Some events can happen in more than one mutually exclusive way, such as a dice sum of 7 or 11. This lesson splits such events into disjoint cases, finds each case's probability separately, and adds them together using the addition rule for events that cannot both occur.
What you will learn
- Split a probability problem into disjoint cases
- Compute each case's probability separately
- Add case probabilities to find the total
Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.
Recognizing when a problem needs cases
An event needs casework when it can occur through more than one distinguishable route, and no single outcome can satisfy more than one of those routes at once. For example, "the sum of two dice is 7 or 11" is really two separate events — sum equals 7, and sum equals 11 — and a single roll can produce only one sum, so the two routes never overlap.
The signal to look for is the word "or" connecting two conditions that describe different, non-overlapping outcomes. If the two conditions could both be true at the same time for some outcome, the event is not a clean case split and needs a different technique to avoid double-counting.
Computing each case's probability separately
Once the cases are identified, each one is treated as its own probability problem: build or reuse the sample space, count the favorable outcomes for that specific case, and divide by the total. The methods from the previous two lessons — listing an explicit sample space and labeling identical-looking objects individually — apply within each case exactly as before.
It helps to keep the total sample space size the same across all cases being compared, since the cases will eventually be combined over that same total. For two dice, every case uses the same 36-outcome sample space; only the favorable count changes from case to case.
Adding case probabilities: the addition rule for disjoint events
Because the cases cannot overlap, the total probability of the combined event is simply the sum of the individual case probabilities: . This works because adding disjoint favorable counts and then dividing by the shared total gives the same result as dividing each case first and then adding.
This addition rule fails if the cases are not actually disjoint, since an outcome belonging to both cases would then be counted twice. Before adding, it is worth double-checking that no single outcome could satisfy more than one case, the same exhaustive, non-overlapping standard used for casework in counting problems.
See the reasoning unfold
Pause after each step and predict what comes next.
Worked example 1
Red or green from a mixed bag
A single marble is drawn from a bag of 3 red, 2 blue, and 1 green marble, all individually distinguishable objects. What is the probability that the marble is red or green?
- The event "red or green" splits into two disjoint cases: drawing a red marble or drawing a green marble; a single marble cannot be both.
1 of 3 steps revealed
Worked example 2
Dice sum of 7 or 11
Two fair six-sided dice are rolled. What is the probability that the sum is 7 or 11?
- Sum equal to 7 and sum equal to 11 are disjoint cases, since a single roll produces only one sum.
1 of 3 steps revealed
Worked example 3
A combined coin-and-die event
A fair coin is flipped and a fair six-sided die is rolled. What is the probability that the result is either heads together with an even number, or tails together with a number greater than 4?
- These two descriptions are disjoint cases, since a single trial shows either heads or tails, never both.
1 of 3 steps revealed
Common mistakes to catch
- Splitting into cases that actually overlap, which double-counts some outcomes when the probabilities are added.
- Missing a case entirely, which undercounts the true probability.
- Forgetting to recompute the total sample space size for a combined experiment, such as a coin and a die together.
Try with support
Guided practice
Choose a strategy, take a first step, and use the hints when you need direction.
Guided problem 1
Difficulty 2/10A bag has 4 red, 3 blue, and 2 yellow marbles, all individually distinguishable. One marble is drawn at random. What is the probability that the marble is blue or yellow?
Guided problem 2
Difficulty 3/10Two fair six-sided dice are rolled. What is the probability that the sum is 2 or 12?
Put it into practice
Independent practice
Work on your own first. If you reveal a solution, work through it and mark it reviewed.
Independent problem 1
Difficulty 3/10A drawer has 5 black socks, 3 white socks, and 2 gray socks, all individually distinguishable. One sock is drawn at random. What is the probability that the sock is black or gray?
Independent problem 2
Difficulty 3/10A fair coin is flipped twice. What is the probability that the two flips show the same result, that is, both heads or both tails?
Independent problem 3
Difficulty 4/10Two fair six-sided dice are rolled. Which fraction gives the probability that the sum is 5 or 9?
Independent problem 4
Difficulty 3/10A bag has 6 distinguishable chips numbered 1 through 6. One chip is drawn at random. What is the probability that the chip shows a number less than 3 or greater than 5?
Bring it together
- Casework in probability splits an event into mutually exclusive routes and adds each route's probability.
- Every case must be checked to be truly disjoint from every other case, or the sum overcounts.
- The same casework idea applies to single experiments, combined experiments, and multi-object draws alike.
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