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Lesson 6 of 8

Probability · Lesson 6 of 8

The Addition Rule for Events

“Correct for overlapping outcomes when calculating the probability that at least one event occurs.”

Learning Objectives

• Explain why adding overlapping event probabilities counts some outcomes twice. • Derive the two-event addition rule by splitting sets into disjoint regions. • Use the mutually exclusive special case and solve for an unknown intersection. • Check the consistency of proposed event probabilities. • Derive and interpret the three-event inclusion–exclusion formula.

Where a simple sum goes wrong

The probability axiom allows direct addition when events are mutually exclusive. When they overlap, the union still includes each elementary outcome only once, but the sum of their separate probabilities includes shared outcomes twice. We need a correction equal to the probability of the overlap.

This is a question about membership, so it works for unequal as well as equal outcome weights. Each shared point contributes its full probability to both event sums. Removing one copy of its weight leaves the single copy required by the union.

Example — Two overlapping three-toss events

Problem
Three fair tosses have eight equally likely sequences. Let A={HHT,HTH,THH} and B={HTH,THH,HHH}. Compare P(A∪B) with P(A)+P(B).

  1. 1.Each set has three members, so P(A)=P(B)=3/8.
  2. 2.The union is {HHT,HTH,THH,HHH}, with four members. Thus P(A∪B)=4/8=1/2.
  3. 3.The simple sum gives 3/8+3/8=6/8. It counts HTH and THH twice.
  4. 4.The intersection is {HTH,THH}, with probability 2/8. Subtracting one copy gives 6/8−2/8=4/8, matching the direct union count.

The event A or B means A alone, B alone, or both. It does not mean exactly one. The addition rule retains one copy of the overlap; it does not remove the overlap altogether.

Addition rule for two eventsLaTeX
P(A∩B) is the probability of both events. Subtract it once to correct the two copies already present in P(A)+P(B).
SABA − BA ∩ BB − ANeither A nor B
The regions behind the addition rule— The union consists of A−B, A∩B, and B−A. The middle region is included in both separate event totals.

Deriving the rule from three disjoint pieces

A Venn diagram suggests the rule, but the axioms can prove it. Split the union into its A-only, shared, and B-only parts. These three sets are pairwise disjoint, so the probability of their union is the sum of their probabilities.

A disjoint decompositionLaTeX
No outcome belongs to two of these three pieces, and together they contain every point of the union.

Let x=P(A−B), y=P(A∩B), and z=P(B−A). Disjoint additivity gives P(A∪B)=x+y+z. Also A is the union of its first two pieces, so P(A)=x+y; similarly P(B)=y+z. Adding these last two equations gives x+2y+z. Subtract y to recover x+y+z.

The double-counting proofLaTeX
x,y,z name the weights of the three disjoint Venn regions; they need not be equal.

A second proof uses only one extra region. Write the union as A together with B−A, which is disjoint from A. Also split B into its shared part A∩B and B−A. Both equations follow immediately from disjoint additivity.

An alternative derivationLaTeX
Subtracting the second equation from the first cancels the common B−A term. Rearrangement gives the same addition rule.

The disjoint case and missing values

When A and B are mutually exclusive, their intersection is the empty event. Its probability is zero, so the correction disappears. This confirms that the general rule agrees with the original axiom rather than replacing it.

The rule can also be rearranged to find an intersection, a union, or one event probability from the other three quantities. Identify the meaning of each given number before substituting, because a percentage for both is not interchangeable with a percentage for either.

Mutually exclusive special caseLaTeX
The intersection probability vanishes. The statement follows from the sets being disjoint, not from their probabilities being equal.
Example — Two school subjects

Problem
In a group, 40% study Mathematics, 30% study Biology, and 10% study both. For a uniformly chosen student, find the probability of studying at least one of these subjects.

  1. 1.Let M and B denote the two subject events. Convert percentages to probabilities: P(M)=0.40, P(B)=0.30, P(M∩B)=0.10.
  2. 2.Use the addition rule: P(M∪B)=0.40+0.30−0.10=0.60.
  3. 3.The shared students contributed to both percentages, so one copy of their 0.10 must be removed. The answer is 60%, not 70%.
Example — Finding the shared part

Problem
A student qualifies in the first assessment with probability 0.8 and in the second with probability 0.7. The probability of qualifying in at least one is 0.95. Find the probability of qualifying in both.

  1. 1.Let A and B represent the two qualifications. At least one is the union, so P(A∪B)=0.95.
  2. 2.Rearrange: P(A∩B)=P(A)+P(B)−P(A∪B).
  3. 3.Substitute to obtain 0.8+0.7−0.95=0.55.
  4. 4.The shared probability 0.55 is no larger than either individual probability, which is a necessary consistency check.
Example — Completing an addition-rule table

Problem
Complete: (i) P(A)=1/3, P(B)=1/5, P(A∩B)=1/15; (ii) P(A)=0.35, P(A∩B)=0.25, P(A∪B)=0.60; (iii) P(A)=0.50, P(B)=0.35, P(A∪B)=0.70.

  1. 1.For (i), the union is 1/3+1/5−1/15=(5+3−1)/15=7/15.
  2. 2.For (ii), P(B)=P(A∪B)−P(A)+P(A∩B)=0.60−0.35+0.25=0.50.
  3. 3.For (iii), the intersection is 0.50+0.35−0.70=0.15.
  4. 4.In every row, the union contains each event and the intersection is contained in each. The resulting probabilities respect these relationships.
CaseP(A)P(B)P(A∩B)P(A∪B)
i1/31/51/157/15
ii0.350.500.250.60
iii0.500.350.150.70

Do the proposed probabilities fit together?

Numbers can individually lie between zero and one while describing an impossible combination of event relationships. Because A∩B is contained in A and B, it cannot carry more weight than either. Because A∪B contains both, it cannot carry less weight than either. The unshared pieces and the outside region must also have nonnegative weights.

Example — An inconsistent proposal and a consistent one

Problem
Check (i) P(A)=0.5, P(B)=0.7, P(A∩B)=0.6; (ii) P(A)=0.5, P(B)=0.4, P(A∪B)=0.8.

  1. 1.In (i), A∩B⊆A, but its proposed weight 0.6 exceeds P(A)=0.5. This would make the A-only region have weight 0.5−0.6=−0.1, so the proposal is invalid.
  2. 2.In (ii), derive P(A∩B)=0.5+0.4−0.8=0.1.
  3. 3.The A-only, shared, and B-only regions have weights 0.4,0.1,0.3. The outside region has the remaining weight 0.2.
  4. 4.All four weights are nonnegative and total 1, so a model with those region weights is consistent.
Common mistake

Do not replace P(A∩B) by P(A)P(B) without a justified additional assumption. This chapter’s addition rule uses the intersection probability supplied or derived from the event information.

Extending the same idea to three events

For three events, an outcome may occur in one event, two events, or all three. Add the three event weights, then subtract the three pairwise overlaps. A point belonging to all three was initially counted three times and then subtracted three times. It has disappeared, so one copy of the triple overlap must be added back.

Three-event inclusion–exclusionLaTeX
Each pairwise intersection includes any points shared by all three. The final positive term corrects their excessive subtraction.
SA onlyB onlyAB onlyAC onlyBC onlyC onlyABCABCTriple-region copies: 3 − 3 + 1 = 1
Seven disjoint regions of three events— AB only means A and B but not C; the central ABC region belongs to all three. Pairwise intersection terms include that centre.
Example — Deriving the three-event formula

Problem
Prove the three-event formula using the two-event rule and set distributivity.

  1. 1.Put E=B∪C. Then P(A∪B∪C)=P(A)+P(E)−P(A∩E).
  2. 2.Use the two-event rule on E: P(E)=P(B)+P(C)−P(B∩C).
  3. 3.Distribute intersection: A∩E=(A∩B)∪(A∩C). Apply the two-event rule again to obtain P(A∩E)=P(A∩B)+P(A∩C)−P(A∩B∩C).
  4. 4.Substitute both expressions into the first equation. Subtracting the last expression changes its final minus sign into a plus.
  5. 5.The result is the sum of the three single-event probabilities, minus the three pairwise intersections, plus the triple intersection.
Example — Three overlapping interests

Problem
Let P(A)=0.40, P(B)=0.35, P(C)=0.30; pairwise intersections are 0.10,0.08,0.06 for AB,AC,BC; the triple intersection is 0.02. Find the probability of at least one interest.

  1. 1.The initial sum is 0.40+0.35+0.30=1.05. A sum above one is possible at this intermediate stage because overlaps are counted repeatedly.
  2. 2.Subtract the pairwise terms: 1.05−0.10−0.08−0.06=0.81.
  3. 3.Add back the triple intersection: 0.81+0.02=0.83.
  4. 4.The resulting union probability is 0.83. Omitting the triple correction would lose all outcomes lying in all three events.

Quiz

Quick check

If P(A)=0.6, P(B)=0.55, and P(A∪B)=0.9, what is P(A∩B)?

Quick check

Why subtract P(A∩B) once in the union rule?

Quick check

Which proposed values are inconsistent?

Quick check

For mutually exclusive events of probabilities 3/5 and 1/5, what is the union probability?

Quick check

In the three-event formula, what is the sign of the triple-intersection term?

Quick check

The sample-space weights of A-only, overlap, and B-only are 0.2,0.3,0.1. What is P(A∪B)?

Practice Problems

Practice Problems
  1. Check whether these assignments are consistent: (i) P(A)=0.5, P(B)=0.7, P(A∩B)=0.6; (ii) P(A)=0.5, P(B)=0.4, P(A∪B)=0.8. Justify each decision using region weights.
  2. Complete the missing entries in the table: (i) P(A)=1/3, P(B)=1/5, P(A∩B)=1/15, find the union; (ii) P(A)=0.35, P(A∩B)=0.25, P(A∪B)=0.6, find P(B); (iii) P(A)=0.5, P(B)=0.35, P(A∪B)=0.7, find the intersection.
  3. For mutually exclusive A,B with probabilities 3/5 and 1/5, find P(A or B). Explain why the overlap correction is zero.
  4. In a class, 40% study Mathematics, 30% Biology, and 10% both. A student is selected uniformly. Find the probability of studying at least one of the subjects.
  5. The chances of qualifying in the first and second assessments are 0.8 and 0.7, while qualifying in at least one has chance 0.95. Find the chance of both.
  6. Reconstruct the three-event addition formula from the two-event formula by putting E=B∪C. Explain the positive triple-intersection term with a sample point shared by all three.
  7. Create a consistent example with unequal event probabilities and a nonzero overlap. Label all four Venn-region weights and verify the union formula.

Key Takeaways

Key Takeaways

• The union keeps one copy of every shared outcome. • P(A∪B)=P(A)+P(B)−P(A∩B), for equal or unequal outcome weights. • Disjoint sets give a zero correction and recover direct addition. • Probability assignments must respect subset relationships and nonnegative region weights. • For three events, subtract pairwise overlaps and add the triple intersection back once.

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Counting Arrangements and Selections

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Complements and Multi-Step Probability