Probability Addition Rule

\( P(A\cup B)=P(A)+P(B)-P(A\cap B) \)
Probability GCSE

\( P(A)=0.15, P(B)=0.5, P(A∩B)=0.05. Find P(A∪B). \)

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
Use addition rule.

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)

  1. Find \(P(A)\).
  2. Find \(P(B)\).
  3. Find \(P(A \cap B)\) if there is overlap.
  4. 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

  1. Given: \(P(A)=0.3\), \(P(B)=0.4\), \(P(A \cap B)=0.1\).
    Answer: \(0.3+0.4-0.1=0.6\).
  2. 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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  1. \( P(A)=0.3, P(B)=0.4, P(A∩B)=0.1. Find P(A∪B). \)
    1. P(A)+P(B)-P(A∩B)
    2. \( 0.3+0.4-0.1=0.6 \)
    Answer: 0.6
  2. \( P(A)=0.5, P(B)=0.2, P(A∩B)=0.1. Find P(A∪B). \)
    1. \( 0.5+0.2-0.1=0.6 \)
    Answer: 0.6
  3. \( P(A)=0.4, P(B)=0.3, P(A∩B)=0.2. Find P(A∪B). \)
    1. \( 0.4+0.3-0.2=0.5 \)
    Answer: 0.5
  4. \( P(A)=0.6, P(B)=0.1, P(A∩B)=0.05. Find P(A∪B). \)
    1. \( 0.6+0.1-0.05=0.65 \)
    Answer: 0.65
  5. \( P(A)=0.7, P(B)=0.4, P(A∩B)=0.2. Find P(A∪B). \)
    1. \( 0.7+0.4-0.2=0.9 \)
    Answer: 0.9
  6. \( P(A)=0.25, P(B)=0.5, P(A∩B)=0.05. Find P(A∪B). \)
    1. \( 0.25+0.5-0.05=0.7 \)
    Answer: 0.7
  7. \( P(A)=0.8, P(B)=0.3, P(A∩B)=0.25. Find P(A∪B). \)
    1. \( 0.8+0.3-0.25=0.85 \)
    Answer: 0.85
  8. \( P(A)=0.55, P(B)=0.35, P(A∩B)=0.15. Find P(A∪B). \)
    1. \( 0.55+0.35-0.15=0.75 \)
    Answer: 0.75
  9. \( P(A)=0.9, P(B)=0.4, P(A∩B)=0.3. Find P(A∪B). \)
    1. \( 0.9+0.4-0.3=1.0 \)
    Answer: 1.0
  10. \( P(A)=0.65, P(B)=0.5, P(A∩B)=0.3. Find P(A∪B). \)
    1. \( 0.65+0.5-0.3=0.85 \)
    Answer: 0.85