Using Dice Outcomes to Understand a Union
A two-dice experiment gives a concrete way to see how the addition rule works. For example, what is the probability of rolling doubles or getting a sum greater than 9? Some outcomes satisfy both descriptions, so simply adding the two event probabilities would count those outcomes twice.
As in Sample Spaces and Outcomes, describe each complete result of rolling two dice as an ordered pair \((a,b)\), where \(a\) is the result on die 1 and \(b\) is the result on die 2. The pair \((2,5)\) is distinct from \((5,2)\), because the results on the two dice are recorded in separate positions. With two fair six-sided dice, there are \(6\times 6=36\) equally likely ordered pairs.
An organized 6-by-6 outcome grid is a useful counting tool: one die labels the rows, the other labels the columns, and each cell represents one ordered pair. Doubles lie on the diagonal, where the two die results match. Outcomes with a specified sum lie along diagonal bands running the other way. Marking both events in this grid makes their overlap visible.
The general addition rule, introduced in The General Addition Rule, applies whether the events overlap or not. The key counting step is to identify the outcomes in both events. In dice problems, listing those pairs is often a quick way to check that the overlap has been counted correctly.
Worked Example: Doubles or a Sum Greater Than 9
Worked Example: Doubles or a Sum Greater Than 9
Roll two fair six-sided dice. Let \(A\) be the event “the dice show doubles” and let \(B\) be the event “the sum is greater than 9.” Find the probability of \(A\cup B\).
State: The requested event is rolling doubles, getting a sum greater than 9, or doing both. The dice can show both doubles and a sum greater than 9, so the events overlap.
Plan: Use the general addition rule. Count outcomes in \(A\), outcomes in \(B\), and outcomes in the intersection \(A\cap B\), using the 36 equally likely ordered pairs as the sample space.
Do: The doubles are \((1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\), so \(A\) has 6 outcomes. Thus \(P(A)=6/36\). As a check on the count, there is one double for each possible face value, giving 6 doubles altogether.
A sum greater than 9 is 10, 11, or 12. A sum of 10 has 3 ordered pairs: \((4,6),(5,5),(6,4)\). A sum of 11 has 2: \((5,6),(6,5)\). A sum of 12 has 1: \((6,6)\). Therefore, \(B\) has \(3+2+1=6\) outcomes, and \(P(B)=6/36\). The count is also checked by listing all six pairs across the three sums.
The intersection consists of doubles whose sum is greater than 9: \((5,5)\) and \((6,6)\). So \(P(A\cap B)=2/36\). There are exactly two such pairs in the doubles list, which independently verifies the intersection count.
Substitute these probabilities into the general addition rule:
A direct count checks the result: the union contains the four doubles with sums no greater than 9, namely \((1,1),(2,2),(3,3),(4,4)\), along with the six outcomes whose sums are greater than 9. That is \(4+6=10\) distinct outcomes, matching the addition-rule count. The fraction \(10/36\) simplifies to \(5/18\), and \(5\div18\) rounds to \(0.2778\).
Conclude: The probability of rolling doubles or getting a sum greater than 9 is \(5/18\), or about 0.2778. The overlap consists of \((5,5)\) and \((6,6)\), so it must be subtracted once after adding the two event counts.
When the Dice Events Are Disjoint
The same counting method also helps determine whether the simpler addition rule for mutually exclusive events applies. As explained in What Mutually Exclusive Events Mean, two events are mutually exclusive if no outcome can belong to both. A sum of 7 cannot occur on a roll of doubles: the sum of two matching face values is even, while 7 is odd. Therefore, “doubles” and “sum is 7” are mutually exclusive events.
Do not decide that events are disjoint just because their descriptions sound different. Check whether a single ordered pair can satisfy both descriptions. If the intersection is empty, its probability is zero and the general addition rule reduces to adding the two probabilities.
Worked Example: Doubles or a Sum of 7
Roll two fair six-sided dice. Find the probability of rolling doubles or getting a sum of exactly 7.
State: Let \(A\) be rolling doubles and \(B\) be getting a sum of 7. The union \(A\cup B\) includes outcomes in either event.
Plan: Count each event in the 36 equally likely ordered pairs and check whether any outcome belongs to both. If there is no overlap, add the two event probabilities.
Do: There are 6 doubles, so \(P(A)=6/36\). A sum of 7 occurs with \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\), giving 6 outcomes and \(P(B)=6/36\). The six listed pairs all have unequal dice, so none is a double. Thus \(P(A\cap B)=0/36=0\).
The rule gives:
As a direct check, the 6 doubles and the 6 sum-of-7 outcomes are distinct groups, so together they contain 12 outcomes. This confirms \(12/36\), and \(12\div36=1/3\), which is about 0.3333.
Conclude: The probability of rolling doubles or a sum of 7 is \(1/3\), or about 0.3333. These events are mutually exclusive because no ordered pair is both a double and a sum of 7.
Find the Overlap Before You Add
Some dice events overlap in a less obvious way than doubles and a high sum. For example, “at least one die shows 6” and “the sum is greater than 9” can occur together. Listing the intersection first keeps the event definitions separate and prevents either omission or double counting.
For “at least one die shows 6,” remember that \((6,2)\) and \((2,6)\) are different outcomes. But \((6,6)\) should be counted only once within that event. One way to count the event is to take the six outcomes with die 1 equal to 6 and the six outcomes with die 2 equal to 6, then remove their shared outcome \((6,6)\). The complement offers a check: 25 outcomes have no 6, leaving \(36-25=11\) with at least one 6.
Worked Example: A Sum Greater Than 9 or at Least One 6
Roll two fair six-sided dice. Let \(A\) be the event “the sum is greater than 9” and \(B\) be the event “at least one die shows 6.” Find \(P(A\cup B)\).
State: The union occurs when the roll has a sum greater than 9, at least one 6, or both.
Plan: Count each event and their shared outcomes among the 36 equally likely ordered pairs. Use the general addition rule, then verify the result by directly counting the union.
Do: As counted earlier, \(A\) has 6 outcomes: the three pairs with sum 10, the two with sum 11, and the one with sum 12. Therefore \(P(A)=6/36\). For \(B\), there are 6 outcomes with die 1 showing 6 and 6 with die 2 showing 6; \((6,6)\) is shared between those lists. So \(B\) has \(6+6-1=11\) outcomes and \(P(B)=11/36\). The complement check gives the same count: \(36-5\times5=36-25=11\).
The intersection \(A\cap B\) includes \((4,6),(5,6),(6,4),(6,5),(6,6)\), so it has 5 outcomes and probability \(5/36\). To check the list, the sum-greater-than-9 outcomes are \((4,6),(5,5),(6,4),(5,6),(6,5),(6,6)\); all but \((5,5)\) have at least one 6.
For a separate direct check, every outcome in \(B\) is already in the union, giving 11 outcomes. Of the sum-greater-than-9 outcomes, \((5,5)\) is not in \(B\), so it adds exactly one new outcome. The union therefore has \(11+1=12\) outcomes, agreeing with the addition-rule calculation. The fraction \(12/36\) simplifies to \(1/3\), or about 0.3333.
Conclude: The probability that the roll has a sum greater than 9 or includes at least one 6 is \(1/3\), or about 0.3333. The overlap has 5 outcomes and is subtracted once so those outcomes are counted only once in the union.
A Reliable Dice-Union Method
A short, repeatable process makes these questions easier. Start from the complete outcomes, not from a guess about whether the events overlap. Then check the rule’s result against the actual distinct outcomes in the union.
Translate each phrase into a clear event, such as “the dice match” or “the sum is greater than 9.”
For two fair six-sided dice, count favorable pairs from the 36 equally likely outcomes. Keep the die positions distinct.
Find the pairs that satisfy both event definitions. If there are none, the events are mutually exclusive.
Use \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), then verify the union by listing or otherwise recounting its distinct outcomes.
Common Mistakes and AP Exam Tips
- Counting 21 outcomes instead of 36. Two dice produce ordered pairs, not just unordered pairs. For example, \((2,5)\) and \((5,2)\) are separate equally likely outcomes.
- Adding without checking the overlap. “Doubles” and “sum greater than 9” overlap at \((5,5)\) and \((6,6)\). Add both event probabilities, then subtract the intersection once.
- Counting \((6,6)\) more than once in “at least one 6.” It has a 6 on both dice but is still one outcome. Count the shared pair once within the event.
- Assuming an “or” means exactly one event. A union includes outcomes in \(A\), in \(B\), and in both. If the question asks for exactly one, that is a different event.
- Using the disjoint-events rule without evidence. Check the intersection. A doubles event and a sum-of-7 event are disjoint, but doubles and a sum greater than 9 are not.
- Giving only a number. A strong response identifies the events, shows the overlap count, uses the addition rule, and states what the probability means in context.
A useful final check is to compare the union count with each event count: the union must contain every outcome in either event, so it cannot have fewer outcomes than either one. It also cannot exceed 36. These checks will not establish that your event setup is correct by themselves, but they can reveal an arithmetic or counting error.
Check Your Understanding
Use the 36 equally likely ordered pairs for each question. Show how you count the event or intersection before finding the probability.
- Roll two fair dice. Let \(A\) be rolling doubles and \(B\) be getting a sum of 8. How many outcomes are in \(A\), \(B\), and \(A\cap B\)? Find \(P(A\cup B)\).
- How many ordered pairs give a sum greater than 10? List them and find the probability.
- Let \(A\) be getting a sum of 6 and \(B\) be getting at least one 1. Find the number of outcomes in \(A\cap B\), then use the addition rule to find \(P(A\cup B)\).
- A student says that “at least one die shows 5” has 12 outcomes because there are 6 choices for each die. Identify the double-counted outcome and give the correct count.
- Are “sum of 9” and “at least one die shows 6” mutually exclusive? List their intersection and explain your decision.