Chapter 14 · Probability

14.1 Events and Their Algebra

List every possible result of an experiment as a set $S$, and every event becomes a subset of $S$. Then “not $A$”, “$A$ or $B$”, “$A$ and $B$” and “$A$ but not $B$” are just $A'$, $A\cup B$, $A\cap B$ and $A-B$.

Before an IPL match the two captains walk out for the toss; in a game of Ludo you need a six to bring a counter out; a quiz app picks one question at random from a bank of two hundred. In each case we cannot say in advance what will happen — but we can say exactly what might happen. Writing that list down carefully is the first step of every probability calculation, and most mistakes in probability come from skipping it: counting $(1,2)$ and $(2,1)$ as the same result when they are not, or forgetting a case altogether.

In this section we describe the possible results of an experiment as a set, the sample space, and the things we are interested in ("a six", "at least two heads", "a red king") as its subsets, the events. Because events are sets, all the operations of Section 1.3 — union, intersection, difference, complement and De Morgan's laws — carry over at once, and they turn out to be exactly the words "or", "and", "but not" and "not". We finish with two properties that decide how probabilities combine: events that cannot happen together, and collections of events that between them cover every possibility. No probabilities yet — those come in Section 14.2.

The Sample Space

Tossing a coin, throwing a die, drawing a card from a well-shuffled pack, choosing a student at random from a class: each is an experiment that can be repeated under the same conditions, whose possible results are all known in advance, but whose actual result on any one occasion cannot be predicted. Such an experiment is called a random experiment, and each possible result is an outcome. (Class 10 Mathematics, Chapter 14, used these words informally; we now make them into sets.)

📘 Definition 14.1 — Sample space and sample points

The sample space of a random experiment is the set $S$ of all its possible outcomes, listed so that every trial of the experiment produces exactly one element of $S$. Each element of $S$ is called a sample point. If $S$ is finite, $n(S)$ denotes the number of sample points.

The two requirements matter. "Every trial gives an element of $S$" means nothing is forgotten; "exactly one" means no two listed outcomes can happen at the same time. For a die, $\{1,2,3,4,5,6\}$ is a sample space, but $\{\text{even},\ \text{odd},\ \text{six}\}$ is not: a throw of $6$ is both "even" and "six".

✏️ Example 1 — Writing down sample spaces

Write a sample space and give $n(S)$ for each experiment.
(a) A coin is tossed three times.
(b) Two dice, one red and one blue, are thrown together.
(c) A family with two children is chosen, and the genders are recorded in order of birth. What if only the number of girls is recorded?
(d) A coin is tossed. If it shows a head, a die is thrown; if it shows a tail, the coin is tossed once more.

💡 Show solution▶

(a) Record the three results in order: $S=\{HHH,\,HHT,\,HTH,\,THH,\,HTT,\,THT,\,TTH,\,TTT\}$. Each toss has $2$ results, so by the multiplication principle (Section 6.1) $n(S)=2^3=8$.

(b) Write $(r,b)$ for "red shows $r$, blue shows $b$": $S=\{(r,b): r,b\in\{1,2,3,4,5,6\}\}$, the $6\times6$ grid of ordered pairs, $n(S)=36$. The pairs $(1,2)$ and $(2,1)$ are different outcomes — the red die did different things.

(c) In birth order: $S=\{BB,\,BG,\,GB,\,GG\}$, $n(S)=4$. Recording only the number of girls gives $S=\{0,1,2\}$, $n(S)=3$. Both are correct sample spaces — what you record decides the sample space — but in the second one the outcome $1$ contains two of the first list ($BG$ and $GB$).

(d) After a head, six results of the die; after a tail, two results of the second toss: $S=\{H1,H2,H3,H4,H5,H6,\,TH,\,TT\}$, $n(S)=8$. (Not $2\times6$: the second stage depends on the first, so we add the two branches, as in the addition principle.)

✏️ Example 2 — Counting sample points without listing them

Find $n(S)$.
(a) Four coins are tossed. (b) Two students are chosen together from five, $A,B,C,D,E$, to represent the class. (c) Two cards are drawn one after the other, without replacement, from four cards numbered $1,2,3,4$, and the order is recorded. (d) A three-digit code is set using the digits $0$–$9$, repetition allowed. (e) A coin is tossed until the first head appears.

💡 Show solution▶

(a) $2\times2\times2\times2=2^4=16$.

(b) The pair is unordered, so each sample point is a $2$-element subset: $n(S)={}^5C_2=10$ (Section 6.3). They are $AB, AC, AD, AE, BC, BD, BE, CD, CE, DE$.

(d) $10\times10\times10=1000$ codes, $000$ to $999$.

(e) $S=\{H,\,TH,\,TTH,\,TTTH,\dots\}$: there is no largest possible number of tosses, so $S$ is an infinite set. In this chapter the sample spaces we measure will be finite.

Check In (c) each unordered pair of cards $\{a,b\}$ appears twice, as $(a,b)$ and $(b,a)$. If the order were not recorded we would get $\tfrac{12}{2}=6={}^4C_2$ sample points — the same relation ${}^nP_r=r!\,{}^nC_r$ as in Chapter 6.

Events

We are rarely interested in one exact outcome. A Ludo player wants "a six"; a batsman's team wants "an even number of runs off the last ball"; a gambler wants "at least one double six". Each such description picks out some of the sample points and not others — that is, it defines a subset of $S$.

📘 Definition 14.2 — Event

An event is a subset $E$ of the sample space $S$. When the experiment is performed and the outcome is $\omega$, we say that the event $E$ has occurred if $\omega\in E$, and that $E$ has not occurred if $\omega\notin E$.

For a die, $S=\{1,2,3,4,5,6\}$ and the event "an even number" is $E=\{2,4,6\}$. If the die shows $4$, then $E$ has occurred; if it shows $5$, it has not. Many different events can occur on the same throw: a $4$ is also "more than $3$" and "not a six".

📘 Definition 14.3 — Types of events

Let $S$ be a sample space.

The empty set $\phi$ is the impossible event: no outcome lies in it, so it never occurs.
$S$ itself is the sure event: every outcome lies in it, so it always occurs.
An event with exactly one sample point, $\{\omega\}$, is a simple event (also called an elementary event).
An event with more than one sample point is a compound event.

So an event is decided by its sample points, not by the words used to describe it. "A number greater than $6$ on a die" and "a number that is both odd and even" are both $\phi$. A sample space with $n$ points has exactly $n$ simple events, one for each point.

✏️ Example 3 — Events of three tosses

A coin is tossed three times (the sample space of Example 1(a)). Write each event as a set and say what type it is.
$A$: exactly two heads. $B$: at least two heads. $C$: no head. $D$: more than three heads. $F$: at most three heads. $G$: the first and last tosses agree.
If the coin actually falls $HTH$, which of these events have occurred?

💡 Show solution▶

A, B $A=\{HHT,\,HTH,\,THH\}$, compound ($3$ points). "At least two" means two or three: $B=\{HHT,\,HTH,\,THH,\,HHH\}$, compound ($4$ points). Note $A\subset B$.

C, D $C=\{TTT\}$: a simple event. Three tosses cannot give more than three heads, so $D=\phi$, the impossible event.

F, G Every outcome has at most three heads, so $F=S$, the sure event. $G=\{HHH,\,HTH,\,THT,\,TTT\}$ (the middle toss is free): compound.

HTH $HTH$ lies in $A$, $B$, $F$ and $G$, so those four events have occurred; $C$ and $D$ have not. One outcome makes several events occur at once.

Choose an experiment: every tile is one sample point of $S$. Pick a ready-made event from the list, or click tiles to build your own event $E$ (purple); the tiles left white make up the complementary event $E'$. Press Play to step through the ready-made events. The readouts count $n(E)$ and $n(E')$, name the type of event, and show the share $\tfrac{n(E)}{n(S)}$ of the sample space that $E$ covers — in Section 14.2 this share becomes the probability of $E$ when all outcomes are equally likely.

experimentevent

$n(S)$—

$n(E)$—

$n(E')$—

type of $E$—

share $n(E)/n(S)$—

The Algebra of Events

Since events are subsets of $S$, we can combine them with the set operations of Chapter 1, taking $S$ as the universal set. Each operation has a plain-English meaning, and learning to translate in both directions is most of the work in a probability problem.

📘 Definition 14.4 — Complementary event; “or”, “and”, “but not”

Let $A$ and $B$ be events of a sample space $S$.

The complementary event of $A$, written $A'$ and read "not $A$", is $A'=S-A=\{\omega\in S:\omega\notin A\}$. It occurs exactly when $A$ does not.
The event "$A$ or $B$" is the union $A\cup B$: it occurs when at least one of $A$, $B$ occurs (possibly both).
The event "$A$ and $B$" is the intersection $A\cap B$: it occurs when both occur.
The event "$A$ but not $B$" is the difference $A-B=A\cap B'$: it occurs when $A$ occurs and $B$ does not.

in words	as a set	occurs when the outcome $\omega$…
not $A$	$A'$	$\omega\notin A$
$A$ or $B$ (at least one of them)	$A\cup B$	$\omega\in A$ or $\omega\in B$
$A$ and $B$ (both)	$A\cap B$	$\omega\in A$ and $\omega\in B$
$A$ but not $B$ (only $A$)	$A-B=A\cap B'$	$\omega\in A$, $\omega\notin B$
neither $A$ nor $B$	$(A\cup B)'=A'\cap B'$	$\omega\notin A$ and $\omega\notin B$
exactly one of $A$, $B$	$(A-B)\cup(B-A)$	$\omega$ is in one of them but not both

The word "or" in probability is always the inclusive or — "a king or a heart" includes the king of hearts. When only one of the two is wanted, say "exactly one".

📐 Theorem 14.1 — Laws of the algebra of events

For any events $A$, $B$ of a sample space $S$:

(i) $(A')'=A$, $\ A\cup A'=S$, $\ A\cap A'=\phi$, $\ S'=\phi$, $\ \phi'=S$.

(ii) De Morgan's laws: $(A\cup B)'=A'\cap B'$ and $(A\cap B)'=A'\cup B'$. In words: "not ($A$ or $B$)" is "neither $A$ nor $B$", and "not both" is "at least one of them fails".

(iii) Splitting an event: $A=(A\cap B)\cup(A-B)$, and the two pieces have no sample point in common. Likewise $A\cup B=(A-B)\cup(A\cap B)\cup(B-A)$, three pieces with no point in common.

Proof. Events are subsets of the universal set $S$, so (i) and (ii) are the laws of complements and De Morgan's laws already proved for sets in Section 1.3. For (iii), take any $\omega\in A$. Either $\omega\in B$, and then $\omega\in A\cap B$; or $\omega\notin B$, and then $\omega\in A-B$. So $A\subset(A\cap B)\cup(A-B)$; the reverse inclusion holds because both pieces are parts of $A$. No $\omega$ can be in both pieces, since that would need $\omega\in B$ and $\omega\notin B$. The three-piece statement follows in the same way by sorting each $\omega\in A\cup B$ according to whether it lies in $A$ only, in both, or in $B$ only. $\blacksquare$

✏️ Example 4 — Combining events of two dice

Two dice are thrown ($n(S)=36$). Let $A$: "the sum is $8$" and $B$: "both dice show even numbers". List or count $A\cap B$, $A\cup B$, $A-B$, $B-A$ and $A'$.

💡 Show solution▶

List $A=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$, so $n(A)=5$. $B=\{2,4,6\}\times\{2,4,6\}$, so $n(B)=3\times3=9$.

And $A\cap B$ = sum $8$ with both even $=\{(2,6),(4,4),(6,2)\}$, $3$ points.

But not $A-B=\{(3,5),(5,3)\}$ ($2$ points); $B-A$ is $B$ without its three sum-$8$ pairs: $9-3=6$ points, $\{(2,2),(2,4),(4,2),(4,6),(6,4),(6,6)\}$.

Or, not By Theorem 14.1(iii), $n(A\cup B)=n(A-B)+n(A\cap B)+n(B-A)=2+3+6=11$. And $n(A')=36-5=31$.

Check $n(A)+n(B)-n(A\cap B)=5+9-3=11$ ✓ — the counting form of the addition rule you will prove for probabilities in Section 14.2.

✏️ Example 5 — Translating words with De Morgan's laws

One card is drawn from a well-shuffled pack of $52$. Let $K$: "the card is a king" and $H$: "the card is a heart". Describe in words, and count: (a) $K\cup H$ (b) $K\cap H$ (c) $K-H$ (d) $K'\cap H'$ (e) $(K\cap H)'$.

💡 Show solution▶

(a) "A king or a heart": the $13$ hearts and the $3$ other kings, $16$ cards. (Adding $4+13=17$ counts the king of hearts twice.)

(b) "A king and a heart": only the king of hearts, $1$ card.

(d) By De Morgan, $K'\cap H'=(K\cup H)'$: "neither a king nor a heart", $52-16=36$ cards.

(e) "Not the king of hearts", i.e. "not both a king and a heart": $52-1=51$ cards. By De Morgan this is also $K'\cup H'$, "not a king, or not a heart".

Mutually Exclusive and Exhaustive Events

Two questions about a collection of events decide how their probabilities will combine: can two of them happen together? and between them, do they cover everything that can happen?

📘 Definition 14.5 — Mutually exclusive and exhaustive events

Two events $A$ and $B$ are mutually exclusive if $A\cap B=\phi$: they have no sample point in common, so the occurrence of one rules out the other. Events $E_1,E_2,\dots,E_n$ are mutually exclusive if every two of them are: $E_i\cap E_j=\phi$ whenever $i\ne j$.

Events $E_1,E_2,\dots,E_n$ are exhaustive if $E_1\cup E_2\cup\cdots\cup E_n=S$: on every trial at least one of them occurs.

If $E_1,\dots,E_n$ are both mutually exclusive and exhaustive, then on every trial exactly one of them occurs.

Two families are always mutually exclusive and exhaustive. An event and its complement: $A\cap A'=\phi$ and $A\cup A'=S$ (Theorem 14.1(i)). And the simple events of a finite sample space: if $S=\{\omega_1,\dots,\omega_n\}$, the events $\{\omega_1\},\dots,\{\omega_n\}$ never overlap and together make up $S$ — that is just the "exactly one outcome per trial" requirement of Definition 14.1. Section 14.2 builds the whole theory of probability on this second fact.

✏️ Example 6 — Testing collections of events

A die is thrown. Let $A=\{2,4,6\}$ (even), $B=\{1,3,5\}$ (odd), $C=\{2,3,5\}$ (prime), $D=\{1\}$ and $F=\{4,6\}$. Decide whether each collection is mutually exclusive, exhaustive, both, or neither: (a) $A,B$ (b) $A,C$ (c) $C,D$ (d) $C,D,F$ (e) $A,C,D$.

💡 Show solution▶

(a) $A\cap B=\phi$ and $A\cup B=S$: both. (Indeed $B=A'$.)

(b) $A\cap C=\{2\}\ne\phi$: not mutually exclusive ($2$ is even and prime). $A\cup C=\{2,3,4,5,6\}\ne S$: not exhaustive either. Neither.

(d) $C\cap D=\phi$, $C\cap F=\phi$, $D\cap F=\phi$, and $C\cup D\cup F=\{1,2,3,4,5,6\}=S$: both. Whatever the die shows, exactly one of "prime", "one", "four or six" is true.

(e) $A\cup C\cup D=\{1,2,3,4,5,6\}=S$: exhaustive. But $A\cap C=\{2\}$: not mutually exclusive. (One overlapping pair is enough to spoil it.)

The $36$ cells are the outcomes $(r,b)$ of throwing a red and a blue die. Choose two events: cells of $A$ get a red dot, cells of $B$ a blue ring. Then choose an operation — its cells are shaded purple on the grid, and the matching region of the Venn diagram on the right is shaded too. Press Play to run through the operations. The readouts count the cells and test whether $A$ and $B$ are mutually exclusive and whether they are exhaustive.

$A$$B$show

$n(A)$—

$n(B)$—

$n(A\cap B)$—

shaded cells—

mutually exclusive?—

exhaustive?—

⚠️ Common mistakes with sample spaces and events

1. Treating ordered outcomes as unordered. With two dice, "a $1$ and a $2$" is two outcomes, $(1,2)$ and $(2,1)$. Imagine the dice in different colours and the problem disappears. The same goes for three coins: "two heads and a tail" is three outcomes, not one.

2. Mixing up mutually exclusive and complementary. Complementary events are mutually exclusive and exhaustive. "Throw a $1$" and "throw a $2$" are mutually exclusive but not complementary — they leave out $3,4,5,6$.

3. Reading "or" as "exactly one". $A\cup B$ includes the outcomes in both. If you want one but not both, use $(A-B)\cup(B-A)$.

4. Confusing $A-B$ with $B-A$. "$A$ but not $B$" keeps the part of $A$ outside $B$; in Example 4, $n(A-B)=2$ but $n(B-A)=6$.

5. Thinking exhaustive means mutually exclusive. Exhaustive events cover $S$ but may overlap (Example 6(e)); mutually exclusive events never overlap but may leave gaps (Example 6(c)).

6. Negating "at least one" wrongly. The complement of "at least one head in three tosses" is "no head", $\{TTT\}$ — not "at most one head" and not "exactly one head".

Exercises

Sample spaces.

1 A coin is tossed and a die is thrown together. Write the sample space and find $n(S)$.
2 A bag contains two red balls $R_1,R_2$ and one green ball $G$. Two balls are taken out together. Write the sample space.
3 From the same bag, one ball is taken out, its label noted, and it is put back; then a second ball is taken out. Find $n(S)$ if the labels are recorded in order.
4 A die is thrown twice and only the larger of the two numbers is recorded (if they are equal, that number). Write the sample space.
5 Find $n(S)$ when five coins are tossed.
6 Three students are chosen together from a group of eight to form a team. Find $n(S)$.
7 The letters of the word DICE are arranged in a row at random. Find $n(S)$.
8 A coin is tossed. If it shows a head it is tossed again; if it shows a tail a die is thrown. Write the sample space.
9 A family has three children. Write the sample space of their genders in order of birth.

💡 Show solutions▶

1 $S=\{H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6\}$; $n(S)=2\times6=12$.

2 Unordered pairs: $S=\{R_1R_2,\ R_1G,\ R_2G\}$, $n(S)={}^3C_2=3$.

3 Each draw has $3$ results and the ball is replaced: $n(S)=3\times3=9$.

4 $S=\{1,2,3,4,5,6\}$ (the larger number can be any face; for example $1$ only from $(1,1)$).

5 $2^5=32$.

6 ${}^8C_3=\dfrac{8\cdot7\cdot6}{3!}=56$.

7 Four different letters in a row: $4!=24$.

8 $S=\{HH,\ HT,\ T1,\ T2,\ T3,\ T4,\ T5,\ T6\}$, $n(S)=2+6=8$.

9 $S=\{BBB,BBG,BGB,GBB,BGG,GBG,GGB,GGG\}$, $n(S)=2^3=8$.

Describing events.

1 Three coins are tossed. Write the event "at least one tail".
2 Three coins are tossed. Write the event "exactly one head".
3 Three coins are tossed. Write the event "a head on the second toss".
4 Two dice are thrown. Write the event "the sum is $7$".
5 Two dice are thrown. Write the event "the sum is less than $4$".
6 Two dice are thrown. Write the event "the sum is greater than $12$". What type of event is it?
7 Two dice are thrown. Write the event "the product of the numbers is $12$".
8 Classify each event as simple, compound, sure or impossible: (a) a die shows a number at most $6$ (b) a die shows $7$ (c) two dice show a double six (d) two dice show a doublet (the same number on both).
9 One card is drawn from a pack of $52$. How many sample points does the event "a red face card" have? List them.

💡 Show solutions▶

1 Every outcome except $HHH$: $\{HHT,HTH,THH,HTT,THT,TTH,TTT\}$, $7$ points.

2 $\{HTT,\ THT,\ TTH\}$.

3 $\{HHH,\ HHT,\ THH,\ THT\}$.

4 $\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$, $6$ points.

5 Sum $2$ or $3$: $\{(1,1),(1,2),(2,1)\}$.

6 The largest possible sum is $12$, so the event is $\phi$: the impossible event.

7 $12=2\times6=3\times4$ (and $1\times12$ is impossible): $\{(2,6),(6,2),(3,4),(4,3)\}$.

8 (a) sure (it is all of $S$) (b) impossible (c) simple, $\{(6,6)\}$ (d) compound, $\{(1,1),(2,2),\dots,(6,6)\}$ with $6$ points.

9 $6$ points: the jack, queen and king of hearts and of diamonds.

III

The algebra of events. A die is thrown; $A$: "a number less than $4$", $B$: "a prime number", $C$: "a multiple of $3$".

1 Find $A\cup B$.
2 Find $A\cap B$.
3 Find $A-B$ and describe it in words.
4 Find $B'$.
5 Find $A'\cap B'$ and check that it equals $(A\cup B)'$.
6 Find $(A\cap C)'$.
7 Find $B-C$.
8 Find $A\cap B\cap C$.
9 Two dice are thrown; $X$: "the first die shows $6$", $Y$: "the sum is at least $10$". Find $n(X\cap Y)$, $n(X\cup Y)$ and $n(Y-X)$.
10 A student is chosen at random from a school; $A$: "plays cricket", $B$: "plays chess". Write in set notation: (a) plays neither game (b) plays only cricket (c) plays at least one of the games (d) does not play both.

Check: combine two events of one die

$A$ =$B$ =

$A\cup B$—

$A\cap B$—

$A-B$—

$A'$—

$(A\cup B)'$—

💡 Show solutions▶

Sets $A=\{1,2,3\}$, $B=\{2,3,5\}$, $C=\{3,6\}$.

1 $A\cup B=\{1,2,3,5\}$.

2 $A\cap B=\{2,3\}$.

3 $A-B=\{1\}$: "a number less than $4$ that is not prime".

4 $B'=\{1,4,6\}$.

5 $A'=\{4,5,6\}$, $B'=\{1,4,6\}$, so $A'\cap B'=\{4,6\}$; and $(A\cup B)'=S-\{1,2,3,5\}=\{4,6\}$ ✓ (De Morgan).

6 $A\cap C=\{3\}$, so $(A\cap C)'=\{1,2,4,5,6\}$.

7 $B-C=\{2,5\}$.

8 $A\cap B\cap C=\{3\}$.

9 $X=\{(6,1),\dots,(6,6)\}$, $6$ points; $Y=\{(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)\}$, $6$ points. $X\cap Y=\{(6,4),(6,5),(6,6)\}$: $n=3$. $n(X\cup Y)=6+6-3=9$. $Y-X=\{(4,6),(5,5),(5,6)\}$: $n=3$.

10 (a) $A'\cap B'$, equivalently $(A\cup B)'$ (b) $A-B$, i.e. $A\cap B'$ (c) $A\cup B$ (d) $(A\cap B)'$, equivalently $A'\cup B'$.

Mutually exclusive and exhaustive events.

1 A die is thrown; $E=\{1,3,5\}$, $F=\{2,4\}$, $G=\{6\}$. Are $E,F,G$ mutually exclusive? Exhaustive?
2 A die is thrown; $P=\{1,2,3\}$, $Q=\{3,4,5,6\}$. Are $P$ and $Q$ mutually exclusive? Exhaustive?
3 Two dice are thrown. Are the events "the sum is $7$" and "a doublet" mutually exclusive? Are they exhaustive?
4 Three coins are tossed. Show that "at least two heads" and "at least two tails" are complementary events.
5 One card is drawn from a pack. Are "a spade" and "an ace" mutually exclusive?
6 Give three events of a single throw of a die that are mutually exclusive and exhaustive, none of them a simple event.
7 Prove that if $A$ and $B$ are mutually exclusive events then $A\subset B'$.
8 A quality inspector tests three bulbs from a batch and records each as $D$ (defective) or $G$ (good). Write the events $E_0$: "no defective", $E_1$: "exactly one defective" and $E_2$: "at least two defective", and show that they are mutually exclusive and exhaustive.
9 In one over a bowler delivers six legal balls, and each ball is recorded only as a dot ball ($0$) or a scoring ball ($R$). Find $n(S)$, and the number of sample points in the event "at least one scoring ball".

💡 Show solutions▶

1 No two share a face, and $E\cup F\cup G=\{1,2,3,4,5,6\}$: mutually exclusive and exhaustive.

2 $P\cap Q=\{3\}$: not mutually exclusive. $P\cup Q=S$: exhaustive.

3 A doublet $(k,k)$ has the even sum $2k$, never $7$: mutually exclusive. Together they have only $6+6=12$ of the $36$ points: not exhaustive.

4 With three tosses, "at least two heads" means "at most one tail". Its complement is "at least two tails". Explicitly, $\{HHH,HHT,HTH,THH\}$ and $\{TTT,TTH,THT,HTT\}$ have no common point and together make up all $8$ points.

5 No — the ace of spades is in both.

6 For example $\{1,2\}$, $\{3,4\}$, $\{5,6\}$ (any split of the six faces into three groups of at least two faces each).

7 Let $\omega\in A$. If $\omega$ were in $B$, it would be in $A\cap B=\phi$, which is impossible. So $\omega\notin B$, i.e. $\omega\in B'$. Hence $A\subset B'$.

8 $E_0=\{GGG\}$; $E_1=\{DGG,GDG,GGD\}$; $E_2=\{DDG,DGD,GDD,DDD\}$. Each outcome has a definite number of defectives ($0$, $1$, $2$ or $3$), so it lies in exactly one event: they do not overlap, and $1+3+4=8=n(S)$, so they cover $S$.

9 $n(S)=2^6=64$. The complement "no scoring ball" is the single outcome $000000$, so the event has $64-1=63$ points.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask you to write a sample space (often a two-stage experiment), to list an event, its complement or a union, intersection or difference, and to decide whether given events are mutually exclusive or exhaustive (1–3 marks). JEE rarely asks about events alone — but every probability question begins with a correct sample space, and counting its points quickly with ${}^nC_r$ and ${}^nP_r$ is tested directly.

E1. Board · 1 mark Three coins are tossed. The complement of the event "at least one head" is (A) "exactly one tail" (B) $\{TTT\}$ (C) "at most one head" (D) "exactly one head"

💡 Show solution▶

Idea "At least one head" fails only when there is no head at all, i.e. on the outcome $TTT$.

Trap (C) "at most one head" also contains $HTT$, $THT$, $TTH$, which have a head; (A) and (D) are three-point events that have nothing to do with the complement.

Answer (B).

E2. Board · 2 marks Two dice are thrown. Let $A$: "the sum is even" and $B$: "the sum is a multiple of $3$". Write $A\cap B$ and state, with a reason, whether $A$ and $B$ are mutually exclusive.

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List $A\cap B$ = "sum is a multiple of $6$" = sum $6$ or $12$: $\{(1,5),(2,4),(3,3),(4,2),(5,1),(6,6)\}$. (1 mark)

Answer $A\cap B\ne\phi$ (it has $6$ points), so $A$ and $B$ are not mutually exclusive. (1 mark)

E3. Board · 3 marks A coin is tossed. If it shows a head, a ball is drawn from a bag containing $3$ red and $4$ black balls; if it shows a tail, a die is thrown. Describe the sample space. Write the events $A$: "the die shows a number greater than $4$" and $B$: "a black ball is drawn", and decide whether $A$ and $B$ are mutually exclusive and whether they are exhaustive.

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S Label the balls $R_1,R_2,R_3,B_1,B_2,B_3,B_4$. Then $S=\{HR_1,HR_2,HR_3,HB_1,HB_2,HB_3,HB_4,\ T1,T2,T3,T4,T5,T6\}$, $n(S)=7+6=13$. (1 mark)

Events $A=\{T5,T6\}$, $B=\{HB_1,HB_2,HB_3,HB_4\}$. (1 mark)

Answer $A\cap B=\phi$ (a die and a ball are never both used), so they are mutually exclusive; $A\cup B$ has only $6$ of the $13$ points, so they are not exhaustive. (1 mark)

E4. Board · Case study At a blood-donation camp each donor's blood group ($A$, $B$, $AB$ or $O$) and Rh factor ($+$ or $-$) are recorded, for example $AB+$. (i) Write the sample space for one donor. (1 mark) (ii) Let $E$: "the group contains antigen A" (groups $A$ and $AB$) and $F$: "Rh negative". Write $E\cap F$. (1 mark) (iii) Write the event "neither antigen A nor Rh positive" in terms of $E$ and $F$, and list it. (2 marks)

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(i) $S=\{A+,A-,B+,B-,AB+,AB-,O+,O-\}$, $n(S)=4\times2=8$. (1 mark)

(ii) $E=\{A+,A-,AB+,AB-\}$, $F=\{A-,B-,AB-,O-\}$, so $E\cap F=\{A-,\ AB-\}$. (1 mark)

(iii) Set "Not Rh positive" is $F$ itself (each donor is $+$ or $-$), so the event is $E'\cap F$, which by De Morgan is also $(E\cup F')'$. (1 mark)

(iii) List $E'=\{B+,B-,O+,O-\}$, so $E'\cap F=\{B-,\ O-\}$. (1 mark)

E5. JEE Main A die is thrown three times. The number of sample points in the event "the three numbers obtained are all different and appear in increasing order" is (A) $20$ (B) $120$ (C) $216$ (D) $56$

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Method Choose which three different faces appear: ${}^6C_3=20$ ways. For each choice exactly one order is increasing, so the event has $20$ sample points (out of $n(S)=6^3=216$).

Trap (B) $=\,{}^6P_3$ counts all orders of three different faces; (D) $=\,{}^8C_3$ counts non-decreasing triples, which allow repeats such as $(2,2,5)$; (C) is the whole sample space.

⏱ Fast route Increasing triples of different numbers are just $3$-element subsets of $\{1,\dots,6\}$: ${}^6C_3$.

Answer (A).

E6. JEE Advanced One or more options correct. Two dice are thrown. Let $A$: "the first die shows an even number", $B$: "the second die shows an odd number" and $C$: "the sum is odd". Which statements are true? (A) $A$ and $B$ are mutually exclusive (B) $A\cap B\subset C$ (C) $C=(A\cap B)\cup(A'\cap B')$ (D) $n(A\cup B)=27$

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(A) False: $(2,1)$ lies in both $A$ and $B$.

(B), (C) The sum is odd exactly when one die is even and the other odd. $A\cap B$ is "first even, second odd"; $A'\cap B'$ is "first odd, second even". So $C$ is their union (true), and in particular $A\cap B\subset C$ (true).

(D) $n(A)=3\times6=18$, $n(B)=6\times3=18$, $n(A\cap B)=3\times3=9$, so $n(A\cup B)=18+18-9=27$: true. (Or: the complement $A'\cap B'$ has $9$ points, and $36-9=27$.)

Answer (B), (C) and (D).

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. What really can happen in one turn of a game you play — and are the "obvious" outcomes really the right sample space?

Why it matters. Game designers, quiz-app makers and lottery regulators must list every possible outcome before they can promise that a game is fair. Engineers do the same for a machine ("each of three components works or fails") and doctors for a diagnostic test ("positive or negative, with or without the disease"). A forgotten or double-counted outcome makes every later calculation wrong.

What to do.

Choose a game with a random step: two dice in a board game, a Ludo throw, three coins, a spinner, or drawing two cards. Write its sample space carefully, as ordered outcomes, and give $n(S)$ (use the multiplication principle if listing is long).
Write four events that matter in the game (for example "a doublet", "the sum is at least $10$", "a six on either die"), as sets. Mark their types and find one pair that is mutually exclusive and one collection that is exhaustive.
Write a second, "coarser" sample space that records less (for two dice, just the sum $2,\dots,12$). Explain which outcomes of the first space make up each point of the second.
Play $100$ turns (or pool results with classmates) and tally how often each point of the coarse space occurs. Compare with the number of fine outcomes behind it.
Write a paragraph: why is "the sum is $7$" so much more common than "the sum is $2$", even though both are single points of the coarse space?

Tools: Dice, coins or cards; a spreadsheet or tally sheet; squared paper for the outcome grid

Hand in: Both sample spaces, the four events with their types and relations, the tally table and your paragraph.

Going further. Some games use three dice. How many outcomes are there? How many of them give each sum from $3$ to $18$? (Section 14.2 turns these counts into probabilities.)

Instructor note: for two dice the coarse sums $2,3,\dots,12$ come from $1,2,3,4,5,6,5,4,3,2,1$ fine outcomes, so in $100$ turns expect about $17$ sevens and about $3$ twos; counts within roughly $\pm8$ of these are normal. The paragraph should say that the coarse sample points are not equally likely — the key idea of Section 14.2. For three dice there are $216$ outcomes, and the sums $10$ and $11$ each come from $27$ of them.

🧪 Self-check quiz

0/5 correct

Q1. A coin is tossed and two dice are thrown. The number of sample points is

Q2. Two dice are thrown. The event "the sum is $13$" is

Q3. A die is thrown, $A=\{1,2\}$ and $B=\{2,3,4\}$. The event "$A$ but not $B$" is

Q4. For one throw of a die, which pair of events is mutually exclusive but not exhaustive?

Q5. The complement of the event "neither $A$ nor $B$" is

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