MathsCBSE Class 10 · Probability

Properties of Probability and Complementary Events

Q: Can a probability be written as a percentage?

Yes. A probability can be a fraction, decimal or percentage, as long as it lies from 0 to 1, that is from 0% to 100%. A value like 120% cannot be a probability.

Q: What is the difference between a sure event and an impossible event?

A sure event contains every outcome of the experiment, so its probability is 1. An impossible event contains no outcome, so its probability is 0. For a die, getting a number below 7 is sure and getting 7 is impossible.

Q: When should I use P(not E) = 1 − P(E)?

Use it when the opposite event is easier to count, or when one probability is given and the other is asked. Signal words are not, at least one, neither nor, and lose in a game.

Q: Are any parts of this topic removed from the 2026-27 syllabus?

No. The syllabus keeps the classical definition and simple problems on the probability of an event, and these properties are part of that. Everything on this page is within the 2026-27 syllabus.

Updated 13 October 2026 · Fact-checked

Every probability lies between 0 and 1. A sure event has probability 1, an impossible event has probability 0, and an event and its complement always add to 1. The board asks you to reject impossible values, find P(not E) from P(E), use the complement for 'not' and 'at least one' questions, and find unknown counts.

Valid probabilities sit between 0 and 1, so reject -0.2 and 5/4; an event and its complement fill the whole bar of 1.

Illustration drawn for QuestionBuddy.

Understand Properties of Probability and Complementary Events

You already know the classical definition: $P (E)$ is the number of favourable outcomes divided by the total number of equally likely outcomes. This topic is about the rules that follow from that one fraction. They let you check answers, reject wrong options and solve awkward questions quickly.

Range. The favourable count can never be less than $0$ or more than the total. So if there are $n$ outcomes and $m$ of them suit $E$ , then $0 \leq m \leq n$ . Dividing by $n$ gives $0 \leq P (E) \leq 1$ . A probability can be a fraction, a decimal or a percentage, but never negative and never above $1$ (or $100%$ ). So $- 0.2$ , $\frac{5}{4}$ and $110%$ can never be probabilities.

Sure and impossible events. If every outcome suits $E$ , then $m = n$ and $P (E) = 1$ . This is a sure event, like getting a number less than 7 on a die. If no outcome suits $E$ , then $m = 0$ and $P (E) = 0$ . This is an impossible event, like getting 7 on a die. Elementary events, each with one outcome, have probabilities that add to $1$ .

Complementary events. The event 'not $E$ ', written $\overline{E}$ , happens exactly when $E$ does not. Each outcome is in $E$ or in $\overline{E}$ , never both. So if $E$ has $m$ outcomes, $\overline{E}$ has $n - m$ . Then $P (E) + P (\overline{E}) = \frac{m}{n} + \frac{n - m}{n} = 1$ , so $P (\overline{E}) = 1 - P (E)$ .

When to use it. Look for the words 'not', 'at least one', 'neither ... nor' and 'lose' in a game. Counting the opposite is often far easier. For example, 'at least one head' in three coins has seven outcomes, but its opposite, 'no head', has only one. You also use the rule in reverse: if $P (E)$ is given and the total is known, you can find how many items do not suit $E$ .

Conditions. These rules hold whenever $P (E)$ is found from a list of equally likely outcomes, so the sample space must be written correctly first. The complement must cover everything else: the opposite of 'more than 4' on a die is 'at most 4', not 'less than 4'. Remember too that two events that are merely different, such as 'red card' and 'face card', are not complements; only $E$ and 'not $E$ ' add to $1$ .

Properties of Probability and Complementary Events: formulas to remember

Range of probability: $0 \leq P (E) \leq 1$; Reject any negative value or any value above 1 (or above 100%).
Sure event: $P (sure event) = 1$; Every outcome is favourable.
Impossible event: $P (impossible event) = 0$; No outcome is favourable.
Complementary events: $P (E) + P (\overline{E}) = 1 \Rightarrow P (\overline{E}) = 1 - P (E)$; Use for 'not', 'at least one', 'neither ... nor' and win/lose questions.
Sum of elementary events: $P (E_{1}) + P (E_{2}) + \dots + P (E_{k}) = 1$; Holds when the events cover all outcomes with no overlap.
Count of outcomes not favourable: $number not in E = n - n \cdot P (E) = n \cdot P (\overline{E})$; Use to find how many items are defective, not red, and so on.

How to solve Properties of Probability and Complementary Events questions

This method works for any question on the range of probability, sure or impossible events, or complements.

1Read the event and underline signal words: not, at least, at most, neither, nor, lose.
2Write the sample space or the total number of equally likely outcomes $n$ .
3Decide whether $E$ or its opposite is easier to count. Define the opposite carefully so that it covers every other outcome.
4Count the easier one and find its probability as count over $n$ .
5If you counted the opposite, use $P (E) = 1 - P (\overline{E})$ .
6If a probability is given and a count is asked, multiply the probability by the total.
7Simplify the fraction and check that the answer lies between $0$ and $1$ , and that $P (E) + P (\overline{E}) = 1$ .

Properties of Probability and Complementary Events in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Complementary events

Suppose your school canteen packs 200 samosas for the annual sports day. The cook says that if one samosa is picked at random, the probability that it is under-fried is $\frac{3}{40}$ , and the probability that it is over-fried is $\frac{1}{20}$ . The rest are perfectly fried.

Find the probability that a samosa picked at random is perfectly fried, and find how many samosas are not perfectly fried.

Try it first, then show the solutionSolution

Under-fried and over-fried together: $\frac{3}{40} + \frac{1}{20} = \frac{3}{40} + \frac{2}{40} = \frac{5}{40} = \frac{1}{8}$
'Perfectly fried' is the complement of 'not perfectly fried'.
$P (perfect) = 1 - \frac{1}{8} = \frac{7}{8}$
Number not perfectly fried $= \frac{1}{8} \times 200 = 25$
Check: perfectly fried $= \frac{7}{8} \times 200 = 175$ , and $175 + 25 = 200$ .

Answer: Probability of perfectly fried is $\frac{7}{8}$ ; 25 samosas are not perfectly fried.

Memory hook: Good samosas plus bad samosas make the whole tray: the parts add to 1.

Range of probability, sure and impossible events

Imagine your class runs a raffle at the school fete with tickets numbered 1 to 40. One ticket is drawn at random. Later, the organisers think about adding more tickets with higher numbers.

(i) Find the probability that the number drawn is neither a perfect square nor a multiple of 5. (ii) With only tickets 1 to 40, state the probability that the number drawn is less than 41 and name the type of event. (iii) How many extra tickets, all numbered above 40, must be added so that the probability of drawing a number less than 41 becomes $\frac{4}{5}$ ?

Try it first, then show the solutionSolution

Total outcomes $= 40$ .
(i) Perfect squares from 1 to 40: 1, 4, 9, 16, 25, 36, that is 6 numbers.
Multiples of 5 from 1 to 40: 5, 10, 15, 20, 25, 30, 35, 40, that is 8 numbers.
25 is in both lists, so perfect square or multiple of 5: $6 + 8 - 1 = 13$ .
$P (neither) = 1 - \frac{13}{40} = \frac{27}{40}$
(ii) All 40 tickets are less than 41, so $P = \frac{40}{40} = 1$ , a sure event.
(iii) Let $k$ tickets numbered above 40 be added. Tickets less than 41 stay 40; the total becomes $40 + k$ .
$\frac{40}{40 + k} = \frac{4}{5} \Rightarrow 4 (40 + k) = 200 \Rightarrow 40 + k = 50 \Rightarrow k = 10$
Check: $\frac{40}{50} = \frac{4}{5}$ , which lies between 0 and 1, so the event is no longer sure.

Answer: (i) $\frac{27}{40}$ ; (ii) probability 1, a sure event; (iii) 10 tickets must be added.

Memory hook: All tickets win means 1, no ticket wins means 0, and adding new tickets pulls a sure 1 down into the middle.

Tricks to remember Properties of Probability and Complementary Events

Short memory aids. Say each one aloud once, then use it on a question the same day.

The 0-1 fence
Probability lives inside a fence from 0 to 1. Anything outside the fence, negative or bigger than 1 or 100%, is fake.
Use it when: MCQs asking which value cannot be a probability.
At least one = 1 minus none
Say it aloud: 'at least one is one minus none'.
Use it when: Coin, dice and draw questions with 'at least one'.
Complement covers the rest
The opposite of 'more than 4' is 'at most 4'. Flip the words and include the boundary.
Use it when: Defining the complement correctly before you count.

Properties of Probability and Complementary Events: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Which cannot be a probability · 1 mark · MCQBoard level

Which of the following cannot be the probability of an event?

(a) $0.07$

(b) $\frac{2}{3}$

(d) $15%$

Try it first, then show the solutionSolution

A probability must satisfy $0 \leq P (E) \leq 1$ .
$0.07$ , $\frac{2}{3}$ and $15% = 0.15$ all lie between 0 and 1.
$- 1.5$ is negative, so it cannot be a probability.

Answer: (c) $- 1.5$

Example 2: Assertion-reason on complements · 1 mark · Assertion-reasonBoard level

Assertion (A): If $P (E) = 0.4$ and $P (F) = 0.6$ for two events $E$ and $F$ of the same experiment, then $F$ is the complement of $E$ .

Reason (R): For any event $E$ , $P (E) + P (\overline{E}) = 1$ .

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not the correct explanation of A.

(d) A is false but R is true.

Try it first, then show the solutionSolution

R is the complement rule, which is true for every event.
Test A: spin a fair spinner with 10 equal sectors numbered 1 to 10. Let $E$ = 'number at most 4', so $P (E) = \frac{4}{10} = 0.4$ .
Let $F$ = 'number at most 6', so $P (F) = \frac{6}{10} = 0.6$ . The probabilities add to 1, but 1, 2, 3, 4 lie in both $E$ and $F$ , so $F$ is not 'not $E$ '. A is false.
Adding to 1 does not make two events complements; only $E$ and 'not $E$ ' are complements.

Answer: (d) A is false but R is true.

Example 3: Neither ... nor with numbered cards · 3 marks · Short answerHarder board level

Cards numbered 1 to 60 are mixed well and one card is drawn at random. Find the probability that the number on the card is (i) divisible by 4 or 6, (ii) divisible by neither 4 nor 6.

Try it first, then show the solutionSolution

Total outcomes $= 60$ . (½ mark)
Multiples of 4 up to 60: $\frac{60}{4} = 15$ . Multiples of 6: $\frac{60}{6} = 10$ . (½ mark)
Multiples of both 4 and 6 are multiples of 12: $\frac{60}{12} = 5$ , counted twice. (½ mark)
Divisible by 4 or 6: $15 + 10 - 5 = 20$ . (½ mark)
(i) $P = \frac{20}{60} = \frac{1}{3}$ (½ mark)
(ii) 'Neither 4 nor 6' is the complement: $P = 1 - \frac{1}{3} = \frac{2}{3}$ (½ mark)

Answer: (i) $\frac{1}{3}$ , (ii) $\frac{2}{3}$

Example 4: Finding counts from complementary probabilities · 5 marks · Long answerHarder board level

A box has 90 pens, of which some are defective. The probability that a pen drawn at random is not defective is $\frac{13}{15}$ . (i) Find the number of defective pens. (ii) Some defective pens are removed so that the probability of drawing a non-defective pen becomes $\frac{39}{40}$ . How many defective pens were removed? (iii) After the removal, find the probability that a pen drawn is defective, and check that the two probabilities add to 1. (iv) Can the probability of drawing a defective pen be made exactly 1 by removing only defective pens? Explain.

Try it first, then show the solutionSolution

Non-defective pens $= \frac{13}{15} \times 90 = 78$ . (½ mark)
$P (defective) = 1 - \frac{13}{15} = \frac{2}{15}$ , so defective pens $= \frac{2}{15} \times 90 = 12$ . (1 mark)
(ii) Let $x$ defective pens be removed. New total $= 90 - x$ ; non-defective stays 78. (½ mark)
$\frac{78}{90 - x} = \frac{39}{40} \Rightarrow 39 (90 - x) = 3120 \Rightarrow 90 - x = 80 \Rightarrow x = 10$ . (1 mark)
(iii) Defective left $= 2$ , total $= 80$ : $P (defective) = \frac{2}{80} = \frac{1}{40}$ . (½ mark)
Check: $\frac{39}{40} + \frac{1}{40} = 1$ . (½ mark)
(iv) If only defective pens are removed, the 78 non-defective pens always stay, so $P (non-defective) > 0$ and $P (defective) < 1$ . It cannot be made 1; that would need all 78 non-defective pens removed instead. (1 mark)

Answer: (i) 12 defective pens, (ii) 10 removed, (iii) $\frac{1}{40}$ , sum checks to 1, (iv) No, since the 78 non-defective pens remain; it becomes 1 only if all 78 non-defective pens are removed.

Example 5: Case study: the school spinner game · 4 marks · Case-basedHOTS / competency

Suppose at a school fete a fair spinner is divided into 12 equal sectors numbered 1 to 12. You spin it once. You win a prize if the number is a prime; you lose otherwise. A second game uses two fair coins tossed together, and you win a sticker if you get at least one head.

(i) Find the probability of losing the spinner game. (1 mark)

(ii) Find the probability of getting a number greater than 12 on the spinner, and name this type of event. (1 mark)

(iii) Riya says the chance of at least one head with two coins is $\frac{2}{3}$ , because the results are 'no head, one head, two heads' and two of these have a head. Find the correct probability and explain her error. (2 marks)

Try it first, then show the solutionSolution

(i) Primes from 1 to 12: 2, 3, 5, 7, 11, that is 5. $P (win) = \frac{5}{12}$ . (½ mark)
$P (lose) = 1 - \frac{5}{12} = \frac{7}{12}$ . (½ mark)
(ii) No sector shows more than 12, so $P = 0$ ; it is an impossible event. (1 mark)
(iii) Equally likely outcomes: HH, HT, TH, TT, total 4. (½ mark)
'No head' is TT only, so $P (no head) = \frac{1}{4}$ . (½ mark)
$P (at least one head) = 1 - \frac{1}{4} = \frac{3}{4}$ . (½ mark)
Riya's three results are not equally likely: 'one head' happens in two ways (HT, TH), so the classical definition cannot be applied to her list. (½ mark)

Answer: (i) $\frac{7}{12}$ , (ii) 0, impossible event, (iii) $\frac{3}{4}$ ; her outcomes were not equally likely.

Where students lose marks in Properties of Probability and Complementary Events

The slips examiners see most often, and what to write instead.

Mistake: Treating two different events as complements
Students subtract from 1 for any two events, not only for E and not E.
Wrong working
P(red card) $= \frac{1}{2}$ , so P(face card) $= 1 - \frac{1}{2} = \frac{1}{2}$ .
Right working
Face cards are not the opposite of red cards. P(face card) $= \frac{12}{52} = \frac{3}{13}$ .
Do this: Use $1 - P (E)$ only when the second event is exactly 'not E'.
Usually costs: The full answer mark for that part.
Mistake: Wrong boundary in the complement
The opposite of 'more than' is taken as 'less than', dropping the boundary value.
Wrong working
Die: P(more than 4) $= 1 - P (less than 4) = 1 - \frac{3}{6} = \frac{1}{2}$ .
Right working
Opposite is 'at most 4' (1, 2, 3, 4): P $= 1 - \frac{4}{6} = \frac{1}{3}$ .
Do this: Check that the event and its complement together list every outcome exactly once.
Usually costs: 1 to 2 marks in a short answer.
Mistake: Accepting an answer above 1
An arithmetic slip, such as swapping numerator and denominator, is not checked.
Wrong working
P(not defective) $= \frac{90}{78}$ .
Right working
P(not defective) $= \frac{78}{90} = \frac{13}{15}$ , which lies between 0 and 1.
Do this: Always check $0 \leq P (E) \leq 1$ before writing the final answer.
Usually costs: The answer mark and often the next part that uses it.
Mistake: Counting 'at least one' directly and missing cases
Listing many favourable outcomes by hand invites omissions.
Wrong working
Three coins, at least one head: HHH, HHT, HTH, HTT gives $\frac{4}{8}$ .
Right working
Only TTT has no head, so P $= 1 - \frac{1}{8} = \frac{7}{8}$ .
Do this: For 'at least one', count 'none' and subtract from 1.
Usually costs: 1 mark or more.

Exam tips

In MCQs on 'cannot be a probability', convert percentages and decimals mentally and reject anything negative or above 1.
Write the line 'P(not E) = 1 − P(E)' explicitly; examiners give method marks for the stated rule.
When finding a count from a given probability, show the multiplication by the total as a separate step.
In assertion-reason items, test the assertion with a correct sample space; R may be a true rule that does not explain A.
Finish by checking that your P(E) and P(not E) add to 1, especially in case-based parts.

Practice questions on Properties of Probability and Complementary Events

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
The probability that it will rain tomorrow is $0.35$ . What is the probability that it will not rain tomorrow? Also state whether 'it will rain or it will not rain tomorrow' is a sure event, with its probability.
Hint
Rain and no rain are complements; together they cover everything.
Answer
$P (no rain) = 1 - 0.35 = 0.65$ . 'Rain or no rain' always happens, so it is a sure event with probability 1.
Q2 · 3 marks · Short answerHarder board level
A number is chosen at random from 1 to 50. Find the probability that it is not a multiple of 3 and not a multiple of 5.
Hint
Count multiples of 3 or 5, removing multiples of 15, then use the complement.
Answer
Multiples of 3: 16; of 5: 10; of 15: 3. Multiples of 3 or 5: $16 + 10 - 3 = 23$ . Required $P = 1 - \frac{23}{50} = \frac{27}{50}$ .
Q3 · 3 marks · Short answerBoard level
Three fair coins are tossed together. Find the probability of getting (i) at least one tail, (ii) at most two tails, (iii) neither all heads nor all tails.
Hint
List all 8 outcomes; each part has a complement with very few outcomes.
Answer
Total 8. (i) $1 - P (HHH) = 1 - \frac{1}{8} = \frac{7}{8}$ . (ii) $1 - P (TTT) = \frac{7}{8}$ . (iii) $1 - \frac{2}{8} = \frac{3}{4}$ .
Q4 · 5 marks · Long answerHarder board level
A bag contains only red, green and white balls. The probability of a red ball is $\frac{1}{5}$ and of a green ball is $\frac{1}{3}$ . The bag has 21 white balls. (i) Find the total number of balls. (ii) Find the probability that the ball drawn is not green. (iii) How many red balls must be added so that the probability of drawing a red ball becomes $\frac{1}{3}$ ?
Hint
P(white) = 1 − 1/5 − 1/3. Adding red balls changes both the red count and the total.
Answer
(i) $P (white) = 1 - \frac{1}{5} - \frac{1}{3} = \frac{7}{15}$ , so total $= 21 \div \frac{7}{15} = 45$ . (ii) $1 - \frac{1}{3} = \frac{2}{3}$ . (iii) Red $= 9$ ; $\frac{9 + x}{45 + x} = \frac{1}{3} \Rightarrow 27 + 3 x = 45 + x \Rightarrow x = 9$ . Check: $\frac{18}{54} = \frac{1}{3}$ .
Q5 · 3 marks · Short answerHOTS / competency
Two fair dice are thrown together. Find the probability that (i) the product of the two numbers is not a perfect square, (ii) the sum of the two numbers is neither a prime number nor a multiple of 4.
Hint
Count the perfect-square products, including unequal pairs like (1, 4). For (ii), count sums that are prime or a multiple of 4, then take the complement.
Answer
Total 36. (i) Products that are perfect squares: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (1,4), (4,1), that is 8. $P = 1 - \frac{8}{36} = \frac{7}{9}$ . (ii) Prime sums 2, 3, 5, 7, 11 occur $1 + 2 + 4 + 6 + 2 = 15$ ways; sums 4, 8, 12 occur $3 + 5 + 1 = 9$ ways; no sum is both. So $15 + 9 = 24$ outcomes, and $P = 1 - \frac{24}{36} = \frac{1}{3}$ . Check: sums 6, 9, 10 occur $5 + 4 + 3 = 12$ ways, and $\frac{12}{36} = \frac{1}{3}$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a Class 10 section surveys its students on their favourite sport. The counts are:
- Cricket: $x$ students
- Football: 15 students
- Badminton: 9 students
- Kabaddi: 12 students
One student is chosen at random. The probability that the student's favourite sport is cricket is $\frac{2}{5}$ .
(i) Find $x$ . (1 mark)
(ii) Find the probability that the student's favourite sport is not cricket. (1 mark)
(iii) Find the probability that the favourite sport is neither football nor kabaddi. Also, how many new students who like badminton must join the section so that the probability of choosing a cricket lover becomes $\frac{1}{3}$ ? (2 marks)
Hint
The total is $36 + x$ . Adding badminton lovers increases the total but not the cricket count.
Answer
(i) $\frac{x}{36 + x} = \frac{2}{5} \Rightarrow 5 x = 72 + 2 x \Rightarrow x = 24$ ; total $= 60$ . (ii) $1 - \frac{2}{5} = \frac{3}{5}$ . (iii) Football or kabaddi $= 27$ , so $P (neither) = 1 - \frac{27}{60} = \frac{33}{60} = \frac{11}{20}$ . Let $y$ join: $\frac{24}{60 + y} = \frac{1}{3} \Rightarrow 60 + y = 72 \Rightarrow y = 12$ . Check: $\frac{24}{72} = \frac{1}{3}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Range of any probability?Show answer
$0 \leq P (E) \leq 1$
2.Probability of a sure event?Show answer
1
3.Probability of an impossible event?Show answer
0
4.If $P (E) = 0.4$ , what is $P (\overline{E})$ ?Show answer
$0.6$
5.Sum of probabilities of all elementary events?Show answer
1
6.Complement of 'at least one head'?Show answer
No head at all.

Properties of Probability and Complementary Events: frequently asked questions

Can a probability be written as a percentage?

Yes. A probability can be a fraction, decimal or percentage, as long as it lies from 0 to 1, that is from 0% to 100%. A value like 120% cannot be a probability.

What is the difference between a sure event and an impossible event?

A sure event contains every outcome of the experiment, so its probability is 1. An impossible event contains no outcome, so its probability is 0. For a die, getting a number below 7 is sure and getting 7 is impossible.

When should I use P(not E) = 1 − P(E)?

Use it when the opposite event is easier to count, or when one probability is given and the other is asked. Signal words are not, at least one, neither nor, and lose in a game.

Are any parts of this topic removed from the 2026-27 syllabus?

No. The syllabus keeps the classical definition and simple problems on the probability of an event, and these properties are part of that. Everything on this page is within the 2026-27 syllabus.