MathsISC Class 11 · Probability

Sample Space, Events and the Probability of an Event

Updated 17 October 2026 · Fact-checked

The sample space S is the set of all possible outcomes of an experiment, and an event is a subset of S. If all outcomes are equally likely, P(E) = n(E)/n(S). Probabilities satisfy 0 ≤ P(E) ≤ 1, P(S) = 1 and P(not E) = 1 − P(E); events E and F are mutually exclusive if E ∩ F = ∅.

An event is a subset of S: count its outcomes (3) over all outcomes (6), and everything outside E is the complement E′.

Illustration drawn for QuestionBuddy.

Understand Sample Space, Events and Probability of an Event

Random experiment and sample space. An experiment whose outcome cannot be predicted with certainty but all possible outcomes are known. The sample space S lists them: a coin gives {H, T}; two coins give 4 outcomes; three coins 8; one die {1, ..., 6}; two dice 36 ordered pairs.

Events. Any subset of S. A simple event has one outcome; the sure event is S; the impossible event is ∅. Combined events: $E \cup F$ (E or F), $E \cap F$ (E and F), $E'$ (not E).

Types. *Mutually exclusive:* $E \cap F = \emptyset$ (cannot occur together). *Exhaustive:* $E_{1} \cup E_{2} \cup ... = S$ . Exclusive and exhaustive events partition S, and their probabilities add up to 1.

Probability (equally likely outcomes). $P (E) = \frac{n ( E )}{n ( S )}$ . Axioms: (1) $0 \leq P (E) \leq 1$ ; (2) $P (S) = 1$ ; (3) for mutually exclusive events, $P (E \cup F) = P (E) + P (F)$ . Consequences: $P (\emptyset) = 0$ and $P (E') = 1 - P (E)$ .

Odds. If $P (E) = \frac{a}{a + b}$ then the odds in favour of E are $a : b$ and the odds against are $b : a$ .

Counting outcomes. For draws, use combinations: the probability of drawing r balls of a given kind from a bag is $\frac{^{m} C _{r}}{^{N} C _{r}}$ . For cards: 52 cards, 4 suits of 13, 12 face cards, 26 red and 26 black cards, 4 aces.

Sample Space, Events and Probability of an Event: formulas to remember

Classical probability: $P(E) = \frac{n(E)}{n(S)}$; Equally likely outcomes.
Complement: $$P(E') = 1 - P(E)$$; Also P(E) + P(E′) = 1.
Exclusive and exhaustive: $$E \cap F = \emptyset$ | $E \cup F = S$$; Probabilities of a partition sum to 1.
Odds: $In favour: $P(E) : P(E')$$; If P(E) = a/(a + b), odds are a : b.

Sample Space, Events and Probability of an Event 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.

Drawing balls from a bag

Suppose a bag has 5 red, 4 blue and 3 green balls, and two balls are drawn at random.

Find the probability that both are red, and that the balls are of different colours.

Try it first, then show the solutionSolution

Total ways $=^{12} C_{2} = 66$ .
Both red: $^{5} C_{2} = 10$ , so $P = \frac{10}{66} = \frac{5}{33}$ .
Same colour: $10 + 6 + 3 = 19$ ; different colours $= 1 - \frac{19}{66} = \frac{47}{66}$ .

Answer: 5/33; 47/66.

Memory hook: Use combinations and the complement.

Tricks to remember Sample Space, Events and Probability of an Event

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

1 minus the opposite
For "at least one", compute 1 − P(none).
Use it when: Complement events.
Ordered pairs for two dice
Count 36 outcomes, not 21.
Use it when: Two-dice problems.

Sample Space, Events and Probability of an Event: solved examples

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

Example 1: Sum on two dice · 1 mark · MCQBoard level

Two dice are thrown. The probability that the sum is 8 is:
(a) $\frac{1}{6}$
(b) $\frac{5}{36}$
(c) $\frac{1}{9}$
(d) $\frac{7}{36}$

Try it first, then show the solutionSolution

The favourable pairs are (2,6), (3,5), (4,4), (5,3), (6,2): 5 of 36.

Answer: (b) $\frac{5}{36}$

Example 2: Three coins · 3 marks · Short answerBoard level

A coin is tossed three times. Write the sample space and find the probability of getting at least two heads, and of getting exactly one tail.

Try it first, then show the solutionSolution

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, 8 outcomes. (1 mark)
At least two heads: HHH, HHT, HTH, THH: $\frac{4}{8} = \frac{1}{2}$ . (1 mark)
Exactly one tail: HHT, HTH, THH: $\frac{3}{8}$ . (1 mark)

Answer: 1/2 and 3/8.

Example 3: Two dice · 4 marks · Short answerHarder board level

Two dice are thrown. Find the probability that (i) the sum is divisible by 4, (ii) the product is 12, (iii) the numbers are different.

Try it first, then show the solutionSolution

(i) Sums 4, 8, 12: $3 + 5 + 1 = 9$ outcomes, $P = \frac{9}{36} = \frac{1}{4}$ . (1.5 marks)
(ii) (2,6), (3,4), (4,3), (6,2): $P = \frac{4}{36} = \frac{1}{9}$ . (1.5 marks)
(iii) Doubles are 6, so different numbers $= 1 - \frac{6}{36} = \frac{5}{6}$ . (1 mark)

Answer: 1/4; 1/9; 5/6.

Example 4: Cards · 4 marks · Short answerHarder board level

One card is drawn from a pack of 52. Find the probability that it is (i) a face card, (ii) neither a heart nor an ace.

Try it first, then show the solutionSolution

(i) 12 face cards (J, Q, K in four suits): $\frac{12}{52} = \frac{3}{13}$ . (1.5 marks)
(ii) Hearts 13, aces 4, but the ace of hearts is counted twice: union $= 13 + 4 - 1 = 16$ . (1.5 marks)
Neither: $\frac{52 - 16}{52} = \frac{36}{52} = \frac{9}{13}$ . (1 mark)

Answer: 3/13; 9/13.

Example 5: Classifying events · 5 marks · Long answerHOTS / competency

Three coins are tossed. A: exactly two heads; B: at least one tail; C: the first toss is a head. Find P(A), P(B), P(C), and decide whether A and B are mutually exclusive. Find the odds in favour of B.

Try it first, then show the solutionSolution

$A = {H H T, H T H, T H H}$ : $P (A) = \frac{3}{8}$ . (1 mark)
$B$ = all except HHH: $P (B) = \frac{7}{8}$ . (1 mark)
$C = {H H H, H H T, H T H, H T T}$ : $P (C) = \frac{1}{2}$ . (1 mark)
$A \cap B = A \neq = \emptyset$ , so A and B are not mutually exclusive. (1 mark)
Odds in favour of B $= \frac{7}{8} : \frac{1}{8} = 7 : 1$ . (1 mark)

Answer: 3/8; 7/8; 1/2; not exclusive; odds 7 : 1.

Where students lose marks in Sample Space, Events and Probability of an Event

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

Mistake: Counting 21 outcomes for two dice
Using unordered pairs.
Do this: There are 36 ordered pairs.
Usually costs: 2 marks.
Mistake: Forgetting the overlap in cards
Adding 13 and 4 for hearts or aces.
Do this: Subtract the ace of hearts counted twice.
Usually costs: 1 to 2 marks.
Mistake: Probability above 1
Not simplifying or counting wrongly.
Do this: Check that 0 ≤ P ≤ 1.
Usually costs: 1 mark.

Exam tips

List the sample space for small experiments.
Check that outcomes are equally likely.
Use complements for "at least".
Express answers as simplified fractions.

Practice questions on Sample Space, Events and Probability of an Event

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

Q1 · 1 mark · MCQBoard level
If P(E) = 0.35, then P(E′) is:
(a) 0.35
(b) 0.65
(c) 1.35
(d) 0.7
Hint
Complement rule.
Answer
(b) 0.65
Q2 · 2 marks · Short answerBoard level
A die is thrown once. Find the probability of getting a prime number.
Hint
Primes: 2, 3, 5.
Answer
$\frac{3}{6} = \frac{1}{2}$ .
Q3 · 3 marks · Short answerHarder board level
A fair coin is tossed four times. Find the probability of getting at least one head, and the odds against getting all four tails.
Hint
Use 1 − P(no head).
Answer
Outcomes 16; no head means TTTT: $P = \frac{1}{16}$ . At least one head $= \frac{15}{16}$ . The odds against all tails are $15 : 1$ .
Q4 · 3 marks · Short answerHarder board level
A bag contains 4 white and 6 black balls. Two balls are drawn. Find the probability that both are of the same colour.
Hint
Combinations.
Answer
$\frac{^{4} C _{2} + ^{6} C _{2}}{^{10} C _{2}} = \frac{6 + 15}{45} = \frac{7}{15}$ .
Q5 · 3 marks · Short answerHarder board level
Two dice are thrown. Find the probability that the sum is greater than 9.
Hint
Sums 10, 11, 12.
Answer
Outcomes: $3 + 2 + 1 = 6$ ; $P = \frac{6}{36} = \frac{1}{6}$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a lucky draw has 100 tickets numbered 1 to 100.
(i) Find P(a multiple of 7). (1 mark)
(ii) Find P(a number ending in 5). (1 mark)
(iii) Find P(a perfect square). (1 mark)
(iv) Find P(not a multiple of 7). (1 mark)
Hint
Count the favourable tickets.
Answer
(i) Multiples of 7: 14 (7 to 98): $\frac{14}{100} = 0.14$ . (ii) 10 tickets: $0.1$ . (iii) 1, 4, ..., 100: 10 numbers: $0.1$ . (iv) $1 - 0.14 = 0.86$ .

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.P(S)?Show answer
1.
2.P(E′)?Show answer
1 − P(E).
3.Number of outcomes for two dice?Show answer
36.
4.Mutually exclusive events?Show answer
E ∩ F = ∅.

Sample Space, Events and Probability of an Event: frequently asked questions

What is an equally likely outcome?

Outcomes with no reason to prefer one over another, such as the faces of a fair die.

Can probability be negative?

No. It is a number between 0 and 1 inclusive.

What is the sure event?

The whole sample space S, with probability 1.