Compound events as unions and intersections, disjoint events whose probabilities add, the general addition rule for overlapping events, rearranging that rule to find the probability of an intersection, complements, and using a complement to answer an at-least question.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 10 — Counting Methods and Probability
Find Probabilities of Disjoint and Overlapping Events
Objectives
Five outcomes. One addition rule, and one subtraction that keeps it honest.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 707-711 — the lesson these objectives are drawn from
Warm-up
Lesson 10.3 found probabilities by counting outcomes.
Discussion prompt
From a deck of 52, how many cards are hearts? How many are threes? Add those two counts. Now count the cards that are hearts or threes. Do the two answers agree?
Hint: Is any card on both lists?
Answer:
\[ 13 \text{ hearts} + 4 \text{ threes} = 17 \]
But there are only 16 cards that are hearts or threes. The three of hearts appears on both lists and was counted twice.
\[ 13+4-1 = 16 \]
That single subtraction is the whole of this lesson. Adding probabilities works only when the two events share nothing.
Concept
The probability that either of two events occurs is the sum of their probabilities minus the probability that both occur. When the events share no outcomes, that last term is zero and the probabilities simply add.
disjoint events — Two events with no outcomes in common, also called mutually exclusive. Overlapping events share at least one outcome, and their union's probability requires subtracting the shared part.
\[ P(A \text{ or } B) = P(A)+P(B)-P(A \text{ and } B) \]
There is really only one formula. The disjoint case is the general rule with a zero in it, so checking for overlap is the only decision to make.
Figure (svg): Two columns comparing disjoint events with overlapping ones
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 707-707
Section
Section 1
Concept
The union of two events holds every outcome in either one; the intersection holds only the outcomes in both. Two events are disjoint when their intersection is empty and overlapping when it is not.
\[ A \text{ or } B; \qquad A \text{ and } B \]
Deciding whether two events can happen at once is the first step of every problem in this lesson, and it cannot be skipped.
Figure (svg): Three Venn diagrams showing a union, an intersection, and two disjoint events
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 707-707 — Probability of Compound Events
Picture it
A union, an intersection, and a disjoint pair.
Figure (svg): Three Venn diagrams showing a union, an intersection, and two disjoint events
In the third picture the two circles do not touch, so nothing is in both and the subtraction term vanishes.
Worked example
Applying the definitions to card events.
\[ \text{Which pairs are disjoint? A } 10 \text{ and a face card; a heart and a three; a spade and a club; a king and a diamond.} \]
A 10 and a face card
Why: Face cards are jacks, queens and kings only.
A heart and a three
Why: The three of hearts is both.
A spade and a club
Why: No card has two suits.
A king and a diamond
Why: The king of diamonds is both.
Figure (svg): Three Venn diagrams showing a union, an intersection, and two disjoint events
\[ \text{disjoint}, \; \text{overlap}, \; \text{disjoint}, \; \text{overlap} \]
Verify: notice the pattern
Why: Two events about the same attribute — rank against rank, or suit against suit — are disjoint, since a card has only one rank and one suit. Two events about DIFFERENT attributes almost always overlap, since a card has both. That observation settles most card questions without any counting.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 707-707
Sorting
Can one card satisfy both?
Sort into buckets
Sort each pair of card events.
The disjoint pairs all compare like with like — rank with rank or suit with suit — while the overlapping ones mix a rank with a suit.
Worked example
Distinguishing a union from an intersection.
\[ \text{For hearts and threes, describe the union and the intersection and count each.} \]
Describe the union
Why: Every card that is a heart, a three, or both.
Count the union
Why: Thirteen hearts plus the three other threes.
\[ 16\text{ cards} \]
Describe the intersection
Why: Cards that are both a heart and a three.
Count the intersection
Why: Exactly one card.
\[ 1\text{ card} \]
Figure (svg): Two overlapping card events with the shared card subtracted once
\[ |A \cup B| = 16; \quad |A \cap B| = 1 \]
Verify: check the counts add up
Why: Thirteen plus 4 is 17, which exceeds the union's 16 by exactly the size of the intersection. That relationship — the sum of the parts minus the shared part equals the union — is what the addition rule states in probability terms.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 708-708
Trap
\[ P(\text{heart or three}) = \tfrac{13}{52}+\tfrac{4}{52} \]
Add the two probabilities
Why: The events are treated as if they could not both happen.
\[ = \tfrac{17}{52} \quad \text{(wrong)} \]
Only 16 cards are hearts or threes. The three of hearts was counted twice, once as a heart and once as a three.
\[ P = \tfrac{13}{52}+\tfrac{4}{52}-\tfrac{1}{52} = \tfrac{16}{52} \]
Check for shared outcomes before adding
Why: Subtract the intersection whenever there is one.
\[ \text{the } 3 \text{ of hearts is in both events} \]
The safest habit is to use the general rule always, computing the intersection even when you expect it to be zero — which is the case for a 10 and a face card.
Fill the middle
Hearts and threes.
Fill in the blanks
13+4-1 = 16
Why: Thirteen plus 4 is 17, and subtracting the one shared card gives 16. The subtraction removes the double count, not the card itself.
Matching
Or means union; and means intersection.
Match the pairs
Why: The last needs no subtraction, since a card cannot be both a spade and a club. Reading which word appears — or, and, not — identifies the compound event before any arithmetic begins.
Prediction
Commit before reasoning.
Predict first
Can an event and its complement share an outcome?
Correct: No — the complement holds exactly the outcomes the event does not, so they are always disjoint.
\[ P(A)+P(\overline{A}) = 1, \quad P(A \text{ and } \overline{A}) = 0 \]
Why: An outcome is either in an event or not in it, never both, so the intersection is empty by construction. They are also exhaustive: together they cover every outcome. That is why their probabilities add to exactly 1, which is the complement rule of the fourth idea. This is lesson exercise 2's question, and its answer is a definition rather than a computation.
Section
Section 2
Concept
When two events cannot both occur, the probability that either occurs is simply the sum of their probabilities. Nothing has to be subtracted, because nothing was counted twice.
\[ P(A \text{ or } B) = P(A)+P(B) \]
This is the general rule with the intersection term equal to zero, so it is a special case rather than a separate formula.
Figure (svg): Two disjoint card events whose probabilities simply add
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 707-707 — Find probability of disjoint events
Picture it
Example 1: a 10 or a face card.
Figure (svg): Two disjoint card events whose probabilities simply add
Four tens and 12 face cards, with no card in both, gives 16 out of 52 — about 0.308.
Worked example
Example 1.
\[ \text{From a } 52 \text{-card deck, find } P(10 \text{ or face card}). \]
Count each event
Why: Four tens; 12 face cards.
\[ 4\text{ and } 12 \]
Check for overlap
Why: A 10 is not a jack, queen or king.
Add the probabilities
Why: Four over 52 plus 12 over 52.
\[ \frac{16}{52} \]
Simplify
Why: Divide both by 4.
\[ \frac{4}{13},\text{ about } 0.308 \]
Figure (svg): Two disjoint card events whose probabilities simply add
\[ \tfrac{4}{52}+\tfrac{12}{52} = \tfrac{16}{52} = \tfrac{4}{13} \]
Verify: count the union directly
Why: Four tens plus 12 face cards is 16 cards in the union, and 16 over 52 is 4 over 13 — matching. Because the events are disjoint, adding the counts and adding the probabilities give the same answer, which is exactly what disjoint means.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 707-707
Fill the middle
Example 1.
Fill in the blanks
\frac16___+\frac______ = \frac___}___
Why: Four plus 12 is 16, and no subtraction is needed since no card is both a 10 and a face card. The result simplifies to 4 over 13.
Worked example
Guided Practice 1 and lesson exercise 22.
\[ \text{Find } P(\text{ace or eight}) \text{ and } P(\text{spade or club}). \]
Aces and eights: count
Why: Four of each, with no card both.
\[ 4 + 4 \]
Aces and eights: add
Why: Eight over 52.
\[ \frac{2}{13} \]
Spades and clubs: count
Why: Thirteen of each, no card both.
\[ 13 + 13 \]
Spades and clubs: add
Why: Twenty-six over 52.
\[ \frac{1}{2} \]
Figure (svg): The solution to Worked example two more disjoint unions shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{8}{52} = \tfrac{2}{13}; \qquad \tfrac{26}{52} = \tfrac{1}{2} \]
Verify: sanity-check the second
Why: Spades and clubs are exactly the black cards, which are half the deck — so one half is right without any arithmetic. Recognising a familiar event behind a compound description is a useful shortcut and a good check.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 708-710
Error analysis
A student computes the probability of drawing a king or a diamond.
Annotate
On: \( P = \frac{4}{52}+\frac{13}{52} = \frac{17}{52} \)
A rank and a suit almost always overlap. The habit of asking whether one card could satisfy both takes a second and prevents this every time.
Sorting
Check for shared outcomes first.
Sort into buckets
Sort each probability question.
Three of five need no subtraction here, but the two that do would each be wrong by one card — a small error that a habit of checking removes entirely.
Matching
Add, subtracting only when they overlap.
Match the pairs
Why: The first and last give the same probability by different routes: one adds 4 and 12 with no overlap, the other adds 13 and 4 and subtracts 1. Equal answers do not mean equal reasoning.
Prediction
Commit before reasoning.
Predict first
Adding P(A) equal to 0.6 and P(B) equal to 0.7 gives 1.3. What does that tell you?
Correct: The events must overlap, and the intersection has probability at least 0.3.
\[ P(A \text{ and } B) \geq P(A)+P(B)-1 \]
Why: Since P of A or B cannot exceed 1, the addition rule forces P of A and B to be at least 0.6 plus 0.7 minus 1, which is 0.3. Two events with probabilities summing above 1 cannot possibly be disjoint — there is not enough room. The impossible sum is a signal about the events rather than an error in the arithmetic.
Section
Section 3
Concept
When two events share outcomes, adding their probabilities counts the shared outcomes twice. Subtracting the probability of the intersection once restores the correct total.
\[ P(A \text{ or } B) = P(A)+P(B)-P(A \text{ and } B) \]
The rule covers disjoint events too, since the subtracted term is zero there — so it is the only formula that needs remembering.
Figure (svg): Two overlapping card events with the shared card subtracted once
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 708-708 — Probability of compound events
Picture it
Example 2: a heart or a three.
Figure (svg): Two overlapping card events with the shared card subtracted once
Thirteen plus 4 is 17, but the union holds only 16 cards. Subtracting the three of hearts once fixes the count.
Worked example
Example 2, a multiple-choice item.
\[ \text{From a } 52 \text{-card deck, find } P(\text{heart or three}). \]
Count each event
Why: Thirteen hearts; 4 threes.
\[ 13\text{ and } 4 \]
Find the intersection
Why: Only the three of hearts is both.
\[ 1\text{ card} \]
Apply the rule
Why: Thirteen over 52 plus 4 over 52 minus 1 over 52.
\[ \frac{16}{52} \]
Simplify
Why: Divide both by 4.
\[ \frac{4}{13} \]
Figure (svg): Two overlapping card events with the shared card subtracted once
\[ \tfrac{13}{52}+\tfrac{4}{52}-\tfrac{1}{52} = \tfrac{4}{13} \]
Verify: list the union
Why: The 13 hearts plus the three of spades, clubs and diamonds gives 16 cards — matching. Listing the union directly is a good check whenever the counts are small enough, and it is exactly what the rule is designed to compute when they are not.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 708-708
Fill the middle
Example 2.
Fill in the blanks
\frac1___+\frac______-\frac___}___ = \frac______
Why: Exactly one card, the three of hearts, is both — so 1 over 52 is subtracted. Subtracting more would remove the card from the union entirely.
Worked example
Guided Practice 2 and lesson exercise 21.
\[ \text{Find } P(10 \text{ or diamond}) \text{ and } P(\text{king or diamond}). \]
Ten or diamond: count
Why: Four tens, 13 diamonds, and the ten of diamonds in both.
\[ 4, 13, 1 \]
Ten or diamond: apply the rule
Why: Four plus 13 minus 1.
\[ \frac{16}{52} = \frac{4}{13} \]
King or diamond: count
Why: Four kings, 13 diamonds, one shared.
\[ 4, 13, 1 \]
King or diamond: apply the rule
Why: The same arithmetic.
\[ \frac{16}{52} = \frac{4}{13} \]
Figure (svg): The solution to Worked example two more overlapping unions shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{16}{52} = \tfrac{4}{13} \text{ in both cases} \]
Verify: explain why the two agree
Why: Any single rank paired with any single suit gives 4 plus 13 minus 1, which is always 16 — because exactly one card has both that rank and that suit. So every such question has the same answer, which is worth knowing and worth being able to explain.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 708-710
Trap
\[ P(\text{heart or three}) = \tfrac{13}{52}+\tfrac{4}{52}-\tfrac{2}{52} \]
Remove the shared card from both events
Why: The overlap is subtracted once for each event it appears in.
\[ = \tfrac{15}{52} \quad \text{(wrong)} \]
The card should be counted once in the union, not zero times. Subtracting twice removes it entirely.
\[ 13+4-1 = 16 \]
Subtract the overlap exactly once
Why: It was counted twice and should be counted once, so one copy is removed.
\[ P = \tfrac{16}{52} = \tfrac{4}{13} \]
The three of hearts IS in the union — it is a heart, after all. The subtraction corrects a double count rather than excluding the card.
Comparison
Fill the blanks. One overlaps and one does not.
Comparison matrix
| Question | 10 or face card | Heart or three |
|---|---|---|
| Counts | 4 and 12 | 13 and 4 |
| Overlap | none | 1 card |
| Computation | 4 + 12 | 13 + 4 - 1 |
| Probability | 16/52 = 4/13 | 16/52 = 4/13 |
Both come to 4 over 13, but only one needed a subtraction. Equal answers arriving by different routes is a coincidence of these particular numbers.
Fill the middle
Lesson exercise 9.
Fill in the blanks
P(A)=0.5, \; P(B)=0.35, \; P(A\text0.65B)=0.2 \;\Longrightarrow\; P(A\text___B) = ___
Why: Point five plus 0.35 is 0.85, minus 0.2 gives 0.65. The rule works with probabilities given directly, with no counting at all.
Prediction
Commit before reasoning.
Predict first
You know P(A) and P(B) but not P(A and B). What can you say about P(A or B)?
Correct: It is at most P(A) + P(B), and at least the larger of the two.
\[ \max(P(A),P(B)) \leq P(A \text{ or } B) \leq P(A)+P(B) \]
Why: The intersection is at least zero, giving the upper bound, and at most the smaller of the two probabilities, giving the lower bound. So a union is never larger than the sum and never smaller than either event alone — which makes sense, since the union contains each event entirely. Those bounds are often enough to check an answer even without the intersection.
Section
Section 4
Concept
The addition rule connects four quantities, so knowing any three determines the fourth. If the union and both individual probabilities are known, the intersection follows by rearranging.
\[ P(A \text{ and } B) = P(A)+P(B)-P(A \text{ or } B) \]
The rearranged form has a plain reading: the amount by which the two counts overshoot the union is exactly the amount counted twice.
Figure (svg): The addition rule rearranged to find the probability of the intersection
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 708-708 — Use a formula to find P(A and B)
Picture it
Example 3: seniors who are athletes or on the honour roll.
Figure (svg): The addition rule rearranged to find the probability of the intersection
Seventy-four plus 51 is 125, which exceeds 113 by 12 — so 12 seniors are both, a probability of 0.06.
Worked example
Example 3.
\[ \text{Of } 200 \text{ seniors, } 74 \text{ are athletes, } 51 \text{ are on the honour roll, and } 113 \text{ are either. Find } P(\text{both}). \]
Write the rule
Why: The union equals the sum minus the intersection.
Substitute
Why: One hundred thirteen over 200 equals 74 over 200 plus 51 over 200 minus the unknown.
\[ \frac{113}{200} = \frac{125}{200} - x \]
Rearrange
Why: The unknown is 125 over 200 minus 113 over 200.
\[ \frac{12}{200} \]
Simplify
Why: Twelve over 200 is 3 over 50.
\[ 0.06 \]
Figure (svg): The addition rule rearranged to find the probability of the intersection
\[ P(A \text{ and } B) = \tfrac{12}{200} = 0.06 \]
Verify: check the counts add up
Why: Twelve are both, so 62 are athletes only and 39 are on the honour roll only, totalling 113 in the union — matching the given figure. And 200 minus 113 leaves 87 seniors who are neither, which is a sensible remainder rather than a negative number.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 708-708
Fill the middle
Example 3.
Fill in the blanks
P(A\text113B) = \frac______+\frac______-\frac___}___
Why: The union count, 113, is subtracted from the sum of the two individual counts. The result, 12, is exactly the amount by which the sum overshot.
Worked example
Guided Practice 3 and lesson exercise 10.
\[ \text{With } 32 \text{ in band and } 64 \text{ in band or honour roll out of } 200; \text{ and } P(A)=0.6, P(B)=0.2, P(A\text{ or }B)=0.7. \]
First: substitute
Why: Thirty-two plus 51 minus 64, all over 200.
\[ \frac{19}{200} \]
First: simplify
Why: Nineteen over 200.
\[ 0.095 \]
Second: substitute
Why: Point six plus 0.2 minus 0.7.
\[ 0.1 \]
Second: interpret
Why: Ten percent of outcomes lie in both events.
Figure (svg): The solution to Worked example two more rearrangements shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{19}{200} = 0.095; \qquad 0.1 \]
Verify: check both for plausibility
Why: An intersection can never exceed either individual probability: 0.095 is below both 0.16 and 0.255, and 0.1 is below both 0.6 and 0.2. A computed intersection larger than one of the events would signal an arithmetic error or inconsistent data.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 708-710
Error analysis
A student solves the addition rule for the intersection.
Annotate
On: \( P(A \text{ and } B) = \frac{113}{200}-\frac{74}{200}-\frac{51}{200} = -\frac{12}{200} \)
Isolating the unknown one step at a time, rather than rearranging by eye, prevents this — and the sign check catches it if it happens anyway.
Ranking
Finding an intersection from a union.
Put in order
Why: Step five is the check worth building in: an intersection cannot exceed either event, since everything in both is in each. A value outside that range means either an arithmetic slip or data that could not have come from a real population.
Sorting
The rule has four slots.
Sort into buckets
Sort each problem by what it asks for.
The same equation serves both directions, which is why it is worth writing out in full before substituting anything.
Prediction
Commit before reasoning.
Predict first
Seventy-four athletes plus 51 honour students is 125, but only 113 seniors are either. What is the 12?
Correct: The number of seniors counted twice, who are both.
\[ 125-113 = 12 \text{ counted twice} \]
Why: Every senior who is both an athlete and on the honour roll appears in both the 74 and the 51, so the sum counts them twice while the union counts them once. The difference is therefore exactly how many such seniors there are. Reading the rearranged rule this way turns it from a formula into a sentence, and makes the result checkable against the situation.
Section
Section 5
Concept
The complement of an event holds every outcome not in it. Since the two together cover everything and share nothing, their probabilities add to 1, so the complement's probability is 1 minus the event's.
\[ P(\overline{A}) = 1-P(A) \]
This turns an at-least or a not question into whichever of the two is easier to count, and for many questions the complement is far the shorter list.
Figure (svg): A grid of the thirty-six dice outcomes with a complement highlighted
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 709-709 — Probability of the Complement of an Event
Picture it
Example 4: two dice, with the sums above 9 shaded.
Figure (svg): A grid of the thirty-six dice outcomes with a complement highlighted
Counting the six shaded cells and subtracting from 1 is far quicker than counting the thirty unshaded ones.
Worked example
Example 4, both parts.
\[ \text{Rolling two dice, find } P(\text{sum is not } 6) \text{ and } P(\text{sum at most } 9). \]
Count the sums equal to 6
Why: Five of the 36 cells.
\[ \frac{5}{36} \]
Take the complement
Why: One minus five 36ths.
\[ \frac{31}{36},\text{ about } 0.861 \]
Count the sums above 9
Why: Three for 10, two for 11, one for 12.
\[ \frac{6}{36} \]
Take the complement
Why: One minus six 36ths.
\[ \frac{30}{36} = \frac{5}{6},\text{ about } 0.833 \]
Figure (svg): A grid of the thirty-six dice outcomes with a complement highlighted
\[ \tfrac{31}{36}; \qquad \tfrac{30}{36} = \tfrac{5}{6} \]
Verify: check the second by direct count
Why: Counting the cells with sums at most 9 gives 30, matching 36 minus 6. Both routes agree, and the complement route needed six cells counted rather than thirty — a saving that grows with the size of the grid.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 709-709
Fill the middle
Example 4a.
Fill in the blanks
P(\text31 6) = 1-\frac______ = \frac___}___
Why: Thirty-six minus 5 is 31. Counting the five cells that give a sum of 6 is far quicker than counting the thirty-one that do not.
Worked example
Guided Practice 4 to 7.
\[ \text{Find } P(\overline{A}) \text{ when } P(A) = 0.45, \; \tfrac{1}{4}, \; 1, \; 0.03. \]
First
Why: One minus 0.45.
\[ 0.55 \]
Second
Why: One minus one quarter.
\[ \frac{3}{4} \]
Third
Why: One minus 1.
\[ 0 \]
Fourth
Why: One minus 0.03.
\[ 0.97 \]
Figure (svg): The solution to Worked example four complements shown as a ladder of expressions, one row per algebraic move
\[ 0.55, \; \tfrac{3}{4}, \; 0, \; 0.97 \]
Verify: check the two extremes
Why: A certain event has an impossible complement, and the third case shows exactly that: 1 gives 0. And a very unlikely event has a nearly certain complement, as the fourth shows. Both extremes behave sensibly, which is a sign the rule is stated correctly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 709-709
Trap
\[ P(\text{sum at most } 9) \]
Subtract the probability of a sum of 9
Why: The boundary value alone is treated as the complement.
\[ 1-\tfrac{4}{36} = \tfrac{32}{36} \quad \text{(wrong)} \]
The complement of at most 9 is MORE than 9, which covers sums of 10, 11 and 12 — six cells, not four.
\[ 1-P(\text{sum} > 9) = 1-\tfrac{6}{36} = \tfrac{30}{36} \]
Complement the whole condition, not one value
Why: Everything failing at most 9 must be included.
\[ \text{sums } 10, 11, 12: \; 3+2+1 = 6 \text{ cells} \]
Writing the complement out in words before counting anything prevents this: the opposite of at most 9 is greater than 9, not equal to 9.
Matching
Everything not in the event.
Match the pairs
Why: The second row is the one that catches people: the opposite of at most 9 is more than 9, with the boundary value staying on the original side. Writing the complement in words first settles it.
Sorting
Count the shorter list.
Sort into buckets
Sort each event by whether the event or its complement is quicker to count.
At least almost always signals the complement route, since the opposite of at least one is none — a single case rather than many.
Prediction
Commit before reasoning.
Predict first
Why is the probability of an event plus the probability of its complement always exactly 1?
Correct: Because together they contain every outcome and share none, so their counts total the whole.
\[ P(A)+P(\overline{A}) = \frac{|A|+|\overline{A}|}{|\text{total}|} = 1 \]
Why: Every outcome is in exactly one of the two, so the two counts add to the total and the two probabilities add to 1. Being disjoint means the addition rule applies with no subtraction; being exhaustive means the sum is the whole. This holds for experimental and geometric probabilities as well, since it depends only on the two events partitioning everything.
Comparison
Fill the blanks. One rule with three faces.
Comparison matrix
| Situation | Rule | Why |
|---|---|---|
| Any two events | P(A or B) = P(A) + P(B) - P(A and B) | the shared outcomes were counted twice |
| Disjoint events | P(A or B) = P(A) + P(B) | nothing is shared |
| The intersection is unknown | P(A and B) = P(A) + P(B) - P(A or B) | the same rule rearranged |
| The complement | P(not A) = 1 - P(A) | the two together cover everything and share nothing |
All four rows are consequences of one idea: count every outcome exactly once.
Pattern
Read the wording, then choose the rule.
Two events about the same attribute are usually disjoint; two events about different attributes usually overlap.
OpenStax Algebra and Trigonometry 2e, §13.7 Probability §13.7
Check
Check for overlap first.
Check your understanding
What is the probability of drawing a heart or a three?
Answer: A
Why: It is 13/52 plus 4/52 minus 1/52, which is 16/52.
Check
Rearrange the rule.
Check your understanding
Of 200 seniors, 74 are athletes, 51 are on the honour roll and 113 are either. What is P(both)?
Answer: A
Why: 74 plus 51 minus 113 is 12, and 12 over 200 is 0.06.
Check
Complement the whole condition.
Check your understanding
Rolling two dice, what is the probability that the sum is at most 9?
Answer: A
Why: The complement is a sum above 9, which happens in 6 of 36 outcomes.
Real world
A restaurant hides one of 500 different messages in each fortune cookie, at random. Five guests each receive one.
Discussion prompt
Find the probability that at least two of the five get the same message, and explain why the answer is so much larger than intuition suggests.
Hint: The complement is that all five differ.
Answer:
\[ \text{all outcomes: } 500^5 \]
\[ \text{all different: } 500 \cdot 499 \cdot 498 \cdot 497 \cdot 496 \]
\[ P(\text{at least two match}) = 1-\frac{500\cdot 499\cdot 498\cdot 497\cdot 496}{500^5} \approx 0.0199 \]
About 2 percent — small, but far larger than the naive guess of five chances in five hundred, which would suggest one percent, and far larger still than most people's instinct.
The reason is that five people give ten PAIRS of people, not five, and each pair has one chance in 500 of matching. Ten chances of one in 500 is about 2 percent, which matches. With only 100 messages the probability rises to about 9.7 percent, and with 23 people and 365 birthdays the same calculation gives just over one half — the famous birthday problem. In every version the complement does the work, because all different is one clean condition while at least two match is a tangle of overlapping cases.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the probability of drawing a heart or a three equal to 13 over 52 plus 4 over 52?
Correct: No — the three of hearts is in both events, so 1 over 52 must be subtracted.
\[ 13+4-1 = 16, \text{ not } 17 \]
Why: Adding gives 17 over 52, but only 16 cards are hearts or threes. The three of hearts appears once in the count of hearts and again in the count of threes, and a card that can be drawn once should be counted once. Subtracting the intersection exactly once fixes it. Adding without checking would be correct only for disjoint events, such as a 10 or a face card — which is why the check comes before the arithmetic, not after.
Explain it
They keep adding probabilities without checking anything.
Discussion prompt
In four sentences or fewer, explain when adding two probabilities is wrong.
Hint: Think about counting people in two clubs.
Answer:
Suppose 20 students are in the chess club and 15 in the choir. Adding gives 35, but if 5 students are in both, only 30 different students are involved.
The five in both got counted twice, so you subtract them once. The same happens with probabilities: add them, then subtract the probability of the outcomes that belong to both events.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For overlap, ask whether one outcome could satisfy both descriptions. For the subtraction, remember it corrects a double count rather than excluding anything. For rearranging, isolate the unknown one step at a time and check the sign. For complements, write the opposite condition in words before counting anything.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a compound-events page. Top: draw the three Venn diagrams — union, intersection, disjoint — and write the general addition rule beneath, noting where the disjoint case fits. Middle left: work Examples 1 and 2 side by side, marking clearly which needed a subtraction and why. Middle right: work Example 3, and beside it write the sentence that explains what the 12 represents. Bottom left: draw the six-by-six dice grid, shade the sums above 9, and compute both parts of Example 4 by complements. Bottom right: work the fortune cookie problem, showing the complement clearly, and write one sentence on why counting pairs rather than people explains the size of the answer.
If any of your unions came out larger than the sum of the two probabilities, recheck the sign: the overlap is subtracted, never added.
Recap
Five things, and one principle: count each outcome once.
| If you see | Then |
|---|---|
| The word or | A union, so use the addition rule |
| The word and | An intersection, the term that gets subtracted |
| No shared outcomes | The subtraction term is zero |
| The union given instead | Rearrange to find the intersection |
| The words not or at least | Try the complement |
| A probability outside 0 to 1 | A rule was applied in the wrong direction |
Lesson 10.5 asks a different question: not whether two events overlap, but whether one of them changes the chance of the other.
McDougal Littell Algebra 2 (Texas Edition), Ch. 10 Counting Methods and Probability — Lesson 10.4 Find Probabilities of Disjoint and Overlapping Events §10.4, pp. 707-711 — everything on these slides traces back here
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