Probability · Lesson 1 of 9

Sample spaces and counting

Almost every probability question begins with the same step: list what can happen, then count. This lesson builds that habit from nothing, and ends with the counting rules that turn most exam questions into a single fraction.

← Probability moduleLesson 1 of 9 · about 35 minutes

A question to hold on to

Six fair dice are rolled together. What is the probability that every face shows a different number?

UGC NET Economics, December 2024Solved at the end ↓

The idea

What a probability is

A probability is a number between 0 and 1 that measures how likely something is. To give it a precise meaning we need four words, and every question in this module uses them.

An experiment is anything with an uncertain result: tossing a coin, rolling two dice, drawing three cards.
An outcome is one complete result of the experiment. For two coin tosses, HT is one outcome.
The sample space $S$ is the list of every possible outcome. For two tosses, $S = {H H, H T, T H, T T}$ .
An event is any collection of outcomes, the thing we ask about. “At least one head” is the event ${H H, H T, T H}$ . CUET asked exactly this definition in 2019.

When every outcome is equally likely, the probability of an event is simply the share of outcomes it contains. This is the classical (or a priori) definition, the one used in most exam questions.

Classical probability

P (A) = \frac{number of outcomes in A}{number of outcomes in S} = \frac{∣ A ∣}{∣ S ∣}

Valid only when the outcomes are finite in number and equally likely.

Pause and predict

A fair coin is tossed twice. What is the probability of at least one head? (CUET 2025)

There are two other ways to read a probability, and papers ask you to name them. The relative frequency view says a probability is the share of times an event happens over many repetitions. The subjective view treats it as a degree of belief. Modern probability rests on three axioms that all three readings satisfy; they are the subject of Lesson 2.

The habit to build

Before any formula, write down what one outcome looks like. “An ordered pair of dice faces”, “a set of 4 students”, “a seating order”. Most wrong answers come from counting outcomes that are not equally likely, or from mixing ordered and unordered counts.

Interactive

Two dice: 36 outcomes, not 11

Two dice can show sums from 2 to 12, eleven sums in all. But the sums are not equally likely, so they cannot be the sample space for the classical formula. The equally likely outcomes are the ordered pairs (first die, second die): 6 × 6 = 36 of them. Pick an event below and count.

Event

Show the complement (“not” the event)

Second die

123456

First die

P(sum is 7)

636

= 1/6 ≈ 0.167

6 of the 36 equally likely outcomes are in the event. Its complement has the other 30.

Pause and predict

Which is more likely with two dice: a sum of 7, or a sum of 2?

Why ordered pairs, when the dice look identical? Imagine one die is red and one blue. Colouring them changes nothing about how they fall, but now (1,2) and (2,1) are plainly different results. Each of the 36 pairs has probability 1/36. Treating {1, 2} as a single outcome would give it the same weight as {1, 1}, when it is really twice as likely.

Counting rule 1

Counting step by step

Most sample spaces are too big to list, so we count them. The first rule does most of the work: if a choice is made in stages, multiply the number of options at each stage.

The multiplication principle

stage 1 has n_{1} options, stage 2 has n_{2}, \dots \Rightarrow n_{1} \times n_{2} \times \dots \times n_{k} ways in all

Each stage's count must not depend on which option was taken earlier, only on how many.

3 shirts and 4 trousers make 3 × 4 = 12 outfits.
Two dice: 6 × 6 = 36 outcomes. Four dice: 6⁴ = 1,296.
Three multiple-choice questions with 4 options each can be answered in 4 × 4 × 4 = 64 ways.
A function from a set of m elements to a set of n elements chooses an image for each of the m elements: n × n × ⋯ × n = $n^{m}$ functions. JNU asked this in 2013.
A 4-digit number with no digit repeated: the first digit cannot be 0, so 9 choices; then 9, 8 and 7 for the rest: 9 × 9 × 8 × 7 = 4,536. Fill the most restricted position first.

Counting rule 2

Does order matter? Can things repeat?

Choosing r things from n comes in four versions, depending on two questions. Does order matter? A race result does; a committee does not. Can an item be chosen twice? A PIN digit can; a person on a committee cannot. Play with both switches: the lab lists every possibility, so you can see each formula rather than memorise it.

Items to choose from, n = 4How many you pick, r = 2

^{n} P_{r} = \frac{n !}{( n - r )!} = \frac{4 !}{2 !} = 12

Gold, silver and bronze: who comes first matters, and no one wins twice.

12 orderings, grouped by which items they use. Each group has 2! = 2 orderings, so there are 12 ÷ 2 = 6 selections. That is

^{n} P_{r} =^{n} C_{r} \times r!

{A, B}

ABBA

{A, C}

ACCA

{A, D}

ADDA

{B, C}

BCCB

{B, D}

BDDB

{C, D}

CDDC

	Order matters	Order doesn’t matter
No repetition	$^{n} P_{r} = \frac{n !}{( n - r )!}$ medals, seating, arranging	$^{n} C_{r} = (r n) = \frac{n !}{r ! ( n - r )!}$ committees, hands of cards, samples
Repetition allowed	$n^{r}$ PINs, dice sequences, functions	$(r n + r - 1)$ rarely asked; shown for completeness

The link between the two formulas

Every selection of r items can be put in order in r! ways, so $^{n} P_{r} =^{n} C_{r} \times r!$ . Five runners can take gold, silver and bronze in 5 × 4 × 3 = 60 ways, but only 60 ÷ 3! = 10 different trios reach the podium.

Three arrangement tricks the papers love

Identical items. Divide by the orderings of the identical ones that you cannot tell apart. BANANA has 6 letters with three A’s and two N’s: $\frac{6 !}{3 ! 2 !} = 60$ arrangements.
Round a table. Rotating everyone one seat gives the same arrangement, so n people sit in a circle in $(n - 1)!$ ways. Fix one person’s seat and arrange the rest.
Must sit together. Glue them into one block, arrange the blocks, then arrange inside the block. Four people in a row with two together: 3! × 2! = 12 of the 24 orders. Must not sit together: take the complement, 24 − 12 = 12.

Putting it together

Probability by counting

Now count both the event and the sample space in the same way, and divide. Choosing a group without regard to order means combinations top and bottom.

JNU 2018 · standard

A class has 5 boys and 5 girls. Four students are chosen at random. What is the probability that 2 boys and 2 girls are chosen?

One outcome is a group of 4.
Order doesn’t matter, so the sample space is every set of 4 from 10: $(4 10) = 210$ .
Count the event by stages.
Choose 2 of the 5 boys and 2 of the 5 girls: $(2 5) (2 5) = 10 \times 10 = 100$ .
Divide.
$P = \frac{100}{210} = \frac{10}{21}$

Answer

10/21, about 0.476.

This shape, $(k K) (n - k N - K) / (n N)$ , turns up whenever a sample is drawn without replacement from two kinds of object: defective and good bulbs, aces and non-aces, athletes on drugs and clean ones. It has a name, the hypergeometric distribution, which Lesson 6 returns to. Two aces from a pack: $(2 4) / (2 52) = 6/1326 = 1/221$ .

Watch out for

With or without replacement?

JNU 2020 asked: a box of 100 bulbs has 10 defective; what is the chance that a sample of 5 has none defective? The official key gives $(9/10)^{5} = 0.5905$ , which treats the draws as with replacement. Drawn without replacement, the exact answer is $(5 90) / (5 100) = 0.5838$ . In a large box the two are close. If the paper offers only the first, choose it, but know which one is exact.
Drawing “two at once” is the same as drawing one after another without replacement. The count changes (combinations or ordered pairs) but the probability does not, as long as top and bottom are counted the same way.

The complement

“At least one”: count the opposite

“At least one” means one, or two, or three… Counting every case is slow. Its opposite, “none at all”, is a single case. Since an event and its complement share the whole sample space between them:

The complement rule

P (at least one) = 1 - P (none)

A 4 at least once in two throws of a die (asked by JNU three times): of 36 outcomes, 5 × 5 = 25 have no 4. So the answer is 1 − 25/36 = 11/36. Check it in the grid above with “Show the complement”.
The gambler’s puzzle that started probability theory, the Chevalier de Méré’s: is it a good bet that at least one 6 appears in four throws? There are 6⁴ = 1,296 sequences and 5⁴ = 625 have no 6, so P = 671/1296 ≈ 0.518. Slightly better than even.

IIT JAM 2026 · standard

In a year of 366 days, what is the chance that three persons have different birthdays?

One outcome is an ordered triple of birthdays.
Each person has 366 possible days: 366³ outcomes, all equally likely.
Count the triples with no repeat.
The first person has 366 days, the second must avoid that one (365), the third avoids both (364).
Divide and simplify.
$P = \frac{366 \times 365 \times 364}{36 6 ^{3}} = \frac{365 \times 364}{366 \times 366} \approx 0.992$

Answer

(365 × 364)/(366 × 366). The distractor 1 − (365 × 364)/366² is the chance that at least two share a birthday.

Carry on to 23 people in a 365-day year and the chance that some two share a birthday passes one half (0.507). It feels too high because we think of matching our own birthday, but with 23 people there are 253 pairs, and any of them can match.

Stretch

When outcomes are points on a line

Sometimes the outcome is a point chosen at random from a line or a region. There are infinitely many outcomes, so we cannot count them, but the same idea works with length or area in place of number: the probability is the share of the region the event covers.

DSE 2014 · advanced

Two restaurants are located at random on a street 10 km long. What is the probability that they are less than 5 km apart?

One outcome is a pair of positions.
(x, y) with both between 0 and 10: every point of a 10 × 10 square, area 100, equally likely.
The event is a band.
|x − y| < 5 is the band around the diagonal. Outside it are two corner triangles, each ½ × 5 × 5 = 12.5.
Share of the area.
$P = \frac{100 - 2 \times 12.5}{100} = \frac{75}{100} = \frac{3}{4}$

Answer

3/4.

The shaded band is 75% of the square.

Worked

Exam problems, worked

The opening question · UGC NET December 2024

Six fair dice are rolled. What is the probability that each face shows a different number?

One outcome is an ordered list of six faces.
Each die has 6 faces: 6⁶ = 46,656 equally likely outcomes.
All different means every number 1 to 6 appears once.
So the six dice show a rearrangement of 1, 2, …, 6. Order matters (die 1 shows which number), no repeats: 6! = 720 ways.
Divide.
$P = \frac{6 !}{6 ^{6}} = \frac{720}{46656} \approx 0.0154$

Answer

6!/6⁶, about 1.5%. Rarer than most people guess.

DSE 2012 · advanced

Two women and four men sit at random around a circular table. What is the probability that the women are not next to each other?

Count all seatings.
Six people round a table: (6 − 1)! = 120.
Count seatings with the women together.
Glue them into a block: 5 units round a table, (5 − 1)! = 24, times 2 orders inside the block: 48.
Take the complement.
$P (not together) = 1 - \frac{48}{120} = 1 - \frac{2}{5} = \frac{3}{5}$
Check with a shortcut.
Seat the first woman anywhere. The second has 5 seats left, 2 of them beside the first: P(together) = 2/5. Same answer, no counting.

Answer

3/5.

Watch out for

The mistakes examiners are fishing for

Unequal outcomes in the denominator. The sums of two dice, “number of heads”, or “boy or girl” groups are not equally likely. Go back to the ordered outcomes.
Ordered on top, unordered below (or the reverse). Pick one way of counting and use it for both the event and the sample space.
“At least one” counted directly, with some cases counted twice. Use 1 − P(none).
Circular tables counted as rows. n people in a circle: (n − 1)!, not n!.

Revise

The lesson on one screen

Sample space: Every possible outcome, listed so that each is equally likely. Two dice: 36 ordered pairs.
Classical probability: P(A) = |A| / |S|, only for finite, equally likely outcomes.
Multiplication principle: Count choices made in stages by multiplying the options at each stage. Fill restricted positions first.
Order and repetition: Ordered, no repeats: nPr. Unordered: nCr. With repeats, ordered: nʳ. And nPr = nCr × r!.
Arrangements: n! in a row; (n − 1)! in a circle; divide by k! for k identical items; glue items that must sit together.
Without replacement: Two kinds of object: C(K, k)·C(N − K, n − k) / C(N, n).
At least one: 1 − P(none).
Continuous outcomes: Replace counts with lengths or areas.

Practice

Check yourself

0 of 8 correct so far

Count and compute

Type a fraction like 3/8 or a decimal.

Coins

Three fair coins are tossed. What is the probability of exactly two heads?

Seating

Five people sit in a row at random. What is the probability that two particular people sit next to each other?

Committee

A committee of 3 is chosen at random from 4 men and 3 women. What is the probability that it has at least one woman?

Digits

How many 4-digit numbers have no repeated digit?

Count

Quick questions

1In how many distinct ways can the letters of LEVEL be arranged?

2In how many ways can 8 people sit around a round table?

3Three cards are drawn from a pack without replacement. What is the probability that all three are red?

4The classical definition P(A) = |A|/|S| needs the outcomes to be

Real questions

Past papers, by level

Every counting and classical-probability question in the bank, sorted by level. Start at Foundation; if you get most of a level right, move up.

Foundation

Definitions and one-step problems. Most CUET and UGC NET questions sit here.

26 questionsPractise →

Standard

Two or three steps, up to the hardest JAM, JNU and Hyderabad questions.

24 questionsPractise →

Delhi School level

This topic’s hardest questions, from the DSE entrance, are worked through in Lesson 9: Counting and the rules.

18 questionsLesson 9 →

All 50 past-paper questions for this lesson, easiest first: practise the full set →

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

← The moduleOverview and diagnostic Lesson 2 →The rules of probability

Sample spaces and counting

What a probability is

Two dice: 36 outcomes, not 11

Counting step by step

Does order matter? Can things repeat?

Three arrangement tricks the papers love

Probability by counting

With or without replacement?

“At least one”: count the opposite

When outcomes are points on a line

Exam problems, worked

The mistakes examiners are fishing for

The lesson on one screen

Check yourself

Count and compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next

Sample spaces and counting

What a probability is

Two dice: 36 outcomes, not 11

Counting step by step

Does order matter? Can things repeat?

Three arrangement tricks the papers love

Probability by counting

With or without replacement?

“At least one”: count the opposite

When outcomes are points on a line

Exam problems, worked

The mistakes examiners are fishing for

The lesson on one screen

Check yourself

Count and compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next