
Count the shared outcomes: The results in both lists are 4 and 6. Thus, \(A\cap B=\{4,6\}\), and there are 2 outcomes in the intersection. There are 6 outcomes in the full sample space.
Conclude: The probability that one roll of a fair die is both even and greater than 3 is \(\frac{1}{3}\). The numerator counts only results meeting both conditions, not all even results or all results greater than 3.
Ordered Pairs Keep Two-Die Outcomes Straight
For two dice, an outcome records the result on each die. We can write an outcome as an ordered pair \((\text{first die},\text{second die})\). There are \(6\times6=36\) possible ordered pairs, and for two fair dice each pair is equally likely. Order matters in the sample space: \((2,5)\) records a different result from \((5,2)\), even though both have the same sum.
When the events refer to a sum and to one specific die, use the full ordered pairs to check both conditions. Counting only the sums would lose information about which die showed which number.
Worked Example: A Large Sum and an Even First Die
Two fair six-sided dice are rolled. Let \(A\) be the event that the sum is at least 9, and let \(B\) be the event that the first die shows an even number. Find \(P(A\cap B)\).
Define the sample space: Each outcome is an ordered pair, such as \((4,5)\). There are \(6\times6=36\) equally likely ordered pairs.
List the outcomes in both events: The pairs with sums at least 9 are shown below. From these, keep the pairs whose first coordinate is even, because those are the ones in \(B\).
| Sum | Pairs with this sum | Pairs in both \(A\) and \(B\) |
|---|---|---|
| 9 | (3,6), (4,5), (5,4), (6,3) | (4,5), (6,3) |
| 10 | (4,6), (5,5), (6,4) | (4,6), (6,4) |
| 11 | (5,6), (6,5) | (6,5) |
| 12 | (6,6) | (6,6) |
There are \(2+2+1+1=6\) ordered pairs in the intersection. Divide by all 36 equally likely outcomes.
Conclude: The probability that the sum is at least 9 and the first die is even is \(\frac{1}{6}\). The ordered-pair list ensures that the first-die condition is checked on the correct coordinate.
Notice that the numerator counts complete outcomes, not just the possible sums 9, 10, 11, and 12. Those four sums do not have equal probabilities, so dividing the number of matching sums by the number of possible sums would not give the desired probability.
Worked Example: Face Card and Heart
Worked Example: Face Card and Heart
One card is drawn at random from a well-shuffled standard 52-card deck. Let \(A\) be the event that the card is a face card (a jack, queen, or king), and let \(B\) be the event that it is a heart. Find \(P(A\cap B)\).
Define the outcomes: Each outcome is one particular card. The 52 cards are equally likely to be drawn, so the denominator is 52.
Find the overlap: A card in both events must be a heart and be a jack, queen, or king. The shared outcomes are \(J\heartsuit\), \(Q\heartsuit\), and \(K\heartsuit\), for a total of 3.
Conclude: The probability that the drawn card is both a heart and a face card is \(\frac{3}{52}\), or about 0.0577. The numerator counts the three cards satisfying both descriptions, out of all 52 equally likely cards.
Why Direct Counting Helps
Direct counting is especially useful when the sample space is small enough to list or organize, and the problem tells you which outcomes are equally likely. It makes the meaning of the intersection visible: every outcome in the numerator has passed both event checks.
The calculation does not require you to know \(P(A)\) or \(P(B)\) separately. It also does not require you to assume that \(A\) and \(B\) are independent. For the die examples, the overlap count itself provides the probability. Do not replace that count with a product of event probabilities unless the problem gives a justified reason and the method is appropriate.
A quick check is to ask whether each numerator outcome satisfies both definitions, and whether the denominator counts all possible complete outcomes exactly once. The probability must be between 0 and 1. If there are no outcomes in the overlap, the probability is 0; if every outcome lies in both events, it is 1.
Common Mistakes and AP Exam Tips
- Counting outcomes in either event instead of both. “In \(A\cap B\)” means an outcome must satisfy both event definitions. Read the word “and” as a check of both conditions.
- Using the wrong denominator. The denominator is the total number of equally likely outcomes in the complete sample space, not the number in \(A\), in \(B\), or in their union.
- Counting the same shared outcome twice. An outcome may be in both events, but it is still one outcome in the sample space and belongs in the numerator once.
- Treating two-dice sums as equally likely. The ordered pairs are equally likely; the sums are not. Use pairs when the events involve the individual dice or when counting ways to obtain a sum.
- Reporting only a fraction without explaining it. For full-credit communication, show the number of outcomes in both events, the total number of equally likely outcomes, and a sentence interpreting the probability in context.
- Multiplying event probabilities without checking the model. The direct method asks for the number of outcomes in the overlap. Do not assume that multiplying \(P(A)\) and \(P(B)\) gives that probability.
On an AP response, a concise explanation might say: “There are 2 outcomes in \(A\cap B\) among 6 equally likely die results, so \(P(A\cap B)=2/6=1/3\).” This names the overlap, states why the denominator is appropriate, and gives the probability. If the sample space needs explanation—especially for two dice—describe what one outcome records.
Check Your Understanding
For each question, identify the outcomes in both events and use the equally likely sample space.
- A fair die is rolled. Let \(A\) be the event “the result is odd” and \(B\) be the event “the result is greater than 2.” Find \(P(A\cap B)\).
- Two fair dice are rolled. How many ordered pairs have a sum of 8 and an even result on the first die? What is the probability of both events?
- A card is drawn at random from a standard deck. Let \(A\) be “the card is a spade” and \(B\) be “the card is a queen.” Find \(P(A\cap B)\).
- Explain why the sums 2 through 12 are not an equally likely sample space for rolling two fair dice.
- In a fair-die problem, an outcome is counted in the numerator because it satisfies event \(A\), but it does not satisfy event \(B\). Should it be included in the count for \(P(A\cap B)\)? Explain.