Salvatore Mosaico · 2026
Look at what does not happen
When asked “What is the probability that it happens at least once?”, we may be tempted to list every way the event can occur. Sometimes the better route is different: describe precisely the case in which it never occurs.
Teacher: In four die rolls we want at least one 6. Which cases would we need to add?
Student: Exactly one 6, two sixes, three sixes and four sixes.
Teacher: What happens when our requirement is not met?
Student: No 6 appears. That is just one condition to describe!
This dialogue is a teaching simulation. The idea is not to guess a shortcut, but to choose the simplest description of the same experiment.
A simple formula, an important choice
Let Ω denote all possible outcomes and A the required event. Its complement Aᶜ contains exactly the outcomes in Ω that do not belong to A. The two events are disjoint and together exhaust all possibilities.
P(A) + P(Aᶜ) = 1
P(A) = 1 − P(Aᶜ)
This identity always holds, even without independence or equally likely outcomes. Independence may be needed when calculating P(Aᶜ), if we want to multiply probabilities from separate trials. Confusing these two levels is a frequent mistake.
Before the formula: negate the question correctly
| Required event | Complement |
|---|---|
| At least one success | Zero successes |
| At least two successes | Zero or exactly one success |
| All colours appear | At least one colour is missing |
| At least two equal values | All values are different |
“None”, “not all” and “at most one” do not mean the same thing. Always write the complement as a sentence before calculating.
The tools we will use
For sampling without replacement we will use C(N,n), the number of groups of n objects selected from N without considering order. The symbol n! denotes the product of the integers from 1 to n, with 0! = 1. When n > N, set C(N,n) = 0.
C(N,n) = N! / [n! (N − n)!], 0 ≤ n ≤ N
The ratio “favourable outcomes / possible outcomes” is valid when the counted outcomes are equally likely. A product, meanwhile, can come from independence or from a chain of conditional probabilities: we will encounter both.
Twenty-five problems, increasingly demanding ideas
The collection retains all 25 problems from the original conversation. They are not all advanced: some establish the basic mechanism; others introduce sampling without replacement, collisions, conditional probabilities and inclusion-exclusion. Problem 21 requires particular care with overlapping events.
Try each problem yourself first. You can open a hint about the complement and, separately, the complete solution. Percentages are rounded to two decimal places; the preceding expressions are exact under the stated model.
Problem index
- At least one 6 with four dice
- At least one head in ten tosses
- At least one ace in five cards
- At least one double six
- Matching birthdays in a class
- At least one red ball
- At least one even number
- At least one heart
- A sum of at least 3
- A rare success in twenty trials
- At least one woman in the group
- At least one 1 or 2
- At least one face card
- At least one win
- At least one defective item
- At least one value greater than 4
- Five socks, four colours
- A repeated digit in a code
- At least one ace or king
- Two codes coincide
- All three colours appear
- The same birth month
- At least one multiple of 5
- A collision in the boxes
- A repetition with six dice
1. At least one 6 with four dice
Four fair, independent dice are rolled. What is the probability of getting at least one 6?
Show the hint
Complementary event: None of the four dice shows 6.
Show the solution
Complementary event: None of the four dice shows 6.
Each die has five outcomes other than 6 out of six. Independence lets us multiply 5/6 four times. Subtract from 1: this avoids separating the cases with one, two, three or four sixes, which would require more work.
P(A) = 1 − (5/6)⁴ = 671/1296
Result: ≈ 51.77%
2. At least one head in ten tosses
A fair coin is tossed ten times independently. What is the probability of at least one head?
Show the hint
Complementary event: All ten tosses are tails.
Show the solution
Complementary event: All ten tosses are tails.
Each toss has probability 1/2 of tails. Just one sequence, all tails, contains no heads; there are 2¹⁰ equally likely sequences. The complement therefore has probability 1/1024, which we subtract from 1. A very high probability is still not certainty.
P(A) = 1 − (1/2)¹⁰ = 1023/1024
Result: ≈ 99.90%
3. At least one ace in five cards
Five cards are drawn without replacement from a well-shuffled standard 52-card deck without jokers. What is the probability of at least one ace?
Show the hint
Complementary event: All five cards are non-aces.
Show the solution
Complementary event: All five cards are non-aces.
There are 48 non-aces. All five-card hands are equally likely: C(52,5) in total and C(48,5) without aces. Their ratio is the probability of the complement. We do not use (48/52)⁵: without replacement, the deck changes after each draw.
P(A) = 1 − C(48,5) / C(52,5)
Result: ≈ 34.12%
4. At least one double six
Two fair, independent dice are rolled in ten independent rounds. What is the probability of seeing (6,6) at least once?
Show the hint
Complementary event: No round produces (6,6).
Show the solution
Complementary event: No round produces (6,6).
There are 36 equally likely ordered pairs in a round; only one is (6,6). Not getting double six has probability 35/36. Require this in all ten rounds, then subtract from 1. The question is not merely whether either die shows a 6.
P(A) = 1 − (35/36)¹⁰
Result: ≈ 24.55%
5. Matching birthdays in a class
In a class of 25 students, model birthdays as independent and uniform over 365 days, ignoring 29 February. What is the probability that at least two match?
Show the hint
Complementary event: All 25 birthdays are different.
Show the solution
Complementary event: All 25 birthdays are different.
The first birthday is unrestricted. The second must avoid one day, the third two, up to the twenty-fifth, which must avoid 24. Multiply the conditional factors 365/365, 364/365, …, 341/365. The conditions “different from the earlier birthdays” are not independent events: the product uses the correct conditional probabilities. The result concerns a match between any pair, not with a fixed date; the model simplifies real birthdays.
P(A) = 1 − (365 × 364 × ⋯ × 341) / 365²⁵
Result: ≈ 56.87%
6. At least one red ball
An urn contains 7 red and 13 blue balls, all equally likely to be drawn. Five are selected without replacement. What is the probability of at least one red ball?
Show the hint
Complementary event: All five balls are blue.
Show the solution
Complementary event: All five balls are blue.
Of the C(20,5) possible selections, C(13,5) are entirely blue. Subtract their ratio from 1. Following the draws gives the same complement: (13/20) × (12/19) × (11/18) × (10/17) × (9/16). Both numerators and denominators decrease.
P(A) = 1 − C(13,5) / C(20,5)
Result: ≈ 91.70%
7. At least one even number
An integer from 1 to 10 is chosen six times, uniformly, independently and with replacement. What is the probability of at least one even number?
Show the hint
Complementary event: All six numbers are odd.
Show the solution
Complementary event: All six numbers are odd.
Five of the ten numbers are odd, so each choice has probability 1/2 of being odd. Replacement keeps this probability unchanged; independence permits the power. Subtract the probability of six odd numbers from 1.
P(A) = 1 − (1/2)⁶ = 63/64
Result: ≈ 98.44%
8. At least one heart
Four cards are drawn without replacement from a well-shuffled standard 52-card deck without jokers. What is the probability of at least one heart?
Show the hint
Complementary event: No card is a heart.
Show the solution
Complementary event: No card is a heart.
There are 13 hearts and 39 cards of other suits. Hands without hearts number C(39,4), while all hands number C(52,4). Their ratio gives the complement: there is no need to distinguish one, two, three and four hearts.
P(A) = 1 − C(39,4) / C(52,4)
Result: ≈ 69.62%
9. A sum of at least 3
Two fair, independent dice are rolled. What is the probability that their sum is at least 3?
Show the hint
Complementary event: The sum is less than 3.
Show the solution
Complementary event: The sum is less than 3.
The smallest sum is 2. The only excluded outcome is therefore (1,1), one of 36 equally likely ordered outcomes. Sums from 2 to 12 are not equally likely: count pairs, not possible sum values.
P(A) = 1 − 1/36 = 35/36
Result: ≈ 97.22%
10. A rare success in twenty trials
An experiment succeeds with probability 0.08 on each trial. Twenty independent trials are performed. What is the probability of at least one success?
Show the hint
Complementary event: All twenty trials fail.
Show the solution
Complementary event: All twenty trials fail.
Failure has probability 0.92 on each trial. No successes has probability 0.92²⁰. The shortcut 20 × 0.08 is not the probability of at least one success: it is the expected number of successes, and here it even exceeds 1.
P(A) = 1 − (92/100)²⁰
Result: ≈ 81.13%
11. At least one woman in the group
A set of 30 people consists of 12 women and 18 men. A group of five distinct people is selected uniformly. What is the probability that it includes at least one woman?
Show the hint
Complementary event: All five selected people are men.
Show the solution
Complementary event: All five selected people are men.
There are C(30,5) possible groups. Groups without women are selected entirely from the 18 men, giving C(18,5). The model concerns randomly selected groups, not five independent draws with replacement.
P(A) = 1 − C(18,5) / C(30,5)
Result: ≈ 93.99%
12. At least one 1 or 2
A fair die is rolled eight times independently. What is the probability that at least one roll shows 1 or 2?
Show the hint
Complementary event: Every roll shows only 3, 4, 5 or 6.
Show the solution
Complementary event: Every roll shows only 3, 4, 5 or 6.
On each roll four of the six outcomes satisfy the complement, giving 2/3. For eight rolls the probability is (2/3)⁸. The question asks for at least one of the two values, not for both values to occur at least once: these are different events.
P(A) = 1 − (2/3)⁸
Result: ≈ 96.10%
13. At least one face card
Six cards are drawn without replacement from a well-shuffled standard 52-card deck without jokers. What is the probability of at least one face card, meaning jack, queen or king?
Show the hint
Complementary event: All six cards are non-face cards.
Show the solution
Complementary event: All six cards are non-face cards.
There are 3 face cards per suit, so 12 in total. Aces are not face cards under this definition. That leaves 40 non-face cards: count C(40,6) hands without face cards out of C(52,6) total hands, then subtract from 1.
P(A) = 1 − C(40,6) / C(52,6)
Result: ≈ 81.15%
14. At least one win
In the problem's model, a team wins each match with probability 0.4 and the outcomes of five matches are independent. What is the probability of at least one win?
Show the hint
Complementary event: The team wins none of the five matches.
Show the solution
Complementary event: The team wins none of the five matches.
Not winning has probability 0.6. If draws are possible, not winning includes both drawing and losing: the complement must not be called “losing every match”. Independence and constant probability are assumptions of the model, not automatic properties of real sport.
P(A) = 1 − (6/10)⁵ = 2882/3125
Result: ≈ 92.22%
15. At least one defective item
Fifty items each have probability 0.02 of being defective, independently of the others. What is the probability of finding at least one defective item?
Show the hint
Complementary event: All fifty items are good.
Show the solution
Complementary event: All fifty items are good.
One item is good with probability 0.98. All items being good has probability 0.98⁵⁰. The answer exceeds 60% even though an individual defect is rare. If defects were correlated, for example through a shared production fault, this power would not be justified.
P(A) = 1 − (98/100)⁵⁰
Result: ≈ 63.58%
16. At least one value greater than 4
A fair die is rolled twelve times independently. What is the probability of obtaining a value greater than 4 at least once?
Show the hint
Complementary event: Every result is at most 4.
Show the solution
Complementary event: Every result is at most 4.
Avoiding both 5 and 6 leaves four faces out of six, a probability of 2/3 per roll. Requiring this twelve times gives (2/3)¹². The complement of “greater than 4” is “less than or equal to 4”: include the boundary value.
P(A) = 1 − (2/3)¹²
Result: ≈ 99.23%
17. Five socks, four colours
A drawer contains at least five socks, each in one of four colours. Five socks are drawn without replacement. What is the probability that at least two have the same colour?
Show the hint
Complementary event: All five socks have different colours.
Show the solution
Complementary event: All five socks have different colours.
Five different colours cannot appear when only four exist. The complement is impossible and has probability zero. The result is certain regardless of how many socks have each colour or how the five are selected. This is the pigeonhole principle: here the complement leads to a proof, not a power.
P(A) = 1 − 0 = 1
Result: = 100.00%
18. A repeated digit in a code
A code has six digits, each chosen uniformly and independently from 0 to 9. The first may also be 0. What is the probability of at least one repetition?
Show the hint
Complementary event: All six digits are different.
Show the solution
Complementary event: All six digits are different.
There are 10⁶ possible codes. A code without repetitions has 10 choices for the first digit, then 9, 8, 7, 6 and 5. Divide that product by 10⁶ and subtract from 1. This is a six-digit code, not necessarily a six-digit integer, so leading zeroes are allowed.
P(A) = 1 − (10 × 9 × 8 × 7 × 6 × 5) / 10⁶ = 1061/1250
Result: ≈ 84.88%
19. At least one ace or king
Five cards are drawn without replacement from a well-shuffled standard 52-card deck without jokers. What is the probability of finding at least one card that is an ace or a king?
Show the hint
Complementary event: There is neither an ace nor a king.
Show the solution
Complementary event: There is neither an ace nor a king.
Aces and kings are disjoint categories: 4 + 4 = 8 cards. This leaves 44 cards allowed by the complement. Hands avoiding all eight cards number C(44,5). The event does not require both an ace and a king: one card from either category is enough.
P(A) = 1 − C(44,5) / C(52,5)
Result: ≈ 58.21%
20. Two codes coincide
Four three-digit codes are generated uniformly and independently from 000 to 999. What is the probability that at least two are equal?
Show the hint
Complementary event: All four codes are different.
Show the solution
Complementary event: All four codes are different.
The first code is unrestricted. To avoid matches, the second has 999 available values, the third 998 and the fourth 997, out of 1000 possible values at every generation. Multiply these conditional factors and subtract from 1. This models collisions between random codes, not the security of real passwords.
P(A) = 1 − (999 × 998 × 997) / 1000³
Result: ≈ 0.60%
21. All three colours appear
A fair die has two red faces, two green faces and two blue faces. It is rolled ten times independently. What is the probability that each of the three colours appears at least once?
Show the hint
Complementary event: At least one of the three colours is missing.
Show the solution
Complementary event: At least one of the three colours is missing.
Call R, V and B the events “red missing”, “green missing” and “blue missing”. Each has probability (2/3)¹⁰. Adding the three probabilities counts monochromatic sequences twice: for example, ten blues belongs to both R and V. Each intersection of two absences has probability (1/3)¹⁰, and there are three pairs. All three colours being absent is impossible. By inclusion-exclusion, the complement is 3(2/3)¹⁰ − 3(1/3)¹⁰. Subtracting from 1 reverses both signs: this is the collection's genuinely advanced step.
P(A) = 1 − 3(2/3)¹⁰ + 3(1/3)¹⁰
Result: ≈ 94.80%
22. The same birth month
For eight people, model birth months as independent and uniform over the twelve months. What is the probability that at least two share a month?
Show the hint
Complementary event: All eight months are different.
Show the solution
Complementary event: All eight months are different.
The first month is unrestricted; then 11, 10, 9, 8, 7, 6 and 5 months remain that avoid matches with earlier ones. Divide the product of these counts by 12⁸. Equal month probabilities are an explicit model assumption, not an exact fact about real births or month lengths.
P(A) = 1 − (12 × 11 × 10 × 9 × 8 × 7 × 6 × 5) / 12⁸
Result: ≈ 95.36%
23. At least one multiple of 5
Seven integers from 1 to 100 are chosen uniformly and independently, with replacement. What is the probability that at least one is divisible by 5?
Show the hint
Complementary event: None of the seven numbers is divisible by 5.
Show the solution
Complementary event: None of the seven numbers is divisible by 5.
There are 20 multiples of 5, so 80 of the 100 numbers are not multiples. On each choice the local complement has probability 4/5; for all seven choices it is (4/5)⁷. Replacement distinguishes this problem from sampling seven distinct numbers.
P(A) = 1 − (4/5)⁷
Result: ≈ 79.03%
24. A collision in the boxes
Five distinguishable balls are assigned to five distinguishable boxes. Each ball chooses a box uniformly and independently. What is the probability that some box contains at least two balls?
Show the hint
Complementary event: No box contains more than one ball.
Show the solution
Complementary event: No box contains more than one ball.
With five balls and five boxes, no collisions means exactly one ball in each box. There are 5⁵ total assignments, since each ball has five choices. Collision-free assignments number 5! = 120. Mere occupancy profiles are not treated as equally likely: the equally likely outcomes are assignments of the distinguishable balls.
P(A) = 1 − 5! / 5⁵ = 601/625
Result: ≈ 96.16%
25. A repetition with six dice
Six fair, independent dice, distinguishable by position or colour, are rolled. What is the probability that at least two show the same value?
Show the hint
Complementary event: All six results are different.
Show the solution
Complementary event: All six results are different.
With six dice and six values, all different means exactly 1, 2, 3, 4, 5 and 6 in some order. There are 6! = 720 orders out of 6⁶ equally likely ordered outcomes. The complement avoids classifying pairs, triples and every other repetition pattern.
P(A) = 1 − 6! / 6⁶ = 319/324
Result: ≈ 98.46%
Three structures to recognize
Independent trials: if each of n trials has success probability p, the probability of at least one success is:
P(A) = 1 − (1 − p)ⁿ
Without replacement: among N objects, K are of the desired type; select a group of n uniformly, with 1 ≤ n ≤ N. The probability of at least one desired object is:
P(A) = 1 − C(N − K,n) / C(N,n)
Collisions: choose n times uniformly and independently from m values. For 1 ≤ n ≤ m, the probability of at least one repetition is:
P(A) = 1 − [m × (m − 1) × ⋯ × (m − n + 1)] / mⁿ
If n > m, a repetition is certain by the pigeonhole principle. In the three-colour problem 21, however, the complement is a union of overlapping events: inclusion-exclusion is also needed.
A better question before calculating
Do not turn the complement into a formula applied blindly. First describe the experiment, then the event and its negation, check the assumptions, and only then choose the calculation. “At least one” is a clue, not evidence of independence.
The complement need not consist of a single case: it should be a set of cases easier to describe or count. If it becomes more complicated than the original event, you are not obliged to use it.
The key question is: “Is it easier to describe what I want, or what prevents it?” This change of perspective is the method’s real value.