Chapter 14 · Probability

14.2 The Axiomatic Approach to Probability

Three axioms — $P(E)\ge0$, $P(S)=1$, and additivity for mutually exclusive events — are all that a probability must obey. From them: $P(A')=1-P(A)$, $P(E)=\tfrac{n(E)}{n(S)}$ for equally likely outcomes, and $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

In Class 10 you found probabilities by counting: a fair die has six equally likely faces, so the chance of a six is $\tfrac16$. But many of the probabilities people actually use cannot be found that way. A weather service says there is a $70\%$ chance of rain tomorrow; an insurer prices a policy on the chance that a $40$-year-old driver has an accident this year; a die carved slightly unevenly lands on some faces more often than others. None of these comes from a list of equally likely cases. Is there one theory that covers all of them — so that the rules you use for fair dice are also guaranteed to work for a loaded die or a weather forecast?

In 1933 the Russian mathematician Andrey Kolmogorov gave the answer that every modern textbook uses. Instead of saying how to compute a probability, he listed three simple properties that every sensible assignment of probabilities must have, and proved everything else from them. This is the axiomatic approach. In this section we state the three axioms, see how they let us assign probabilities to the sample points of Section 14.1, derive the familiar rules — $P(\phi)=0$, $0\le P(E)\le1$, $P(A')=1-P(A)$ — and recover the Class 10 formula $\tfrac{n(E)}{n(S)}$ as the special case of equally likely outcomes, now with the counting tools of Chapter 6 behind it. The section ends with the most used rule of all, the addition rule for "$A$ or $B$".

Three Axioms

📘 Definition 14.6 — Probability (axiomatic definition)

Let $S$ be the finite sample space of a random experiment. A probability on $S$ is a rule $P$ that assigns to every event $E\subset S$ a real number $P(E)$, called the probability of $E$, so that

Axiom 1. $P(E)\ge0$ for every event $E$;
Axiom 2. $P(S)=1$;
Axiom 3. if $E$ and $F$ are mutually exclusive events, then $P(E\cup F)=P(E)+P(F)$.

Axiom 3 extends at once to more than two events. If $E_1,E_2,E_3$ are mutually exclusive, then $E_1\cup E_2$ has no point in common with $E_3$, so $P(E_1\cup E_2\cup E_3)$ $=P(E_1\cup E_2)+P(E_3)$ $=P(E_1)+P(E_2)+P(E_3)$; repeating the step gives $P(E_1\cup\cdots\cup E_k)$ $=P(E_1)+\cdots+P(E_k)$ for any number of mutually exclusive events. The axioms do not tell us which numbers to use for a particular coin or die — that comes from symmetry, from experiment or from a model — but they tell us which assignments are allowed, and how probabilities of different events must fit together.

📐 Theorem 14.2 — Probabilities of sample points

Let $S=\{\omega_1,\omega_2,\dots,\omega_n\}$ and let $P$ be a probability on $S$. Write $p_i=P(\{\omega_i\})$. Then

(i) $p_i\ge0$ for each $i$; (ii) $p_1+p_2+\cdots+p_n=1$; (iii) for every event $E$, $P(E)$ is the sum of the $p_i$ over the sample points $\omega_i$ in $E$.

Conversely, if numbers $p_1,\dots,p_n$ satisfy (i) and (ii), then defining $P(E)$ by (iii) (with $P(\phi)=0$, the empty sum) gives a probability on $S$.

Proof. (i) is Axiom 1 applied to the simple event $\{\omega_i\}$. For (iii), an event $E$ is the union of the simple events $\{\omega_i\}$ for $\omega_i\in E$, and these are mutually exclusive, so the extended Axiom 3 gives $P(E)=\sum_{\omega_i\in E}p_i$. (The case $E=\phi$ is covered by Theorem 14.3(i) below.) Taking $E=S$ in (iii) and using Axiom 2 gives (ii).

For the converse: each $P(E)$ is a sum of non-negative numbers, so Axiom 1 holds; $P(S)=p_1+\cdots+p_n=1$, so Axiom 2 holds; and if $E\cap F=\phi$, every point of $E\cup F$ lies in exactly one of $E$, $F$, so the sum over $E\cup F$ splits into the sum over $E$ plus the sum over $F$ — Axiom 3. $\blacksquare$

So on a finite sample space, a probability is nothing more than a list of non-negative weights on the sample points that add up to $1$. The probability of any event is then found by adding the weights of its points.

✏️ Example 1 — Valid assignments and a loaded die

(a) $S=\{\omega_1,\omega_2,\omega_3,\omega_4\}$. Which of these assignments of $(p_1,p_2,p_3,p_4)$ are valid? (i) $(0.1,\,0.2,\,0.3,\,0.4)$ (ii) $\left(\tfrac12,\tfrac14,\tfrac18,\tfrac18\right)$ (iii) $(0.5,\,0.4,\,0.3,\,-0.2)$ (iv) $\left(\tfrac13,\tfrac13,\tfrac13,\tfrac13\right)$ (v) $(0,\,0,\,1,\,0)$
(b) A die is loaded so that each even face is twice as likely as each odd face. Find the probability of each face, and the probabilities of "a number greater than $3$" and "a prime number".

💡 Show solution▶

(a) Test Theorem 14.2: every $p_i\ge0$ and the sum is $1$.

(a) Verdicts (i) all non-negative, sum $1.0$: valid. (ii) sum $\tfrac{4+2+1+1}{8}=1$: valid. (iii) the sum is $1.0$, but $p_4=-0.2\lt0$: not valid. (iv) sum $\tfrac43\ne1$: not valid. (v) sum $1$: valid — a sample point may have probability $0$, and here the experiment always gives $\omega_3$.

(b) Weights Let each odd face have probability $p$; each even face has $2p$. By Theorem 14.2(ii), $3p+3(2p)=9p=1$, so $p=\tfrac19$: $P(1)=P(3)=P(5)=\tfrac19$ and $P(2)=P(4)=P(6)=\tfrac29$.

(b) Events $P(\{4,5,6\})=\tfrac29+\tfrac19+\tfrac29=\tfrac59$ and $P(\{2,3,5\})=\tfrac29+\tfrac19+\tfrac19=\tfrac49$. For a fair die both would be $\tfrac12$: on a loaded die, counting faces is not enough.

Check $\tfrac39+\tfrac69=1$ ✓, and the complementary events $\{1,2,3\}$ and $\{1,4,6\}$ get $\tfrac49$ and $\tfrac59$, which add to $1$ with the answers above ✓.

First Consequences of the Axioms

Everything you knew about probability from Class 10 can now be proved from the three axioms — and the proofs show exactly where each rule comes from.

📐 Theorem 14.3 — Basic rules of probability

Let $P$ be a probability on a finite sample space $S$, and let $A$, $B$ be events. Then:

(i) $P(\phi)=0$.
(ii) Complement rule: $P(A')=1-P(A)$.
(iii) $0\le P(A)\le1$.
(iv) If $A\subset B$, then $P(B-A)=P(B)-P(A)$, and so $P(A)\le P(B)$.
(v) $P(A-B)=P(A)-P(A\cap B)$, that is, $P(A\text{ but not }B)=P(A)-P(A\text{ and }B)$.

Proof. (i) $S$ and $\phi$ are mutually exclusive and $S\cup\phi=S$, so Axiom 3 gives $P(S)=P(S)+P(\phi)$, hence $P(\phi)=0$.

(ii) $A$ and $A'$ are mutually exclusive and $A\cup A'=S$ (Theorem 14.1). By Axioms 3 and 2, $P(A)+P(A')=P(S)=1$.

(iii) $P(A)\ge0$ is Axiom 1. Also $P(A')\ge0$, so by (ii) $P(A)=1-P(A')\le1$.

(iv) If $A\subset B$, then $B=A\cup(B-A)$ with the two pieces mutually exclusive, so $P(B)=P(A)+P(B-A)$. Rearranging gives the formula, and $P(B-A)\ge0$ gives $P(A)\le P(B)$.

(v) By Theorem 14.1(iii), $A=(A\cap B)\cup(A-B)$ with the pieces mutually exclusive, so $P(A)=P(A\cap B)+P(A-B)$. $\blacksquare$

Rule (iii) is a built-in error detector: if a calculation ever gives a probability that is negative or larger than $1$, something has gone wrong. Rule (ii) is the most useful short cut in the subject. Whenever an event is described with "at least one", its complement "none" is usually far easier to count.

✏️ Example 2 — Using the rules

(a) $P(A)=\tfrac38$ and $P(A\cap B)=\tfrac18$. Find $P(A')$ and $P(A\text{ but not }B)$, and explain why $P(B)\ge\tfrac18$.
(b) A student claims that two mutually exclusive events have probabilities $0.6$ and $0.5$. Show that this is impossible.
(c) The probability that a randomly chosen phone in a shop is faulty is $0.04$. What is the probability that it is not faulty?

💡 Show solution▶

(a) $P(A')=1-\tfrac38=\tfrac58$. $P(A-B)=P(A)-P(A\cap B)=\tfrac38-\tfrac18=\tfrac14$. Since $A\cap B\subset B$, rule (iv) gives $P(B)\ge P(A\cap B)=\tfrac18$.

(b) If the events $E$, $F$ were mutually exclusive, Axiom 3 would give $P(E\cup F)=0.6+0.5=1.1\gt1$, contradicting rule (iii). So no such events exist.

Equally Likely Outcomes

When an experiment has a symmetry — a fair coin, a well-made die, a well-shuffled pack, a person "chosen at random" — there is no reason to give one sample point more weight than another. That is the situation of Class 10, and the axioms turn it into a theorem.

📘 Definition 14.7 — Equally likely outcomes

The outcomes of a random experiment with finite sample space $S$ are equally likely if every simple event has the same probability: $P(\{\omega\})$ is the same for every $\omega\in S$.

📐 Theorem 14.4 — Probability with equally likely outcomes

If the $n$ sample points of $S$ are equally likely, then each has probability $\tfrac1n$, and for every event $E$ $$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{\text{number of outcomes in }E}{\text{total number of outcomes}}.$$

Proof. Let the common probability be $p$. By Theorem 14.2(ii), $np=1$, so $p=\tfrac1n$. By Theorem 14.2(iii), $P(E)$ is the sum of $n(E)$ copies of $\tfrac1n$, which is $\tfrac{n(E)}{n}$. $\blacksquare$

The hypothesis is essential. Choose the sample space so that its points really are equally likely — usually the most detailed, ordered description. For two dice that is the $36$ ordered pairs, not the $11$ possible sums: a sum of $7$ comes from six pairs and a sum of $2$ from only one.

✏️ Example 3 — Two dice

Two fair dice are thrown. Find the probability that (a) the sum is $7$ (b) the sum is at least $10$ (c) a doublet appears (d) at least one die shows a six (e) the sum is a prime number.

💡 Show solution▶

Set up The $36$ ordered pairs $(r,b)$ are equally likely, so $P(E)=\tfrac{n(E)}{36}$ (Theorem 14.4).

(a)–(c) (a) six pairs $(1,6),\dots,(6,1)$: $\tfrac6{36}=\tfrac16$. (b) sums $10,11,12$ come from $3+2+1=6$ pairs: $\tfrac16$. (c) six doublets: $\tfrac16$.

(d) Use the complement. "No six" means both dice show $1$–$5$: $5\times5=25$ pairs. So $P(\text{at least one six})=1-\tfrac{25}{36}=\tfrac{11}{36}$. (Direct count: $6+6-1=11$ pairs, the $(6,6)$ counted once ✓.)

(e) Prime sums are $2,3,5,7,11$, coming from $1,2,4,6,2$ pairs: $15$ in all. $P=\tfrac{15}{36}=\tfrac5{12}$.

✏️ Example 4 — Counting with ${}^nC_r$

(a) A committee of $4$ is chosen at random from $6$ boys and $4$ girls. Find the probability that it has exactly $2$ girls, and that it has at least one girl.
(b) Two cards are drawn together from a well-shuffled pack of $52$. Find the probability that both are aces, and that both are of the same suit.

💡 Show solution▶

(a) S "At random" means every $4$-person subset is equally likely: $n(S)={}^{10}C_4=210$.

(a) 2 girls Choose the $2$ girls and, independently of that, the $2$ boys: $n(E)={}^4C_2\times{}^6C_2=6\times15=90$, so $P=\tfrac{90}{210}=\tfrac37$.

(a) ≥ 1 girl The complement is "all boys": ${}^6C_4=15$ committees. $P=1-\tfrac{15}{210}=\tfrac{195}{210}=\tfrac{13}{14}$.

(b) S $n(S)={}^{52}C_2=\tfrac{52\cdot51}{2}=1326$.

(b) aces ${}^4C_2=6$ pairs of aces: $P=\tfrac6{1326}=\tfrac1{221}$.

(b) same suit Choose the suit ($4$ ways), then $2$ of its $13$ cards (${}^{13}C_2=78$): $n(E)=312$, so $P=\tfrac{312}{1326}=\tfrac4{17}$.

Check For the committee, the numbers with $0,1,2,3,4$ girls are $15,\,80,\,90,\,24,\,1$, and $15+80+90+24+1=210$ ✓.

A hand of $r$ cards is dealt from a well-shuffled pack of $52$, so all ${}^{52}C_r$ hands are equally likely. Choose a kind of card ($m$ of the $52$ are "special"). Each bar is the probability that the hand contains exactly $k$ special cards, ${}^mC_k\cdot{}^{52-m}C_{r-k}\,/\,{}^{52}C_r$. Pick an event — exactly, at least or at most $k$ — and its bars are shaded; the readouts give the exact fraction. Press Play to let the hand size grow.

special cardshand size $r$5event$k$1

$n(S)={}^{52}C_r$—

$n(E)$—

$P(E)$—

$P(E')=1-P(E)$—

The Addition Rule

Axiom 3 adds probabilities only for mutually exclusive events. What about "a king or a heart", where the king of hearts belongs to both? Adding $P(\text{king})+P(\text{heart})$ counts that card twice, so we subtract it once.

📐 Theorem 14.5 — Addition rule of probability

For any two events $A$ and $B$ of a sample space, $$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$ In particular, if $A$ and $B$ are mutually exclusive, $P(A\cup B)=P(A)+P(B)$. Moreover $$P(\text{neither }A\text{ nor }B)=1-P(A\cup B),$$ $$P(\text{exactly one})=P(A)+P(B)-2P(A\cap B).$$

Proof. Split $A\cup B$ into $A$ and $B-A$, which are mutually exclusive: every point of $A\cup B$ is either in $A$, or in $B$ but not in $A$. By Axiom 3, $P(A\cup B)=P(A)+P(B-A)$. By Theorem 14.3(v) with the roles of $A$ and $B$ exchanged, $P(B-A)=P(B)-P(A\cap B)$. Substituting gives the rule. If $A\cap B=\phi$ then $P(A\cap B)=0$ by Theorem 14.3(i).

"Neither" is the complement of $A\cup B$ (De Morgan), so the complement rule applies. "Exactly one" is $(A-B)\cup(B-A)$, two mutually exclusive pieces with probabilities $P(A)-P(A\cap B)$ and $P(B)-P(A\cap B)$. $\blacksquare$

With equally likely outcomes, dividing the counting rule $n(A\cup B)=n(A)+n(B)-n(A\cap B)$ by $n(S)$ gives the same formula — but the theorem holds for every probability, loaded dice and weather forecasts included.

✏️ Example 5 — “Or” with an overlap

(a) A card is drawn from a well-shuffled pack. Find the probability that it is a king or a heart.
(b) An integer is chosen at random from $1,2,\dots,100$. Find the probability that it is divisible by $4$ or by $6$.
(c) Two dice are thrown. Find the probability of a doublet or a sum of $7$.

💡 Show solution▶

(a) $P(K)=\tfrac4{52}$, $P(H)=\tfrac{13}{52}$, $P(K\cap H)=\tfrac1{52}$ (the king of hearts). $P(K\cup H)=\tfrac{4+13-1}{52}=\tfrac{16}{52}=\tfrac4{13}$.

(b) Multiples of $4$: $25$; of $6$: $16$ (since $6\times16=96$). A number divisible by both $4$ and $6$ is divisible by their L.C.M. $12$: there are $8$ ($12\times8=96$). $P=\tfrac{25+16-8}{100}=\tfrac{33}{100}$.

Trap In (b), the "both" event is "divisible by $12$", not by $4\times6=24$: the L.C.M., not the product, unless the numbers are co-prime.

✏️ Example 6 — A household survey

In a city, $55\%$ of households subscribe to streaming service $X$, $40\%$ to service $Y$, and $20\%$ to both. A household is chosen at random. Find the probability that it subscribes to (a) $X$ or $Y$ (b) neither (c) $X$ but not $Y$ (d) exactly one of the two services.

💡 Show solution▶

Data $P(X)=0.55$, $P(Y)=0.40$, $P(X\cap Y)=0.20$.

(a), (b) $P(X\cup Y)=0.55+0.40-0.20=0.75$, so $P(\text{neither})=1-0.75=0.25$.

(d) $0.55+0.40-2(0.20)=0.55$. Check with a Venn diagram: the four regions are $X$ only $0.35$, both $0.20$, $Y$ only $0.20$, neither $0.25$; they add to $1$ ✓, and "exactly one" is $0.35+0.20=0.55$ ✓.

🧭 Problem-solving strategy — Finding a probability

Fix the sample space. Decide what one outcome is (ordered or unordered?) and make sure the outcomes are equally likely before using $\tfrac{n(E)}{n(S)}$.
Count, do not list, when $S$ is large: multiplication principle, ${}^nP_r$ for ordered selections, ${}^nC_r$ for unordered ones — and use the same kind of count for $n(E)$ and $n(S)$.
"At least one" → complement. $P(\text{at least one})=1-P(\text{none})$.
"Or" → addition rule. Subtract the overlap unless the events are mutually exclusive.
Given $P(A)$, $P(B)$ and one more number ($P(A\cap B)$ or $P(A\cup B)$): draw the Venn diagram, fill in the four regions, and read off "only", "neither" or "exactly one".
Check that your answer lies in $[0,1]$ and, when you can, that complementary answers add to $1$.

Set $P(A)$, $P(B)$ and $P(A\cap B)$ — the overlap slider only allows values that the axioms permit. The Venn diagram shows the probability of each of the four regions, and the event chosen under "show" is shaded. On the right, the bar for $P(A)+P(B)$ is longer than the bar for $P(A\cup B)$ by exactly $P(A\cap B)$: the overlap has been counted twice. Press Play to slide the overlap from its smallest to its largest possible value.

$P(A)$0.55$P(B)$0.40$P(A\cap B)$0.20show

$P(A\cup B)$—

$P(A-B)$—

$P(B-A)$—

neither—

shaded event—

⚠️ Common mistakes with probability

1. Adding without subtracting the overlap. $P(A\cup B)=P(A)+P(B)$ only for mutually exclusive events. "King or heart" is $\tfrac{16}{52}$, not $\tfrac{17}{52}$.

2. Assuming outcomes are equally likely when they are not. "The sum of two dice is one of $11$ values, so $P(\text{sum }7)=\tfrac1{11}$" is wrong; so is "a match is won, lost or drawn, so each has probability $\tfrac13$". Use a sample space whose points really are symmetric.

3. Mixing ordered and unordered counts. If $n(S)$ counts unordered hands (${}^nC_r$), $n(E)$ must count unordered hands too.

4. Getting "at least one" wrong. $P(\text{at least one six in two throws})$ is $1-\tfrac{25}{36}=\tfrac{11}{36}$, not $\tfrac16+\tfrac16=\tfrac13$ (that counts $(6,6)$ twice).

5. Accepting impossible data. $P(A\cap B)$ can never exceed $P(A)$ or $P(B)$, and $P(A)+P(B)-P(A\cap B)$ can never exceed $1$. If the data break these, say so.

6. Thinking probability $0$ means "logically impossible" in every model. In a finite sample space a sample point may be given weight $0$ (Example 1(a)(v)); the axioms allow it.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main and JEE Advanced 2026 (JEE Advanced lists probability computed with permutations and combinations; JEE Main lists probability of an event and permutations); not in the CBSE year-end exam. Both like counting-based probabilities where the complement and ${}^nP_r$ do all the work. The classic is the birthday problem.

The idea. Suppose each of $n$ people is equally likely to have any of $N$ "birthdays" (days, months, …), all $N^n$ ordered lists being equally likely. The lists in which all $n$ are different number ${}^NP_n=N(N-1)\cdots(N-n+1)$. So $$P(\text{at least two share})=1-\dfrac{{}^NP_n}{N^n}.$$ For $N=365$ days this passes $\tfrac12$ already at $n=23$ (about $0.507$) — far fewer people than most guess, because there are ${}^{23}C_2=253$ pairs, each a chance for a match.

Worked example. Five people are chosen at random. Assuming the $12$ months are equally likely for birthdays, find the probability that at least two of them were born in the same month.
Solution. $n(S)=12^5=248\,832$. All different: ${}^{12}P_5=12\cdot11\cdot10\cdot9\cdot8=95\,040$. So $P=1-\tfrac{95040}{248832}=1-\tfrac{55}{144}=\tfrac{89}{144}\approx0.618$. With only five people a shared month is more likely than not.

Exercises

The axioms and probabilities of sample points.

1 $S=\{a,b,c\}$. Is $P(a)=0.3$, $P(b)=0.3$, $P(c)=0.4$ a valid assignment?
2 Is $P(a)=0.6$, $P(b)=0.5$, $P(c)=-0.1$ valid?
3 Is $P(a)=\tfrac12$, $P(b)=\tfrac13$, $P(c)=\tfrac14$ valid?
4 Is $P(a)=0$, $P(b)=1$, $P(c)=0$ valid? What does it say about the experiment?
5 A die is loaded so that the probability of face $k$ is proportional to $k$. Find $P(k)$ and the probability of an even number.
6 $S=\{\omega_1,\omega_2,\omega_3\}$ with $P(\omega_1)=2P(\omega_2)=3P(\omega_3)$. Find the three probabilities and $P(\{\omega_1,\omega_3\})$.
7 A coin is biased so that a head is three times as likely as a tail. Find $P(H)$ and $P(T)$ for one toss.

💡 Show solutions▶

1 All non-negative and $0.3+0.3+0.4=1$: valid.

2 Not valid: $P(c)\lt0$ breaks Axiom 1 (even though the sum is $1$).

3 Not valid: $\tfrac12+\tfrac13+\tfrac14=\tfrac{13}{12}\ne1$.

4 Valid (non-negative, sum $1$). The experiment always gives $b$.

5 $P(k)=ck$ with $c(1+2+\cdots+6)=21c=1$, so $P(k)=\tfrac{k}{21}$. $P(\text{even})=\tfrac{2+4+6}{21}=\tfrac{12}{21}=\tfrac47$.

6 Let the common value be $6t$: $P(\omega_1)=6t$, $P(\omega_2)=3t$, $P(\omega_3)=2t$, and $11t=1$. So $\tfrac6{11},\tfrac3{11},\tfrac2{11}$, and $P(\{\omega_1,\omega_3\})=\tfrac8{11}$.

7 $P(H)=3P(T)$ and $P(H)+P(T)=1$: $P(T)=\tfrac14$, $P(H)=\tfrac34$.

The basic rules.

1 $P(A)=\tfrac27$. Find $P(\text{not }A)$.
2 $P(A')=0.35$. Find $P(A)$.
3 $P(A)=0.6$ and $P(A\cap B)=0.25$. Find $P(A\text{ but not }B)$.
4 $A\subset B$, $P(A)=0.2$, $P(B)=0.5$. Find $P(B-A)$, $P(A\cap B)$ and $P(A\cup B)$.
5 Can $P(A)=0.7$ and $P(A')=0.4$? Explain.
6 $A$ and $B$ are mutually exclusive with $P(A)=0.3$, $P(B)=0.45$. Find $P(A\cup B)$, $P(A\cap B)$ and $P(A'\cap B')$.
7 Prove that for any events $A$, $B$: $P(A\cap B)\le P(A)\le P(A\cup B)$.

💡 Show solutions▶

1 $1-\tfrac27=\tfrac57$.

2 $1-0.35=0.65$.

3 $P(A-B)=0.6-0.25=0.35$.

4 $P(B-A)=0.5-0.2=0.3$. Since $A\subset B$: $A\cap B=A$, so $P(A\cap B)=0.2$; and $A\cup B=B$, so $P(A\cup B)=0.5$.

5 No: $P(A)+P(A')$ must equal $1$, but $0.7+0.4=1.1$.

6 $P(A\cup B)=0.75$; $P(A\cap B)=P(\phi)=0$; $P(A'\cap B')=1-P(A\cup B)=0.25$.

7 $A\cap B\subset A\subset A\cup B$, so both inequalities follow from Theorem 14.3(iv).

III

Equally likely outcomes and counting.

1 A card is drawn from a well-shuffled pack. Find the probability that it is a red face card.
2 A card is drawn. Find the probability that it is neither a heart nor a king.
3 Two dice are thrown. Find the probability that the sum is $9$.
4 Two dice are thrown. Find the probability that the product of the numbers is even.
5 Two dice are thrown. Find the probability that the numbers differ by exactly $2$.
6 Three fair coins are tossed. Find the probability of at least two heads.
7 An integer is chosen at random from $1$ to $50$. Find the probability that it is prime.
8 The letters of the word ORANGE are arranged at random in a row. Find the probability that the three vowels are next to one another.
9 A committee of $3$ is chosen at random from $5$ men and $4$ women. Find the probability that it has exactly $2$ women.
10 For the committee in 9, find the probability that it has at least one man.
11 In a lottery a player picks $3$ different numbers from $1$ to $20$, and $3$ winning numbers are then drawn at random. Find the probability that the player matches all three, and that the player matches exactly two.
12 A bag holds $5$ red, $3$ blue and $2$ green balls. Two balls are drawn together. Find the probability that they are the same colour.

💡 Show solutions▶

1 $6$ red face cards: $\tfrac6{52}=\tfrac3{26}$.

2 "Heart or king" has $16$ cards (Example 5(a)), so $P=1-\tfrac{16}{52}=\tfrac{36}{52}=\tfrac9{13}$.

3 $(3,6),(4,5),(5,4),(6,3)$: $\tfrac4{36}=\tfrac19$.

4 The product is odd only when both numbers are odd: $3\times3=9$ pairs. $P=1-\tfrac9{36}=\tfrac34$.

5 $\{1,3\},\{2,4\},\{3,5\},\{4,6\}$, each in two orders: $8$ pairs, $P=\tfrac8{36}=\tfrac29$.

6 $\{HHH,HHT,HTH,THH\}$: $\tfrac48=\tfrac12$.

7 The primes up to $50$ are $2,3,5,7,11,13,17,19,23,29,31,37,41,43,47$: $15$ of them. $P=\tfrac{15}{50}=\tfrac3{10}$.

8 $n(S)=6!=720$. Glue O, A, E into one block: $4!$ arrangements of the block with R, N, G, times $3!$ orders inside the block: $144$. $P=\tfrac{144}{720}=\tfrac15$.

9 $n(S)={}^9C_3=84$; $n(E)={}^4C_2\times{}^5C_1=30$. $P=\tfrac{30}{84}=\tfrac5{14}$.

10 Complement: all women, ${}^4C_3=4$. $P=1-\tfrac4{84}=\tfrac{80}{84}=\tfrac{20}{21}$.

11 $n(S)={}^{20}C_3=1140$. All three: $\tfrac1{1140}$. Exactly two: choose $2$ of the player's $3$ numbers and $1$ of the other $17$: $3\times17=51$, so $\tfrac{51}{1140}=\tfrac{17}{380}$.

12 $n(S)={}^{10}C_2=45$. Same colour: ${}^5C_2+{}^3C_2+{}^2C_2=10+3+1=14$. $P=\tfrac{14}{45}$.

The addition rule.

1 $P(A)=0.5$, $P(B)=0.35$, $P(A\cap B)=0.2$. Find $P(A\cup B)$.
2 $P(A)=0.42$, $P(B)=0.48$, $P(A\cup B)=0.7$. Find $P(A\cap B)$.
3 $P(A\cup B)=0.8$, $P(A)=0.5$, $P(A\cap B)=0.3$. Find $P(B)$ and $P(B\text{ but not }A)$.
4 An integer is chosen at random from $1$ to $60$. Find the probability that it is divisible by $3$ or by $5$.
5 A card is drawn. Find the probability that it is an ace or a red card.
6 Two dice are thrown. Find the probability that the sum is $8$ or both numbers are even.
7 In a class, $60\%$ of students passed Mathematics, $70\%$ passed Physics and $45\%$ passed both. A student is chosen at random. Find the probability that the student passed at least one subject, failed both, and passed exactly one.
8 Hospital records show that $12\%$ of patients have diabetes, $30\%$ have high blood pressure and $8\%$ have both. Find the probability that a randomly chosen patient has neither condition, and that the patient has diabetes but not high blood pressure.
9 Prove that for any events, $P(A\cup B)\le P(A)+P(B)$ and $P(A\cap B)\ge P(A)+P(B)-1$.
10 Is it possible to have $P(A)=0.6$, $P(B)=0.7$ and $P(A\cap B)=0.2$? Explain.

Check: the addition rule and the four regions

$P(A)$$P(B)$$P(A\cap B)$

$P(A\cup B)$—

$A$ only—

$B$ only—

neither—

exactly one—

💡 Show solutions▶

1 $0.5+0.35-0.2=0.65$.

2 $P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.42+0.48-0.7=0.2$.

3 $0.8=0.5+P(B)-0.3$, so $P(B)=0.6$; $P(B-A)=0.6-0.3=0.3$ (also $=P(A\cup B)-P(A)=0.3$ ✓).

4 Multiples of $3$: $20$; of $5$: $12$; of $15$: $4$. $P=\tfrac{20+12-4}{60}=\tfrac{28}{60}=\tfrac7{15}$.

5 $\tfrac{4+26-2}{52}=\tfrac{28}{52}=\tfrac7{13}$ (the two red aces are in both events).

6 From Example 4 of Section 14.1: $n=5+9-3=11$, so $P=\tfrac{11}{36}$.

7 At least one: $0.6+0.7-0.45=0.85$. Failed both: $1-0.85=0.15$. Exactly one: $0.6+0.7-2(0.45)=0.4$.

8 Either: $0.12+0.30-0.08=0.34$, so neither: $0.66$. Diabetes but not high blood pressure: $0.12-0.08=0.04$.

9 $P(A\cup B)=P(A)+P(B)-P(A\cap B)\le P(A)+P(B)$ because $P(A\cap B)\ge0$. And $P(A\cup B)\le1$ gives $P(A)+P(B)-P(A\cap B)\le1$, i.e. $P(A\cap B)\ge P(A)+P(B)-1$.

10 No: it would give $P(A\cup B)=0.6+0.7-0.2=1.1\gt1$. By part 9, $P(A\cap B)$ must be at least $0.3$.

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

How this is examined: Board papers ask for probabilities of card, dice and coin events, committee and selection problems with ${}^nC_r$, and the addition rule in every form — "or", "neither", "only", "exactly one" — usually for 1–4 marks, often as a case study built on survey data. JEE Main turns the counting up (arrangements, selections with conditions), and JEE Advanced tests the axioms themselves: which values of $P(A\cap B)$ or $P(A\cup B)$ are possible at all.

E1. Board · 1 mark If $P(A)=0.42$, $P(B)=0.48$ and $P(A\cap B)=0.16$, then $P(\text{not }A\text{ and not }B)$ is (A) $0.26$ (B) $0.74$ (C) $0.10$ (D) $0.58$

💡 Show solution▶

Idea By De Morgan, "not $A$ and not $B$" is $(A\cup B)'$. $P(A\cup B)=0.42+0.48-0.16=0.74$, so the answer is $1-0.74=0.26$.

Trap (B) is $P(A\cup B)$ itself; (C) $=1-0.42-0.48$ forgets that the overlap was subtracted twice; (D) is only $P(A')$.

Answer (A).

E2. Board · 1 mark Assertion–Reason. Assertion (A): There are no two mutually exclusive events $E$, $F$ with $P(E)=0.5$ and $P(F)=0.7$. Reason (R): For mutually exclusive events $P(E\cup F)=P(E)+P(F)$, and every probability is at most $1$. Choose: (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

💡 Show solution▶

Check R R is Axiom 3 together with Theorem 14.3(iii): true.

Check A If such events existed, R would give $P(E\cup F)=1.2\gt1$: impossible. So A is true, and R is exactly the reason.

Answer (A).

E3. Board · 3 marks Two cards are drawn together from a well-shuffled pack of $52$. Find the probability that (i) both are kings (ii) one is red and the other black (iii) at least one is a spade.

💡 Show solution▶

(i) $n(S)={}^{52}C_2=1326$; ${}^4C_2=6$ pairs of kings: $\tfrac6{1326}=\tfrac1{221}$. (1 mark)

(ii) One of $26$ red and one of $26$ black: $26\times26=676$. $P=\tfrac{676}{1326}=\tfrac{26}{51}$. (1 mark)

(iii) Complement: no spade, ${}^{39}C_2=741$. $P=1-\tfrac{741}{1326}=\tfrac{585}{1326}=\tfrac{15}{34}$. (1 mark)

E4. Board · Case study In a class of $40$ students, $22$ play cricket, $15$ play football and $8$ play both. A student is chosen at random for a sports survey. (i) Find the probability that the student plays cricket or football. (1 mark) (ii) Find the probability that the student plays neither game. (1 mark) (iii) Find the probability that the student plays exactly one of the two games. (2 marks)

💡 Show solution▶

(i) $P(C\cup F)=\tfrac{22}{40}+\tfrac{15}{40}-\tfrac8{40}=\tfrac{29}{40}$. (1 mark)

(ii) $1-\tfrac{29}{40}=\tfrac{11}{40}$. (1 mark)

(iii) Method $P(\text{exactly one})=P(C)+P(F)-2P(C\cap F)$, or count: cricket only $22-8=14$, football only $15-8=7$. (1 mark)

(iii) Answer $\tfrac{14+7}{40}=\tfrac{21}{40}$. Check: $14+8+7+11=40$ ✓. (1 mark)

E5. JEE Main Four persons are chosen at random from a group of $3$ men, $2$ women and $4$ children. The probability that exactly two of them are children is (A) $\tfrac{10}{21}$ (B) $\tfrac1{21}$ (C) $\tfrac5{21}$ (D) $\tfrac49$

💡 Show solution▶

Method $n(S)={}^9C_4=126$. Choose $2$ of the $4$ children and $2$ of the $5$ adults: ${}^4C_2\times{}^5C_2=6\times10=60$. $P=\tfrac{60}{126}=\tfrac{10}{21}$.

Trap (B) $=\tfrac6{126}$ chooses the children but forgets to choose the other two people; (C) $=\tfrac{30}{126}$ chooses only one adult; (D) is just the fraction of the group who are children.

Answer (A).

E6. JEE Advanced One or more options correct. Events $A$ and $B$ have $P(A)=0.6$ and $P(B)=0.5$. Which of the following can be the value of $P(A\cap B)$? (A) $0.05$ (B) $0.1$ (C) $0.5$ (D) $0.55$

💡 Show solution▶

Upper bound $A\cap B\subset B$, so $P(A\cap B)\le P(B)=0.5$ (Theorem 14.3(iv)). This rules out (D).

Lower bound $P(A\cup B)=1.1-P(A\cap B)\le1$, so $P(A\cap B)\ge0.1$. This rules out (A).

Both ends occur Every value in $[0.1,\,0.5]$ is possible: on $S=\{\omega_1,\omega_2,\omega_3,\omega_4\}$ give the four Venn regions ($A$ only, both, $B$ only, neither) the weights $0.6-x$, $x$, $0.5-x$, $x-0.1$, which are non-negative and add to $1$ exactly when $0.1\le x\le0.5$ (Theorem 14.2).

Answer (B) and (C).

🔬 Research–Teaching NexusMini project · about 75 minutes

Research question. Is a cuboid "die" — a rectangular eraser or a small box with its six faces numbered — equally likely to land on each face? If not, what probabilities should we give its faces, and do they obey the axioms?

Why it matters. Real objects are rarely perfectly symmetric. Casinos test their dice, manufacturers estimate failure rates, and doctors estimate how often a test gives a false alarm — all by assigning probabilities to outcomes from data rather than from symmetry. The axioms are what make such estimated probabilities safe to calculate with.

What to do.

Take a cuboid eraser (or a matchbox) that is clearly not a cube. Number its faces $1$–$6$ so that opposite faces add to $7$, as on a real die. Measure its three edge lengths.
Throw it $120$ times on the same surface (or pool $120$ throws from a group) and record the face that lands on top. Do the same with an ordinary die for comparison.
For each object, set $p_k=\dfrac{\text{number of times face }k\text{ came up}}{120}$. Check that each $p_k\ge0$ and that $p_1+\cdots+p_6=1$ (Theorem 14.2).
Using your $p_k$, compute $P(\text{even})$, $P(\text{at least }5)$ and $P(\text{prime})$ by adding weights, and compare with the equally likely values $\tfrac12$, $\tfrac13$, $\tfrac12$.
Which faces are most likely for the cuboid? Relate your answer to the areas of the faces.

Tools: A cuboid eraser or matchbox, an ordinary die, a ruler, a tally sheet or spreadsheet

Hand in: The two frequency tables, your two probability assignments with the axiom checks, the three event probabilities for each object, and a paragraph explaining what you found.

Going further. Pool the whole class's throws of identical cuboids. Do the estimated $p_k$ change much from $120$ throws to $1000$? Look up how professional casino dice are made and tested for fairness.

Instructor note: a cuboid comes to rest most often on one of its two largest faces, so the two largest faces (which are opposite each other) also come up most often — for a long, flat eraser they typically take well over half of all throws between them, and the two smallest end faces only a few per cent. With an ordinary die each face count in $120$ throws is usually within about $\pm8$ of $20$. Because the estimates come from relative frequencies, they automatically satisfy the axioms — a useful point to discuss.

🧪 Self-check quiz

0/5 correct

Q1. Which of these can not be the probability of an event?

Q2. $A$ and $B$ are mutually exclusive, $P(A)=0.3$ and $P(B)=0.5$. Then $P(A\cup B)$ is

Q3. Two dice are thrown. The probability that the sum is $11$ is

Q4. $P(A)=0.5$, $P(B)=0.4$ and $P(A\cup B)=0.7$. Then $P(A\cap B)$ is

Q5. Two students are chosen at random from $3$ boys and $2$ girls. The probability that both are girls is

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