Probability Addition Rule
\( P(A\cup B)=P(A)+P(B)-P(A\cap B) \)
Probability
GCSE
∑ π √ ≈
\( P(A)=0.55, P(B)=0.35, P(A∩B)=0.15. Find P(A∪B). \)
Explanation
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Statement
The probability of event \(A\) or event \(B\) happening is given by:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Why it’s true
- \(P(A)\) counts all outcomes in event A.
- \(P(B)\) counts all outcomes in event B.
- But if we just add them, the overlap (\(P(A \cap B)\)) is counted twice.
- So we subtract \(P(A \cap B)\) to avoid double-counting.
Recipe (how to use it)
- Find \(P(A)\).
- Find \(P(B)\).
- Find \(P(A \cap B)\) if there is overlap.
- Apply formula: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
Spotting it
Used whenever you are asked for the probability of "A or B" happening. The overlap must be subtracted if events are not mutually exclusive.
Common pairings
- Venn diagrams and set notation.
- Mutually exclusive events (\(P(A \cap B)=0\)).
- Conditional probability in more complex cases.
Mini examples
- Given: \(P(A)=0.3\), \(P(B)=0.4\), \(P(A \cap B)=0.1\).
Answer: \(0.3+0.4-0.1=0.6\).
- Given: \(P(A)=0.5\), \(P(B)=0.2\), \(P(A \cap B)=0\).
Answer: \(0.5+0.2=0.7\).
Pitfalls
- Forgetting to subtract the overlap.
- Assuming events are always mutually exclusive (when they may overlap).
Exam strategy
- Draw a Venn diagram to visualize overlaps.
- Check if events are independent or mutually exclusive.
- Probabilities must always stay between 0 and 1.
Summary
The addition rule ensures correct calculation of "A or B". It prevents double-counting overlap: \[P(A \cup B) = P(A)+P(B)-P(A \cap B)\].
Worked examples
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\( P(A)=0.3, P(B)=0.4, P(A∩B)=0.1. Find P(A∪B). \)
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P(A)+P(B)-P(A∩B)
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\( 0.3+0.4-0.1=0.6 \)
Answer:
0.6
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\( P(A)=0.5, P(B)=0.2, P(A∩B)=0.1. Find P(A∪B). \)
-
\( 0.5+0.2-0.1=0.6 \)
Answer:
0.6
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\( P(A)=0.4, P(B)=0.3, P(A∩B)=0.2. Find P(A∪B). \)
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\( 0.4+0.3-0.2=0.5 \)
Answer:
0.5
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\( P(A)=0.6, P(B)=0.1, P(A∩B)=0.05. Find P(A∪B). \)
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\( 0.6+0.1-0.05=0.65 \)
Answer:
0.65
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\( P(A)=0.7, P(B)=0.4, P(A∩B)=0.2. Find P(A∪B). \)
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\( 0.7+0.4-0.2=0.9 \)
Answer:
0.9
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\( P(A)=0.25, P(B)=0.5, P(A∩B)=0.05. Find P(A∪B). \)
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\( 0.25+0.5-0.05=0.7 \)
Answer:
0.7
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\( P(A)=0.8, P(B)=0.3, P(A∩B)=0.25. Find P(A∪B). \)
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\( 0.8+0.3-0.25=0.85 \)
Answer:
0.85
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\( P(A)=0.55, P(B)=0.35, P(A∩B)=0.15. Find P(A∪B). \)
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\( 0.55+0.35-0.15=0.75 \)
Answer:
0.75
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\( P(A)=0.9, P(B)=0.4, P(A∩B)=0.3. Find P(A∪B). \)
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\( 0.9+0.4-0.3=1.0 \)
Answer:
1.0
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\( P(A)=0.65, P(B)=0.5, P(A∩B)=0.3. Find P(A∪B). \)
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\( 0.65+0.5-0.3=0.85 \)
Answer:
0.85