Consider one hypothetical uniform draw from six labelled tokens: 1, 2, 3, 4, 5, and 6. Let event A mean drawing an even token, and let event B mean drawing 4, 5, or 6. The question “A or B” includes every outcome in A, every outcome in B, and outcomes belonging to both. Therefore, a shared outcome must be counted once, not twice.
Mark membership before counting
| Token | In A? | In B? | Counted in A or B? |
|---|---|---|---|
| 1 | No | No | No |
| 2 | Yes | No | Yes |
| 3 | No | No | No |
| 4 | Yes | Yes | Yes |
| 5 | No | Yes | Yes |
| 6 | Yes | Yes | Yes |
Here, A = {2, 4, 6}, B = {4, 5, 6}, and the overlap is {4, 6}. The union, meaning the outcomes in A or B, is {2, 4, 5, 6}. It contains four distinct tokens. Because the six elementary outcomes are equally likely, the probability is 4/6, or 2/3.
Why adding two probabilities directly fails
A tempting calculation is 3/6 + 3/6 = 6/6. That result is too large because tokens 4 and 6 appear in both events. Each of them is included once when counting A and again when counting B. The duplicated total is therefore 2/6. Subtracting the overlap gives the inclusion-and-exclusion calculation:
P(A or B) = P(A) + P(B) − P(A and B) = 3/6 + 3/6 − 2/6 = 4/6.
Inclusive OR includes overlap. Reference: OpenStax’s probability terminology reference.
Inclusive “or” is not exclusive “or”
Inclusive “or” allows an outcome to belong to both A and B, so it includes 2, 4, 5, and 6. Exclusive “or” keeps only outcomes in exactly one event. In this example those are 2 and 5, because 4 and 6 are shared. Thus exclusive-or probability is 2/6, whereas inclusive-or probability is 4/6. The practical rule is simple: list the union, remove repeated labels, then divide the number of remaining outcomes by the total.
For a different question about interpreting past observations, see our history-screen reading note. The hypothetical counts above are not observed results.