Probability

This companion follows the chapter's own two-part structure. Section 14.1 rebuilds the vocabulary of random experiments in the language of sets: what an event is (a subset of $S$), when it has occurred, the four types of events (impossible, sure, simple, compound), and the algebra that combines events (or, and, but-not, not), closing with the two relationships — mutually exclusive and exhaustive — that recur in every later formula. Section 14.2 introduces the axiomatic approach to probability, the modern, rigorous way to assign numbers to events, and works outward from the three axioms to the equally-likely-outcomes formula, the addition theorem for two and then three events, and the complement rule. Work through the sections in order: the classification of events in 14.1 is the vocabulary the axioms in 14.2 are stated in, and the addition theorem is a consequence of the axioms, not a separate rule to memorise independently.

Sample space, events, and when an event occurs

A random experiment is one whose outcome cannot be predicted with certainty, and its sample space $S$ is the set of all possible outcomes. For tossing a coin twice, $S = \{HH, HT, TH, TT\}$. Given any question about the experiment — "did exactly one head appear?" — the outcomes that answer "yes" form a subset of $S$. This motivates the formal definition used throughout the chapter.

where: $S$ is the sample space of the experiment and $E$ is the event, a set of sample points. The event $E$ is said to have occurred if the actual outcome $\omega$ of the experiment satisfies $\omega \in E$; if $\omega \notin E$, the event has not occurred. For the coin-toss-twice experiment, "exactly one head" corresponds to $E = \{HT, TH\}$; if the experiment actually produces $HT$, then since $HT \in E$, the event $E$ has occurred.

Types of events — impossible, sure, simple, compound

Every event falls into one of four categories. The impossible event is the empty set $\varnothing$: no outcome satisfies it, so it can never occur (for example, "a multiple of 7" when rolling an ordinary die). The sure event is the whole sample space $S$ itself, occurring on every trial. A simple (or elementary) event contains exactly one sample point; a sample space with $n$ distinct outcomes has exactly $n$ simple events. A compound event contains more than one sample point — "at least one head" in three coin tosses is compound, since $F = \{HTT, THT, TTH, HHT, HTH, THH, HHH\}$ has seven points. Simple events are always pairwise mutually exclusive, since two different single points can never coincide.

The algebra of events

Because events are subsets, every set operation from earlier chapters becomes a way to build new events out of old ones. Let $A$ and $B$ be events associated with a sample space $S$.

where: $A'$ is the complementary event, occurring exactly when $A$ does not; $A \cup B$ occurs when either $A$ or $B$ or both occur; $A \cap B$ occurs only when both occur simultaneously; and $A - B$ occurs when $A$ happens but $B$ does not. As a check, for rolling a die with $A = \{2,3,5\}$ ("prime") and $B = \{1,3,5\}$ ("odd"), $A \cup B = \{1,2,3,5\}$, $A \cap B = \{3,5\}$, $A - B = \{2\}$, and $A' = \{1,4,6\}$ — exactly Example 1 of the NCERT text. These four combinations are the toolkit for translating an everyday probability question into set language before any arithmetic is done.

Mutually exclusive and exhaustive events

Two relationships between events dominate the rest of the chapter, and it is essential not to confuse them. Events $A$ and $B$ are mutually exclusive if the occurrence of one rules out the occurrence of the other — formally, $A \cap B = \varnothing$. Rolling a die, "odd" ($\{1,3,5\}$) and "even" ($\{2,4,6\}$) are mutually exclusive, since no outcome is both. By contrast, events $E_1, \dots, E_n$ are exhaustive if their union is the entire sample space, $\bigcup_{i=1}^{n} E_i = S$ — at least one must occur, regardless of overlap. A family can be exhaustive without being mutually exclusive (overlapping events can still cover $S$ between them), and mutually exclusive without being exhaustive (disjoint events need not cover everything). When a family is both — pairwise disjoint and covering $S$ — it is mutually exclusive and exhaustive, the strongest combination, since then exactly one event occurs on every trial.

The axiomatic approach to probability

Earlier classes assigned probability only by counting equally likely outcomes. The axiomatic approach is more general: it treats probability as a function satisfying a short list of rules, from which every other property — including the counting formula — can be derived. Let $S$ be a sample space; the probability $P$ is a real-valued function on the events of $S$ satisfying three axioms.

where: the first axiom says probability is never negative, the second fixes the total probability of the sure event at exactly $1$, and the third (additivity) says the probabilities of mutually exclusive events simply add. If $S = \{\omega_1, \omega_2, \dots, \omega_n\}$, these axioms force $0 \leq P(\omega_i) \leq 1$ for each outcome, $\sum_{i=1}^n P(\omega_i) = 1$, and for any event $A$, $P(A) = \sum P(\omega_i)$ summed over $\omega_i \in A$. Crucially, the axioms do not require equal probabilities — a coin could validly have $P(H) = \tfrac{1}{3}$ and $P(T) = \tfrac{2}{3}$, or indeed any pair $p$ and $1-p$ with $0 \leq p \leq 1$ — the axioms only demand non-negativity and that the total is $1$. Checking whether a proposed assignment is valid means checking exactly these two things: every value lies in $[0,1]$, and the values sum to $1$.

Probability of equally likely outcomes

The familiar school formula is the special case of the axioms where every simple event is assigned the same probability. If $S = \{\omega_1, \dots, \omega_n\}$ and $P(\omega_i) = p$ for every $i$, then additivity and $P(S)=1$ force $np = 1$, so $p = 1/n$. For any event $E$ with $n(E) = m$ sample points,

where: $n(E)$ is the number of sample points in $E$ and $n(S)$ is the total number of sample points, both counted using the permutation and combination techniques from the previous chapter whenever direct listing is impractical. This formula is a theorem under the equally-likely assumption, not a free-standing definition — using it when outcomes are not equally likely (a biased die, a weighted spinner) silently gives the wrong answer, which is exactly why the axiomatic route is introduced first.

The addition theorem — probability of "A or B"

The single most-used formula in this chapter answers: given $P(A)$ and $P(B)$ separately, what is $P(A \cup B)$? Simply adding $P(A) + P(B)$ over-counts, because any outcome in $A \cap B$ gets counted once inside $P(A)$ and again inside $P(B)$.

where: the subtracted term $P(A \cap B)$ removes exactly the double-counted overlap. When $A$ and $B$ are mutually exclusive, $P(A \cap B) = 0$ and formula (8) collapses back to axiom (iii), $P(A \cup B) = P(A) + P(B)$ — consistent, since with no overlap there is nothing to double-count. The full proof is given below as the first derivation.

Probability of the complementary event "not A"

Because $A$ and $A'$ are always mutually exclusive ($A \cap A' = \varnothing$) and always exhaustive ($A \cup A' = S$), the probability of "not A" is pinned down completely by the probability of $A$.

where: $A'$ is the complementary event. This single line is the fastest tool in the chapter for "at least one" problems — computing $P(\text{at least one of something})$ directly is often messy, while its complement, $P(\text{none of it})$, is frequently a single clean product or ratio; formula (9) then flips the answer in one step. The full derivation, including the corollary $P(\varnothing) = 0$, is given below.

Board level - Probabilities from a well-shuffled deck of cards

One card is drawn from a well-shuffled deck of 52 cards, and every outcome is equally likely. Find the probability that the card is (i) a diamond, (ii) not an ace, (iii) a black card, (iv) not a diamond, and (v) not a black card.

Step 1. Fix the sample space. Since the deck has 52 cards and each is equally likely to be drawn, $n(S) = 52$.

Step 2. Diamond. Let $A$ be "the card is a diamond." There are 13 diamonds in the deck, so $n(A) = 13$ and $P(A) = 13/52 = 1/4$.

Step 3. Not an ace. Let $B$ be "the card is an ace," so $n(B) = 4$ and $P(B) = 4/52 = 1/13$. By the complement rule, $P(\text{not an ace}) = P(B') = 1 - P(B) = 1 - 1/13 = 12/13$.

Step 4. Black card. Let $C$ be "the card is black" (clubs or spades). There are 26 black cards, so $P(C) = 26/52 = 1/2$.

Step 5. Not a diamond and not black, both by the complement rule applied to Steps 2 and 4: $P(A') = 1 - P(A) = 1 - 1/4 = 3/4$, and $P(C') = 1 - P(C) = 1 - 1/2 = 1/2$.

P(diamond) = 1/4, P(not an ace) = 12/13, P(black card) = 1/2, P(not a diamond) = 3/4, P(not a black card) = 1/2.

Competitive level (JEE / NEET) - Kings in a 7-card hand from a 52-card deck

A hand of 7 cards is drawn from a well-shuffled deck of 52 cards. Find the probability that it contains (i) all 4 Kings, (ii) exactly 3 Kings, and (iii) at least 3 Kings.

Step 1. Fix the total sample space. Any 7-card hand is equally likely, and the number of ways to choose 7 cards from 52 is $\binom{52}{7}$, so $n(S) = \binom{52}{7}$.

Step 2. All 4 Kings. A hand with all 4 Kings needs the remaining $7 - 4 = 3$ cards chosen from the other 48 (non-King) cards: $\binom{4}{4}\binom{48}{3}$ favourable hands. So $P(\text{4 Kings}) = \dfrac{\binom{4}{4}\binom{48}{3}}{\binom{52}{7}} = \dfrac{1}{7735}$.

Step 3. Exactly 3 Kings. Choose 3 of the 4 Kings, $\binom{4}{3}$ ways, and the remaining 4 cards from the 48 non-Kings, $\binom{48}{4}$ ways: $P(\text{exactly 3 Kings}) = \dfrac{\binom{4}{3}\binom{48}{4}}{\binom{52}{7}} = \dfrac{9}{1547}$.

Step 4. At least 3 Kings. "At least 3 Kings" means exactly 3 or exactly 4, and these two events are mutually exclusive (a 7-card hand cannot have both exactly 3 and exactly 4 Kings at once), so by axiom (iii), $P(\text{at least 3 Kings}) = P(\text{exactly 3}) + P(\text{exactly 4}) = \dfrac{9}{1547} + \dfrac{1}{7735}$. Converting to a common denominator, $9/1547 = 45/7735$, so the sum is $46/7735$.

P(all 4 Kings) = 1/7735, P(exactly 3 Kings) = 9/1547, P(at least 3 Kings) = 46/7735. Timing note: about three minutes. The trap to avoid is computing $\binom{52}{7}$ or the individual terms by brute multiplication under time pressure — leave the answer in unreduced $\binom{n}{r}$ ratio form as far as possible, and only combine the "exactly 3" and "exactly 4" fractions at the very last step, since "at least 3" is a giveaway that the two exact-count events are mutually exclusive and simply add.

Competitive level (JEE / NEET) - Finishing order in a five-team relay race

In a relay race there are five teams $A, B, C, D, E$, and all finishing orders are equally likely. Find the probability that (a) $A$, $B$ and $C$ finish first, second and third respectively, and (b) $A$, $B$ and $C$ are the first three finishers, in any order.

Step 1. Count the sample space. The sample space consists of all possible orderings of the first three finishing places out of 5 teams, which is a permutation count: $^5P_3 = \dfrac{5!}{(5-3)!} = 5 \times 4 \times 3 = 60$. Since all finishing orders are equally likely, each of these 60 sample points has probability $1/60$.

Step 2. Part (a): exact order $A$ first, $B$ second, $C$ third. There is exactly one ordering of the first three places that matches this description, namely the single sequence $ABC$. So $P(A, B, C \text{ finish 1st, 2nd, 3rd respectively}) = 1/60$.

Step 3. Part (b): $A$, $B$, $C$ occupy the first three places in any order. Now every arrangement of $\{A, B, C\}$ across the first three positions counts as a favourable outcome, and there are $3! = 6$ such arrangements (ABC, ACB, BAC, BCA, CAB, CBA). So the number of favourable sample points is 6.

Step 4. Convert to a probability using the equally-likely-outcomes formula: $P(A, B, C \text{ are the first three finishers}) = \dfrac{6}{60} = \dfrac{1}{10}$.

P(A, B, C finish exactly first, second, third) = 1/60. P(A, B, C are the first three finishers, in any order) = 1/10. Timing note: about two minutes. Recognising that part (b) is simply part (a)'s sample space multiplied by the $3!$ internal orderings of $\{A,B,C\}$ — rather than re-counting from scratch — is the fast route; it is the same idea as scaling a single favourable permutation up to a whole favourable block.

• Forgetting to subtract the overlap. The single most common arithmetic slip is writing $P(A \cup B) = P(A) + P(B)$ for events that are not mutually exclusive — always check $A \cap B$ before adding.
• Assuming "equally likely" without checking. The formula $P(E) = n(E)/n(S)$ only holds when every outcome genuinely has the same chance; a loaded die, a biased coin, or an experiment with a stated non-uniform probability assignment must use the axiomatic definition instead.
• Treating "mutually exclusive" and "exhaustive" as the same idea, or as implying each other. Disjoint events need not cover $S$; events that cover $S$ need not be disjoint. Only when both hold together can you say "exactly one of these occurs."
• Confusing "A but not B" with "not A." $A - B = A \cap B'$ is not the same set as $A'$; check which one a question is actually asking for before writing set notation.
• Mixing up combinations and permutations while counting $n(E)$. When order does not matter (a committee, a hand of cards), use $\binom{n}{r}$; when order does matter (a finishing sequence, a 4-digit code), use $^nP_r$ or direct multiplication — using the wrong one silently changes every downstream probability.
• Misreading "at least" and "at most." "At least $k$" means $k, k+1, \dots$ up to the maximum; "at most $k$" means $0, 1, \dots, k$; swapping them is a frequent and entirely avoidable error, especially under exam time pressure.
• Not verifying a given probability assignment before using it. Before trusting any table of $P(\omega_i)$ values in a problem, check both axioms directly: every value must lie in $[0,1]$, and all the values together must sum to exactly $1$.
• Applying De Morgan's law incorrectly for "neither." "Neither $A$ nor $B$" is $A' \cap B' = (A \cup B)'$, not $A' \cup B'$ — get this backwards and every "neither...nor..." probability computation comes out wrong.

• Define an event as a subset of a sample space, and classify any given event as impossible, sure, simple/elementary, or compound.
• Translate everyday descriptions ("A or B," "A and B," "A but not B," "not A") into the correct set expressions $A \cup B$, $A \cap B$, $A - B$, $A'$, and back again.
• Distinguish mutually exclusive events ($A \cap B = \varnothing$) from exhaustive events (union equals $S$), and identify families of events that are both.
• State the three axioms of probability precisely, and derive $P(\varnothing) = 0$ and $0 \leq P(A) \leq 1$ from them without appealing to counting.
• Verify whether a given assignment of probabilities to sample points is valid, and compute $P(A)$ by summing over sample points when outcomes are not equally likely.
• Apply $P(E) = n(E)/n(S)$ correctly for equally likely outcomes, using permutations or combinations as appropriate to count $n(E)$ and $n(S)$.
• Prove and apply the two-event addition theorem $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, and extend it to three events.
• Apply the complement rule $P(\text{not } A) = 1 - P(A)$ efficiently in "at least one" problems by first computing the probability of the complementary "none" event.
• Solve card, dice, coin, and committee-selection probability problems that combine counting techniques with the addition and complement rules, at both board and JEE/NEET difficulty.

Ask the tutor: Want to work through a few more card, dice, or committee-selection probability problems together, blending the addition theorem and complement rule with permutation and combination counting?