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Lesson 7 of 8

Probability · Lesson 7 of 8

Complements and Multi-Step Probability

“Translate neither, at least one, and only one into events and combine the chapter’s methods.”

Learning Objectives

• Derive and apply P(A′)=1−P(A). • Use De Morgan’s laws to distinguish neither from not both. • Find A-only, B-only, and exactly-one probabilities from overlapping event data. • Combine complements with counting in selections and arrangements. • Interpret applications through a consistent sample space and check all region weights.

The outcomes left outside an event

Sometimes the event asked for is harder to count than the event it excludes. The chance of at least one green marble can be found by subtracting the chance of no green marble. This works because an event and its complement divide the sample space into two disjoint parts with total probability one.

Complement reasoning changes the question without changing the experiment. Always identify exactly which event is being excluded. Not exactly one is not the same as none: it also includes two, three, and any larger allowed count.

Example — A numbered-card complement

Problem
Ten cards are numbered 1 through 10 and one is drawn uniformly. Let A={2,4,6,8}. Find the probability of not A.

  1. 1.The four members of A give P(A)=4/10=2/5.
  2. 2.The remaining cards are A′={1,3,5,7,9,10}, six members. Direct counting gives P(A′)=6/10=3/5.
  3. 3.The two values total 2/5+3/5=1. Thus P(A′)=1−2/5, the total probability minus the weight of A.
  4. 4.Do not replace A′ by the odd numbers: 10 is even but is not in the given event A, so it belongs to A′.

For any event A, not just equally likely cards, A∩A′=∅ and A∪A′=S. Disjoint additivity therefore gives P(A)+P(A′)=P(S)=1. Rearranging proves the complement rule for every probability assignment in the chapter.

Complement ruleLaTeX
A′ contains every allowed outcome outside A. The probability one belongs to the entire sample space.
AP(A)A′1 − P(A)SInside A and outside A partition the total weight 1.
Complement as the remaining weight— The blue region is all of S outside A. Its shape is illustrative; the formula concerns probability weights.

Direct counts and complements together

Cards and coloured objects provide a useful first application. You can count the requested category directly, count its complement and subtract, or use separate disjoint categories. These methods should agree when they describe the same event.

Example — Five questions about one card

Problem
Draw one card uniformly from a standard deck of 52. Find the chances of a diamond, not an ace, a black card, not a diamond, and not a black card.

  1. 1.There are 13 diamonds, so P(diamond)=13/52=1/4.
  2. 2.There are 4 aces, so P(not ace)=1−4/52=1−1/13=12/13.
  3. 3.The two black suits, clubs and spades, contain 26 cards: P(black)=26/52=1/2.
  4. 4.P(not diamond)=1−1/4=3/4.
  5. 5.P(not black)=1−1/2=1/2. The complement is the 26 red cards, providing a direct-count check.
Example — Nine coloured discs

Problem
A bag contains 4 red, 3 blue, and 2 yellow discs of the same shape and size. One disc is drawn uniformly. Find the chances of red, yellow, blue, not blue, and red or blue.

  1. 1.The sample points are the nine individual discs, not the three colour names.
  2. 2.The first three probabilities are 4/9,2/9,3/9=1/3.
  3. 3.Not blue includes red and yellow, so P(not blue)=1−1/3=2/3. Direct counting also gives (4+2)/9=6/9.
  4. 4.Red and blue are disjoint for one disc. Thus P(red or blue)=4/9+3/9=7/9.
  5. 5.This last union does not require an overlap correction because one disc cannot have both of the stated colours.

Neither is different from not both

When two events describe two people or two conditions, the phrase neither means that both events fail. Not both means that they do not occur together: one may occur, or neither may occur. A four-region picture makes the distinction precise.

An outcome fails to be in A∪B exactly when it is outside both A and B. It fails to be in A∩B exactly when it is outside at least one of them. These statements are De Morgan’s laws, expressed as event relationships rather than memorised symbol changes.

De Morgan’s laws for eventsLaTeX
Neither A nor B is the complement of the union. At least one event failing is the complement of the intersection.
Verbal conditionEventProbability method
Neither occursA′∩B′=(A∪B)′1−P(A∪B)
At least one does not occurA′∪B′=(A∩B)′1−P(A∩B)
A occurs but B does notA∩B′P(A)−P(A∩B)
Exactly one occurs(A∩B′)∪(A′∩B)P(A)+P(B)−2P(A∩B)

The subtraction for A but not B follows because A is divided into two disjoint pieces: A∩B and A∩B′. Their weights add to P(A), so removing the shared weight leaves the A-only weight. Exactly one then adds the A-only and B-only weights, which cannot occur together.

Only one of two eventsLaTeX
Unlike the union rule, this formula removes both original copies of the overlap, because exactly one excludes every shared outcome.
Example — Two qualification outcomes

Problem
Anil qualifies in a selection task with probability 0.05 and Ashima with probability 0.10. Both qualify with probability 0.02. Find the chances that neither qualifies, at least one fails to qualify, and exactly one qualifies.

  1. 1.Let E mean Anil qualifies and F mean Ashima qualifies. The union has probability 0.05+0.10−0.02=0.13.
  2. 2.Neither qualifies is E′∩F′=(E∪F)′, so its probability is 1−0.13=0.87.
  3. 3.At least one fails is E′∪F′=(E∩F)′, so its probability is 1−0.02=0.98.
  4. 4.Anil only has weight 0.05−0.02=0.03; Ashima only has weight 0.10−0.02=0.08.
  5. 5.Exactly one has weight 0.03+0.08=0.11. The four complete cases have weights 0.02,0.03,0.08,0.87 and total one.
Four disjoint qualification casesBoth qualify0.02Anil only0.03Ashima only0.08Neither qualifies0.87The boxes organise cases; their areas do not encode the weights.
Neither, not both, and exactly one— Neither is only the bottom-right case. Not both includes all cases except the top-left. Exactly one includes the two single-person cases.
Common mistake

“Both do not occur” can be linguistically ambiguous. Write the intended event explicitly: neither is A′∩B′, whereas not both is (A∩B)′. Their probabilities generally differ.

Combining the data into four region weights

For mixed problems, a reliable approach is to calculate the four disjoint cases first. Start with the shared weight, subtract it from each separate event weight, and give the remainder of the total to neither. Once these regions are known, many questions become simple sums of regions.

Example — Two voluntary programmes

Problem
Of 60 students, 30 choose NCC, 32 choose NSS, and 24 choose both. A student is selected uniformly. Find the chances of at least one programme, neither, and NSS but not NCC.

  1. 1.The shared count is 24. NCC only has 30−24=6 students; NSS only has 32−24=8.
  2. 2.At least one has count 6+24+8=38, giving probability 38/60=19/30.
  3. 3.Neither has count 60−38=22, giving 22/60=11/30.
  4. 4.NSS but not NCC has count 8 and probability 8/60=2/15. The four counts 6,24,8,22 add to 60.
PersonRecorded sexAge (years)
HarishM30
RohanM33
SheetalF46
AlisF28
SalimM41
Example — Choosing a spokesperson

Problem
One of the five people in the table is selected uniformly. Find the chance that the person is male or over 35.

  1. 1.Let M={Harish,Rohan,Salim} and O={Sheetal,Salim}. The counts are 3 and 2.
  2. 2.The overlap contains Salim only, with count 1.
  3. 3.The union contains 3+2−1=4 people, so its probability is 4/5.
  4. 4.Alis is the only person satisfying neither condition. This gives a second check: four favourable people out of five.

Selections: count the easier opposite event

For an at-least-one condition, its opposite is none. A none condition may have one uncomplicated combination count even when the original event contains several possible counts. Keep the selection model fixed: unless otherwise stated in these group-draw problems, objects are drawn without replacement and the unordered sets of the stated size are equally likely.

Example — Five marbles with a colour condition

Problem
A box contains 10 red, 20 blue, and 30 green marbles. Five are selected uniformly without replacement. Find the chances that all are blue and that at least one is green.

  1. 1.There are 60 individual marbles, so the total number of five-marble sets is C(60,5).
  2. 2.All blue sets can be chosen in C(20,5) ways. Therefore P(all blue)=C(20,5)/C(60,5).
  3. 3.No green means selecting all five from the 30 red-or-blue marbles. Its probability is C(30,5)/C(60,5).
  4. 4.At least one green is the complement, giving 1−C(30,5)/C(60,5). This includes one, two, three, four, and five green marbles without counting those cases separately.
Example — Avoiding winning lottery tickets

Problem
A lottery has 10,000 tickets and ten distinct winning tickets. A buyer holds k distinct tickets. Under a uniform winning-ticket model, find the chance of no prize for k=1,2,10.

  1. 1.Treat the buyer’s k-ticket set as a uniform set relative to the ten winning tickets. There are C(10000,k) possible k-ticket sets.
  2. 2.No prize requires all tickets to belong to the 9990 non-winning tickets. The favourable count is C(9990,k).
  3. 3.Thus P(no prize)=C(9990,k)/C(10000,k). For one ticket this is 9990/10000=999/1000.
  4. 4.For two tickets it is (9990×9989)/(10000×9999). For ten it is C(9990,10)/C(10000,10).
  5. 5.The descending counts reflect distinct tickets. Raising 999/1000 to the kth power would describe repeated draws with replacement, not this fixed set of distinct tickets.
Example — Two friends assigned to sections

Problem
One hundred students are divided uniformly at random into sections of sizes 40 and 60. Find the chances that two specified friends are in the same section and in different sections.

  1. 1.Equivalently, place the friends into two uniformly selected distinct positions among 100 labelled positions. There are C(100,2)=4950 unordered pairs of positions.
  2. 2.A same-section pair can be selected in C(40,2)+C(60,2)=780+1770=2550 ways.
  3. 3.The same-section probability is 2550/4950=17/33.
  4. 4.Different sections is the complement, giving 1−17/33=16/33. Directly, the count is 40×60=2400 pairs, and 2400/4950 agrees.

Arrangements with at least one match

The three-letter envelope problem combines a small permutation sample space with an at-least-one condition. It can be solved by listing the arrangements or by the three-event addition rule. Both approaches are useful checks, and neither needs a new probability concept.

Example — Letters and their envelopes

Problem
Letters A,B,C are inserted randomly into envelopes addressed to A,B,C, one letter per envelope. Find the probability that at least one letter is correctly placed.

  1. 1.Represent an arrangement by the letters in envelope order A,B,C. The six equally likely orders are ABC,ACB,BAC,BCA,CAB,CBA.
  2. 2.ABC has three correct letters; ACB has A correct; BAC has C correct; CBA has B correct. BCA and CAB have none.
  3. 3.Four of six orders have at least one correct placement, giving 4/6=2/3. The complementary none event has probability 2/6=1/3.
  4. 4.For a formula check, each one-letter correct event has probability 2/6=1/3. Each pair of correct events forces the third letter correct too, so every pairwise and triple intersection has probability 1/6.
  5. 5.Inclusion–exclusion gives 3(1/3)−3(1/6)+1/6=2/3, matching the list.

A mixed problem is best solved in stages: specify the elementary outcomes, translate the requested condition, and choose whether to count directly or use a probability rule. If you combine methods, keep the event meanings visible at each step. A final four-region or total-weight check often catches a subtraction error that ordinary arithmetic checking misses.

Quiz

Quick check

If P(A)=2/11, what is P(A′)?

Quick check

Which event means neither A nor B?

Quick check

If P(A)=0.54, P(B)=0.69, P(A∩B)=0.35, what is P(A only)?

Quick check

What is the complement of at least one green marble?

Quick check

If the probability that both people qualify is 0.02, what is the probability that at least one does not?

Quick check

For three randomly placed letters, what is the chance of at least one correct envelope?

Quick check

If P(A)=0.42, P(B)=0.48, and P(A∩B)=0.16, what is the probability of exactly one?

Quick check

If P(E′∪F′)=0.25, what can be concluded about E and F?

Practice Problems

Practice Problems
  1. For P(A)=2/11, find P(not A) and explain the total-weight argument.
  2. Given P(E)=1/4, P(F)=1/2, P(E∩F)=1/8, find the probabilities of E or F and of neither E nor F.
  3. If P(E′∪F′)=0.25, decide whether E and F can be mutually exclusive. Use De Morgan’s law in your reasoning.
  4. Given P(A)=0.42, P(B)=0.48, P(A∩B)=0.16, find P(A′), P(B′), and P(A∪B).
  5. The chance of qualifying in both English and Hindi assessments is 0.5, and the chance of neither is 0.1. If the English chance is 0.75, find the Hindi chance.
  6. Of 60 students, 30 choose NCC, 32 NSS, and 24 both. A student is selected uniformly. Find the probabilities of at least one programme, neither programme, and NSS but not NCC.
  7. From 10 red,20 blue,30 green marbles, select five uniformly without replacement. Find the probabilities of all blue and at least one green. Explain the selection counts.
  8. Ten distinct tickets win equal prizes among 10000 tickets. Find the chance of no prize when holding one, two, and ten distinct tickets under the uniform model.
  9. A uniform random division puts 100 students into sections of 40 and 60. Find the chances that two specified friends enter the same section and different sections.
  10. Three letters are placed randomly in three addressed envelopes, one per envelope. Find the chance that at least one is correct, using both a list and inclusion–exclusion.
  11. If P(A)=0.54, P(B)=0.69, and P(A∩B)=0.35, find the union, neither, A only, and B only. Check that the four disjoint region weights total one.
  12. Use the five-person table in this lesson to find the probability that a uniformly chosen spokesperson is male or over 35. Show explicitly which person belongs to both groups.
  13. Create two events for a small finite sample space and show that “neither” and “not both” generally contain different outcomes. State the case in which they could agree.

Key Takeaways

Key Takeaways

• A and A′ partition S, giving P(A′)=1−P(A). • Neither is the complement of the union; not both is the complement of the intersection. • Subtract the overlap from an individual event to get its only region. • Exactly one excludes the overlap entirely, so its probability subtracts that overlap twice. • At-least-one conditions can often be counted efficiently through their none complement. • Mixed questions require a consistent sample space, precise event language, and a check of the resulting region weights.

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The Addition Rule for Events

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Chapter Summary and Practice