Why Divide an Event Into Cases?
Some probability questions are easier to solve when the outcomes of interest are grouped into simpler cases. For example, the event “the sum of two dice is at least 10” can be separated into the cases “the sum is 10,” “the sum is 11,” and “the sum is 12.” Each case can be counted separately, and because no outcome belongs to more than one case, the case probabilities can be added.
The central requirement is that the cases form a partition of the event being studied. If cases overlap, an outcome may be counted more than once. If they fail to cover the event, an outcome may be missed. Checking both properties is as important as doing the arithmetic.
When cases partition an event, the probability of the event is the sum of the probabilities of its cases:
“Pairwise disjoint” means that \(C_i\) and \(C_j\) have no outcomes in common whenever \(i\ne j\). The rule does not require the cases to be equally likely. It requires the probability of each case to be calculated correctly under the stated probability model.
Why the Addition Rule Works
The reason for the rule is that disjoint cases contain separate sets of outcomes. When we combine those sets, no outcome is counted twice. In an equally likely finite sample space, probability is the number of favorable outcomes divided by the total number of outcomes. If each case contains \(m_i\) outcomes and the sample space contains \(N\) outcomes, then a disjoint partition of \(A\) gives:
Since \(m_i/N=P(C_i)\), this gives \(P(A)=\sum_{i=1}^k P(C_i)\). In a general probability model, the same addition rule applies to disjoint events: the probability assigned to their union is the sum of their probabilities. The count argument above shows why the rule is natural when all outcomes are equally likely.
Worked Example: A Sum of at Least 10 on Two Dice
Worked Example: A Sum of at Least 10 on Two Dice
Two fair six-sided dice are rolled. What is the probability that their sum is at least 10?
Model. Treat an outcome as an ordered pair \((d_1,d_2)\), where \(d_1\) is the first die and \(d_2\) is the second. There are \(6\cdot6=36\) equally likely ordered outcomes. This is the standard model for two fair dice: each die has six equally likely faces, and the dice rolls are independent.
The event “sum at least 10” can only happen when the sum is 10, 11, or 12. These three cases are disjoint: a single pair of dice has only one sum. Together they cover the target event. Count the ordered outcomes in each case:
- Sum 10: \((4,6),(5,5),(6,4)\), so there are 3 outcomes.
- Sum 11: \((5,6),(6,5)\), so there are 2 outcomes.
- Sum 12: \((6,6)\), so there is 1 outcome.
Each case probability is its number of outcomes divided by 36. Adding the disjoint cases gives:
The probability of a sum of at least 10 is \(1/6\), or about 16.67%. Notice that the cases are not equally likely: sums of 10, 11, and 12 have 3, 2, and 1 ordered outcomes, respectively. We add their probabilities, not one equal amount for each named case.
Choose Cases That Are Exhaustive and Nonoverlapping
A case is a description that selects some outcomes. A useful way to build cases is to classify outcomes according to a feature that each outcome has exactly one value for. Examples include the sum of two dice, the parity of the first die, or the result of the first coin toss. If the cases use all possible values of that feature that can occur within the target event, they will often be exhaustive. Because each outcome has only one value of the feature, the cases will also be disjoint.
The feature should make the original question simpler, not merely replace it with another question of equal difficulty. Before calculating, check two things: every outcome of interest must fit at least one case, and no outcome of interest may fit two cases. If either check fails, revise the cases.
Worked Example: An Odd Sum on Two Dice
Worked Example: An Odd Sum on Two Dice
Two fair six-sided dice are rolled. Find the probability that their sum is odd.
As before, there are 36 equally likely ordered outcomes. Instead of listing all odd sums, divide the target event according to the parity of the first die:
- Case 1: the first die is odd and the second die is even.
- Case 2: the first die is even and the second die is odd.
An odd sum requires one odd number and one even number. Thus the cases are disjoint, and together they include every outcome with an odd sum. The first die has 3 odd faces and the second die has 3 even faces, so Case 1 contains \(3\cdot3=9\) ordered outcomes. Case 2 also contains \(3\cdot3=9\) ordered outcomes. Therefore:
The chance of an odd sum is \(1/2\). The calculation is valid because the dice outcomes are equally likely and the parity cases are both exhaustive for the target event and nonoverlapping. In particular, an outcome such as \((1,2)\) belongs to the first case only, while \((2,1)\) belongs to the second.
Partition by the Number of Successes
A specified number of successes can also be handled by dividing an event into cases. For example, if several trials are recorded in order, cases can be defined by the result of the first trial. The key is to define the cases so that each possible sequence of results belongs to exactly one case.
In the next example, all sequences are equally likely, so we can count them directly. The calculation uses the model of repeated fair coin tosses: every head-or-tail sequence of a given length is equally likely.
Worked Example: Exactly Two Heads in Four Tosses
A fair coin is tossed four times. Find the probability of getting exactly two heads.
There are \(2^4=16\) equally likely ordered sequences of four results. Partition the event “exactly two heads” according to the first toss:
- Case 1: the first toss is H. To finish with exactly two heads, the last three tosses must contain exactly one head. There are 3 such sequences for the last three tosses: HTT, THT, and TTH. This case has 3 outcomes.
- Case 2: the first toss is T. The last three tosses must contain exactly two heads. There are 3 such sequences: HHT, HTH, and THH. This case also has 3 outcomes.
These cases do not overlap, because the first toss cannot be both H and T. They cover every sequence with exactly two heads, because the first toss must be either H or T. Thus:
The probability of exactly two heads is \(3/8\), or 37.5%. The case counts differ from the counts in the dice examples, but the same reasoning applies: define disjoint cases covering the target event, calculate each case probability, and add.
Common Mistake: Adding Overlapping Cases
Suppose someone tries to find the probability of an odd sum on two dice by adding “the first die is odd” and “the second die is odd.” These are not disjoint cases: an outcome with two odd faces belongs to both descriptions. For example, \((1,3)\) has an odd first die and an odd second die. Adding the two event probabilities would count such outcomes twice, and it would not describe the event “odd sum” in the first place.
The correction is to construct cases that are both relevant and nonoverlapping: first odd with second even, or first even with second odd. More generally, do not add event probabilities just because the events sound like separate possibilities. First check whether an outcome could belong to more than one of them.
Another mistake is to treat the case labels as equally likely. In the dice-sum example, there were three possible sums in the target event, but they were not equally likely. The correct contribution from a case depends on its probability, not on the fact that it has been given a separate name.
Check Your Understanding
For each question, identify the cases and explain why adding their probabilities is appropriate.
- Two fair dice are rolled. What cases based on the sum would you use to find the probability that the sum is less than 5?
- Two fair dice are rolled. Explain why “first die is even” and “second die is even” do not form disjoint cases.
- A fair coin is tossed three times. Describe a disjoint partition of the event “at least one head” based on the first toss.
- In a finite equally likely sample space of 40 outcomes, a target event is partitioned into disjoint cases containing 4, 7, and 9 outcomes. What is the probability of the target event?
- Can a partition of an event have cases with different probabilities? Explain using one of the examples in this tutorial.