MISSION · Data and probability
The Addition Rule and Mutually Exclusive Events
Add the two chances, then take back what you counted twice
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The idea in one sentence
To find the chance of A or B you add their chances and then subtract the overlap, because anything in both was counted twice.
The rule
\[P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\]Draw a Venn diagram and it is obvious. Shading circle A shades the middle. Shading circle B shades the middle again. The middle now has two coats of paint and every other region has one, so subtracting the middle once puts things right.
With \(P(A) = 0.5\), \(P(B) = 0.3\) and \(P(A \text{ and } B) = 0.1\):
\[P(A \text{ or } B) = 0.5 + 0.3 - 0.1 = 0.7\]Note that “or” here is inclusive — it means A, or B, or both. Everyday English often means one or the other but not both, and mathematics does not.
Mutually exclusive events
Two events are mutually exclusive (or disjoint) when they cannot both happen. Rolling a 2 and rolling a 5 on one die. Drawing a card that is both a heart and a spade. A coin landing heads and tails.
Then \(P(A \text{ and } B) = 0\), there is no overlap to subtract, and the rule simplifies:
\[P(A \text{ or } B) = P(A) + P(B)\]This is why the simple version works so often in early problems — the events picked are usually different faces of the same die, which cannot co-occur. It is also why the simple version fails the moment a question involves two different attributes.
Cards make the trap concrete. \(P(\text{heart}) = \frac{13}{52}\) and \(P(\text{king}) = \frac{4}{52}\). Adding gives \(\frac{17}{52}\), which is wrong: the king of hearts got counted twice. The overlap is \(\frac{1}{52}\), so
\[P(\text{heart or king}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} \approx 0.308\]Mutually exclusive is not independent
These two words get swapped constantly, and they are close to opposites.
Independent means one event happening tells you nothing about the other: \(P(A \text{ and } B) = P(A)P(B)\).
Mutually exclusive means one happening tells you the other definitely did not: \(P(A \text{ and } B) = 0\).
So two mutually exclusive events with non-zero probabilities are about as dependent as events get. Independence goes with multiplying for “and”; mutual exclusivity goes with plain adding for “or”.
The complement
The chance of something not happening is whatever is left over:
\[P(\text{not } A) = 1 - P(A)\]If \(P(\text{rain}) = 0.35\) then \(P(\text{no rain}) = 0.65\). An event and its complement are mutually exclusive and between them cover everything, which is why they sum to 1.
This is the most useful line in the topic, because “at least one” questions are almost always easier backwards. The chance of at least one head in five flips is a five-case sum the hard way, or \(1 - P(\text{no heads}) = 1 - \left(\frac12\right)^5 = \frac{31}{32}\) the easy way.
Common mix-ups
- Forgetting to subtract the overlap. Check whether the two events can happen together before adding.
- Getting a probability above 1. A sure sign of double-counting.
- Confusing mutually exclusive with independent.
- Adding for “and”. “And” narrows the outcomes, so its probability is never bigger than either part. “Or” widens, so it is never smaller.
- Reading “or” as exclusive. Unless the question says “but not both”, both counts.
Check yourself
With \(P(A) = 0.5\), \(P(B) = 0.3\) and \(P(A \text{ and } B) = 0.1\), find \(P(A \text{ or } B)\). Then give \(P(\text{heart or king})\) as a decimal, and \(P(\text{no rain})\) when \(P(\text{rain}) = 0.35\).
\(0.5 + 0.3 - 0.1 = 0.7\). For the cards, \(\frac{13 + 4 - 1}{52} = \frac{16}{52} \approx 0.308\). And \(1 - 0.35 = 0.65\).
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