📌 Snapshot
- Why a total of seven beats twelve on two dice, and how events, axioms and addition rules measure chance.
- The language of sample spaces and events: simple, compound, exclusive, exhaustive, and the algebra of events.
- The three axioms of probability, the complement rule and the addition theorem, each with the reason it holds.
- Coins, dice, cards and bags solved step by step, plus the traps that catch students in NTA style questions.
📖 Detailed Notes
Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 14.
Why is 7 the popular total?
In a board game with two dice, you will land on a total of 7 far more often than on 12. There are 11 possible totals, 2 to 12, but they are not equally likely. A total of 12 needs a double six and nothing else. A total of 7 can come as 1 and 6, 2 and 5, 3 and 4, and each of those in either order. Counting the ways, 6 out of 36 equally likely pairs give a 7 and only 1 out of 36 gives a 12. Probability is the tidy language for exactly this kind of counting.
Experiments, outcomes and sample spaces
A random experiment is one whose result cannot be predicted in advance, even though all possible results are known, such as tossing a coin or throwing a die. Each possible result is an outcome, and the set of all outcomes is the sample space $S$.
| Experiment | Sample space $S$ | $n(S)$ |
|---|---|---|
| One coin | $\{H, T\}$ | 2 |
| Two coins | $\{HH, HT, TH, TT\}$ | 4 |
| Three coins | $\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$ | 8 |
| One die | $\{1, 2, 3, 4, 5, 6\}$ | 6 |
| Two dice | Ordered pairs $(a, b)$ with $a, b = 1, \dots, 6$ | 36 |
| One card from a pack | The 52 cards | 52 |
Notice that for two coins the outcomes $HT$ and $TH$ are different, and for two dice $(1, 6)$ and $(6, 1)$ are different. Order matters when you list the sample space, because that is what makes the outcomes equally likely.
A standard pack has 52 cards in 4 suits of 13 each (spades and clubs are black, hearts and diamonds are red). The 13 cards of a suit are an ace, 2 to 10, and the three face cards jack, queen and king. So there are $4 \times 3 = 12$ face cards, 4 aces and 4 kings.
Events and their types
An event is any subset $E$ of the sample space. The event occurs when the outcome of the experiment belongs to $E$.
| Type | Meaning | Example with a die |
|---|---|---|
| Impossible event | The empty set $\varnothing$ | A multiple of 7 |
| Sure event | The whole space $S$ | An odd or an even number |
| Simple (elementary) event | Exactly one outcome | $\{4\}$ |
| Compound event | More than one outcome | $\{2, 4, 6\}$, an even number |
A sample space with $n$ outcomes has exactly $n$ simple events. Events are sets, so they follow the same algebra as sets.
| In words | In symbols | Set meaning |
|---|---|---|
| Not $A$ | $A'$ | $S - A$, everything outside $A$ |
| $A$ or $B$ | $A \cup B$ | At least one of them occurs |
| $A$ and $B$ | $A \cap B$ | Both occur |
| $A$ but not $B$ | $A - B = A \cap B'$ | In $A$ but outside $B$ |
| Neither $A$ nor $B$ | $A' \cap B' = (A \cup B)'$ | De Morgan: outside the union |
"Or" in probability is inclusive. $A \cup B$ includes the case where both happen.
Mutually exclusive and exhaustive events
Two events are mutually exclusive if they cannot happen together, $A \cap B = \varnothing$. A set of events is exhaustive if together they cover the whole sample space, $E_1 \cup E_2 \cup \dots \cup E_n = S$. The two ideas are independent. The cleanest way to see this is a die.
| $A$ | $B$ | Mutually exclusive? | Exhaustive? |
|---|---|---|---|
| $\{1, 2\}$ | $\{5, 6\}$ | Yes | No |
| $\{1, 2, 3, 4\}$ | $\{3, 4, 5, 6\}$ | No, both contain 3 and 4 | Yes |
| $\{1, 3, 5\}$ | $\{2, 4, 6\}$ | Yes | Yes |
| $\{1, 2, 3\}$ | $\{3, 4\}$ | No | No |
Exclusive asks about the overlap and exhaustive asks about the union. Events that are both form a partition of $S$. Simple events of any sample space are always mutually exclusive.
Pause and predict: a coin is tossed twice. What is the probability of getting one head and one tail?
Check your answer: $\dfrac{1}{2}$, because the sample space has four equally likely outcomes and two of them, $HT$ and $TH$, give one of each. A common slip is to count "one head one tail" as a single outcome and answer $\dfrac{1}{3}$. The three cases HH, one of each and TT are not equally likely.
The axioms of probability
We assign each outcome $\omega_i$ a number $P(\omega_i)$, and the probability of an event is the sum of the numbers of its outcomes. The assignment is valid exactly when these hold.
| Axiom | Statement | Why it is natural |
|---|---|---|
| 1 | $P(E) \ge 0$ for every event | A chance cannot be negative |
| 2 | $P(S) = 1$ | Something in the sample space must happen |
| 3 | $P(E \cup F) = P(E) + P(F)$ for mutually exclusive $E$ and $F$ | Chances of non overlapping events simply add |
For a finite sample space this means each $P(\omega_i)$ lies between $0$ and $1$ and the numbers add up to $1$. Everything else is a consequence. Take $F = \varnothing$ in the third axiom: $P(E) = P(E) + P(\varnothing)$, so $P(\varnothing) = 0$. Since $A$ and $A'$ are mutually exclusive with union $S$, $P(A) + P(A') = 1$. This gives the complement rule and also shows $0 \le P(A) \le 1$.
$$P(A') = 1 - P(A)$$
Equally likely outcomes. If all $n$ outcomes are equally likely with chance $p$ each, then $np = 1$ and $p = \dfrac{1}{n}$. For an event $E$ with $n(E)$ favourable outcomes,
$$P(E) = \dfrac{n(E)}{n(S)}$$
This is only valid when the outcomes are equally likely. The axioms themselves allow any assignment that satisfies the three conditions, such as a biased coin with $P(H) = 0.7$ and $P(T) = 0.3$.
Worked example: is it a valid assignment? Which of these can be probabilities of the four outcomes $a, b, c, d$? (i) $\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5}$ (ii) $0.1, 0.2, 0.3, 0.4$ (iii) $0.5, 0.3, 0.4, -0.2$.
| Step | What we do | Result |
|---|---|---|
| 1 | Check every value is between $0$ and $1$ | (i) and (ii) pass; (iii) has $-0.2$, so it fails |
| 2 | Add the values | (i) $\dfrac{30 + 20 + 15 + 12}{60} = \dfrac{77}{60} > 1$, so it fails. (ii) $= 1.0$, so it passes |
| 3 | Conclude | Only (ii) is valid |
Common wrong step. Accepting (iii) because its total is $1.0$. Both conditions are needed, so one negative entry is enough to fail.
Counting with two dice
For two dice the 36 ordered pairs are equally likely, so any question about totals is a count. By symmetry, a total $t$ and a total $14 - t$ occur equally often.
| Total | Ways | Probability |
|---|---|---|
| 2 or 12 | 1 | $\dfrac{1}{36}$ |
| 3 or 11 | 2 | $\dfrac{2}{36}$ |
| 4 or 10 | 3 | $\dfrac{3}{36}$ |
| 5 or 9 | 4 | $\dfrac{4}{36}$ |
| 6 or 8 | 5 | $\dfrac{5}{36}$ |
| 7 | 6 | $\dfrac{6}{36} = \dfrac{1}{6}$ |
Worked example: totals on two dice. Find $P(\text{total} = 7)$ and $P(\text{total} \ge 10)$.
| Step | What we do | Result |
|---|---|---|
| 1 | List the pairs for a total of 7 | $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$, so $6$ pairs |
| 2 | Divide by $n(S) = 36$ | $P(7) = \dfrac{6}{36} = \dfrac{1}{6}$ |
| 3 | A total of at least 10 means 10, 11 or 12 | $3 + 2 + 1 = 6$ pairs |
| 4 | Divide | $P = \dfrac{6}{36} = \dfrac{1}{6}$ |
Common wrong step. Treating the 11 totals $2, 3, \dots, 12$ as equally likely and writing $\dfrac{1}{11}$. The totals are not the equally likely outcomes. The 36 ordered pairs are.
Addition theorem
When you want the chance of $A$ or $B$, adding $P(A) + P(B)$ counts the overlap twice. The fix is to subtract it once.
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Why it holds. Split $A \cup B$ into the two non overlapping pieces $A$ and $B - A$, so $P(A \cup B) = P(A) + P(B - A)$. Also $B$ splits into $A \cap B$ and $B - A$, so $P(B - A) = P(B) - P(A \cap B)$. Substitute and you have the theorem. If $A$ and $B$ are mutually exclusive, $A \cap B = \varnothing$ and the formula shrinks to the third axiom.
For three events, the pattern of add singles, subtract pairs and add back the triple continues.
$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$$
For example, with $P(A) = 0.4$, $P(B) = 0.3$, $P(C) = 0.2$, the pair overlaps $0.1$, $0.05$, $0.05$ and the triple overlap $0.02$, the union is $0.9 - 0.2 + 0.02 = 0.72$.
Worked example: king or heart. One card is drawn from a well shuffled pack. Find $P(\text{king or heart})$.
| Step | What we do | Result |
|---|---|---|
| 1 | Name the events | $K$ = king, $H$ = heart |
| 2 | Single probabilities | $P(K) = \dfrac{4}{52}$, $P(H) = \dfrac{13}{52}$ |
| 3 | The king of hearts is in both | $P(K \cap H) = \dfrac{1}{52}$ |
| 4 | Apply the theorem | $\dfrac{4 + 13 - 1}{52} = \dfrac{16}{52} = \dfrac{4}{13}$ |
Common wrong step. Adding to get $\dfrac{17}{52}$. The king of hearts has been counted once as a king and once as a heart. Check whether the events overlap before you choose the short form.
"At least one", "neither" and "exactly one"
Words in the question decide the set, and the set decides the formula.
| Phrase | Set | Probability |
|---|---|---|
| At least one of $A$, $B$ | $A \cup B$ | $P(A) + P(B) - P(A \cap B)$ |
| Neither $A$ nor $B$ | $A' \cap B'$ | $1 - P(A \cup B)$ |
| Exactly one of them | $(A - B) \cup (B - A)$ | $P(A) + P(B) - 2P(A \cap B)$ |
| Not $A$ | $A'$ | $1 - P(A)$ |
| At least one success in several trials | complement of "none" | $1 - P(\text{none})$ |
Worked example: at least one head. A coin is tossed three times. Find the probability of at least one head.
| Step | What we do | Result |
|---|---|---|
| 1 | "At least one" is awkward to list, so use its complement | The opposite is "no head", the single outcome $TTT$ |
| 2 | Probability of the complement | $P(TTT) = \dfrac{1}{8}$ |
| 3 | Subtract from $1$ | $1 - \dfrac{1}{8} = \dfrac{7}{8}$ |
Common wrong step. Listing only the outcomes with exactly one head and answering $\dfrac{3}{8}$. "At least one" includes two heads and three heads as well.
Worked example: neither event. Two students $E$ and $F$ qualify for a test with $P(E) = 0.05$, $P(F) = 0.10$ and $P(E \cap F) = 0.02$. Find the probability that neither qualifies.
| Step | What we do | Result |
|---|---|---|
| 1 | Chance at least one qualifies | $0.05 + 0.10 - 0.02 = 0.13$ |
| 2 | "Neither" is the complement of the union | $1 - 0.13 = 0.87$ |
Common wrong step. Writing $1 - 0.05 - 0.10 = 0.85$. That ignores the overlap and subtracts the students who qualify together twice.
Pause and predict: could two events with $P(A) = 0.6$ and $P(B) = 0.7$ be mutually exclusive?
Check your answer: No. If they were mutually exclusive, $P(A \cup B) = 0.6 + 0.7 = 1.3$, which is more than $1$ and impossible. So $P(A \cap B)$ must be at least $0.3$. A union can never have a probability greater than $1$.
Counting with combinations
When a question picks several items together, the sample space is a set of combinations. The counts come from $^nC_r$ in Permutations and Combinations.
Worked example: two balls from a bag. A bag has 4 red, 3 blue and 2 yellow balls. Two balls are drawn together. Find the probability that (a) both are red and (b) at least one is yellow.
| Step | What we do | Result |
|---|---|---|
| 1 | Total ways to pick 2 of 9 | $^9C_2 = 36$ |
| 2 | (a) Both red: 2 of the 4 reds | $^4C_2 = 6$, so $P = \dfrac{6}{36} = \dfrac{1}{6}$ |
| 3 | (b) Use the complement: no yellow means 2 of the other 7 balls | $^7C_2 = 21$, so $P(\text{no yellow}) = \dfrac{21}{36}$ |
| 4 | At least one yellow | $1 - \dfrac{21}{36} = \dfrac{15}{36} = \dfrac{5}{12}$ |
Common wrong step. Counting ordered pairs in the numerator and unordered pairs in the denominator. Whether you use ordered or unordered counts, use the same kind for both.
Beyond the textbook. For the exam: $P(E) = \dfrac{n(E)}{n(S)}$ for equally likely outcomes, and the axioms cover everything else. In reality: many real experiments have outcomes that are not equally likely, such as rain tomorrow or a loaded die. Probabilities for these come from long run frequencies, and the three axioms, laid down by Kolmogorov in 1933, still apply to them. Why it matters: this is why a weather forecast can say 70 percent without any counting of equally likely cases, and why insurers can use the same rules on accident data.
Beyond the textbook. For the exam: "at least one" is $1 - P(\text{none})$. In reality: this trick gives the famous birthday result. Among just 23 people, the chance that at least two share a birthday is about 50.7 percent, because it is easier to compute the chance that all 23 birthdays are different, $\dfrac{365}{365} \times \dfrac{364}{365} \times \dots \times \dfrac{343}{365} \approx 0.493$, and subtract it from $1$. Why it matters: the complement turns an impossible long list into a single product, and shows why intuition about chance is often wrong.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| "Exhaustive events are mutually exclusive." | Both words sound like a complete split | Exclusive is about the overlap, exhaustive is about the union. $\{1,2,3,4\}$ and $\{3,4,5,6\}$ are exhaustive but not exclusive |
| $P(A \cup B) = P(A) + P(B)$ always | It is the third axiom | It needs $A \cap B = \varnothing$. Otherwise subtract $P(A \cap B)$ |
| The 11 totals on two dice are equally likely | They look like 11 outcomes | Use the 36 ordered pairs. $P(7) = \dfrac{1}{6}$ and $P(12) = \dfrac{1}{36}$ |
| $HT$ and $TH$ are the same outcome | A person sees one head and one tail | They are two outcomes, so the chance is $\dfrac{1}{2}$, not $\dfrac{1}{3}$ |
| $(A \cup B)' = A' \cup B'$ | The bars just move inside | De Morgan flips the sign: $(A \cup B)' = A' \cap B'$ |
| An assignment is valid if the values add to $1$ | The sum is the first thing you check | Every value must also lie between $0$ and $1$, so no negatives |
| "At least one" means "exactly one" | The words feel alike | "At least one" is $1 - P(\text{none})$ and includes all larger counts |
| $P(E) = \dfrac{n(E)}{n(S)}$ for every experiment | It is the formula you use most | It needs equally likely outcomes. A biased coin has $P(H) \ne \dfrac{1}{2}$ |
Formula sheet
| Quantity | Formula | When to use |
|---|---|---|
| Equally likely outcomes | $P(E) = \dfrac{n(E)}{n(S)}$ | Fair coins, dice, cards, random selection |
| Bounds | $0 \le P(E) \le 1$, $P(\varnothing) = 0$, $P(S) = 1$ | Checking an answer |
| Valid assignment | Each $P(\omega_i) \in [0,1]$ and $\sum P(\omega_i) = 1$ | "Is this a probability distribution?" |
| Complement | $P(A') = 1 - P(A)$ | "Not", "at least one" |
| Addition theorem | $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ | "A or B" |
| Mutually exclusive | $P(A \cup B) = P(A) + P(B)$ | $A \cap B = \varnothing$ |
| Neither | $P(A' \cap B') = 1 - P(A \cup B)$ | "Neither A nor B" |
| Exactly one | $P(A) + P(B) - 2P(A \cap B)$ | "Only one of them" |
| Three events | Singles minus pairs plus the triple | $A \cup B \cup C$ |
| Selecting together | $\dfrac{n(E)}{n(S)}$ with $n(S) = {}^nC_k$ | Balls, cards, committees |
60-second revision
| Topic | Lock this in |
|---|---|
| Sample space | All outcomes, listed in order: two coins give 4, two dice give 36 |
| Events | Subsets; $\varnothing$ is impossible, $S$ is sure; simple means one outcome |
| Event algebra | $A'$, $A \cup B$ (or), $A \cap B$ (and), $A - B = A \cap B'$ |
| Exclusive vs exhaustive | No overlap vs covers $S$; both together is a partition |
| Axioms | $P(E) \ge 0$, $P(S) = 1$, additivity for exclusive events |
| Complement | $1 - P(A)$; the shortcut for "at least one" |
| Addition | Subtract the overlap once |
| Cards | 52 = 4 suits of 13; 12 face cards; 4 aces; 26 red |
| Two dice | 36 pairs; total 7 has 6 ways |
Where this chapter connects
- Before: Sets supplies the union, intersection and complement that event algebra borrows, and Permutations and Combinations supplies the counting for cards and committees.
- Neighbour: Statistics describes data that has already happened, while probability predicts what is yet to happen.
- Class XII: Probability builds conditional probability, independence and Bayes' theorem on the axioms and addition rule you learn here.
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. A fair coin is tossed three times. The number of sample points and the probability of getting exactly two heads are respectively
▸ Show answer & explanation
Answer: A
The sample space has $2^3 = 8$ points. Exactly two heads is the event $\{HHT, HTH, THH\}$, with $3$ points, so the probability is $\dfrac{3}{8}$. Option B would need $2$ favourable points.
Q2. Two fair dice are thrown. The probability that the total is $9$ is
▸ Show answer & explanation
Answer: B
The favourable pairs are $(3,6), (4,5), (5,4), (6,3)$, which is $4$ of the $36$ equally likely pairs. So the probability is $\dfrac{4}{36} = \dfrac{1}{9}$. Option D wrongly treats the $11$ totals as equally likely.
Q3. A die is thrown. Let $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$. Then $A$ and $B$ are
▸ Show answer & explanation
Answer: B
$A \cup B = \{1, 2, 3, 4, 5, 6\} = S$, so they are exhaustive. But $A \cap B = \{3, 4\} \ne \varnothing$, so they are not mutually exclusive.
🔒 9 more practice MCQs
Create a free account to unlock every MCQ in this chapter — answers and explanations included. No payment needed.
Already registered? Just log in and they'll all appear here.
Q4. Which of the following is a valid assignment of probabilities to four outcomes $a, b, c, d$?
▸ Show answer & explanation
Answer: B
Option A adds to $1.1$. Option C adds to $1.0$ but has a negative value. Option D adds to $\dfrac{77}{60}$, which is more than $1$. Only option B has all values between $0$ and $1$ with total exactly $1$.
Q5. If $P(A) = 0.45$, $P(B) = 0.35$ and $P(A \cup B) = 0.6$, then $P(A \cap B)$ is
▸ Show answer & explanation
Answer: B
From $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we get $P(A \cap B) = 0.45 + 0.35 - 0.6 = 0.2$. Option D would hold only if the events were mutually exclusive, which would give a union of $0.8$.
Q6. One card is drawn from a well shuffled pack of 52 cards. The probability that it is a face card or a spade is
▸ Show answer & explanation
Answer: B
There are $12$ face cards, $13$ spades and $3$ cards that are both, the jack, queen and king of spades. So the number favourable is $12 + 13 - 3 = 22$ and the probability is $\dfrac{22}{52} = \dfrac{11}{26}$. Option A forgets to subtract the overlap.
Q7. If $P(E) = 0.3$, $P(F) = 0.4$ and $P(E \cap F) = 0.1$, the probability that neither $E$ nor $F$ occurs is
▸ Show answer & explanation
Answer: B
$P(E \cup F) = 0.3 + 0.4 - 0.1 = 0.6$. By De Morgan, $P(E' \cap F') = 1 - 0.6 = 0.4$. Option A comes from $1 - 0.3 - 0.4$, which ignores the overlap.
Q8. A bag contains $5$ red and $4$ white balls. Two balls are drawn together at random. The probability that they are of different colours is
▸ Show answer & explanation
Answer: B
The total number of ways is $^9C_2 = 36$. One red and one white can be chosen in $5 \times 4 = 20$ ways. So the probability is $\dfrac{20}{36} = \dfrac{5}{9}$. Option C is $\dfrac{5}{9} \times \dfrac{4}{9}$, the chance of red then white with replacement, and it also ignores the other order.
Q9. Assertion (A): For any event $A$, $P(A') = 1 - P(A)$. Reason (R): The events $A$ and $A'$ are mutually exclusive and exhaustive.
▸ Show answer & explanation
Answer: A
Since $A \cap A' = \varnothing$ and $A \cup A' = S$, the axioms give $P(A) + P(A') = P(S) = 1$. So R is exactly the reason for A.
Q10. Consider the statements. I: Simple events of a sample space are always mutually exclusive. II: The sure event has probability $1$ and the impossible event has probability $0$. III: $(A \cup B)' = A' \cup B'$. IV: The event "A but not B" is $A \cap B'$. Which are correct?
▸ Show answer & explanation
Answer: A
Two different simple events are single outcomes with nothing in common, so I is true. II and IV are true by definition. III is false: De Morgan's law gives $(A \cup B)' = A' \cap B'$.
Q11. Match List I with List II. List I: (i) number of sample points when a coin is tossed 4 times (ii) number of face cards in a pack of 52 cards (iii) probability that the total on two dice is 7 (iv) $P(A')$ when $P(A) = 0.35$. List II: (p) $0.65$ (q) $16$ (r) $\dfrac{1}{6}$ (s) $12$.
▸ Show answer & explanation
Answer: A
Four tosses give $2^4 = 16$ points. There are $4 \times 3 = 12$ face cards. A total of $7$ has probability $\dfrac{6}{36} = \dfrac{1}{6}$. The complement has probability $1 - 0.35 = 0.65$.
Q12. For three events, $P(A) = 0.4$, $P(B) = 0.3$, $P(C) = 0.2$, $P(A \cap B) = 0.1$, $P(A \cap C) = 0.05$, $P(B \cap C) = 0.05$ and $P(A \cap B \cap C) = 0.02$. Then $P(A \cup B \cup C)$ is
▸ Show answer & explanation
Answer: B
The singles add to $0.9$, the pairs add to $0.2$, and the triple is $0.02$. So $P(A \cup B \cup C) = 0.9 - 0.2 + 0.02 = 0.72$. Option A forgets to add back the triple, and option C adds the singles only.
📊 Previous-Year Questions
Practise with real CUET Mathematics previous-year papers — every question solved, with the correct answer and a step-by-step explanation.
View solved CUET PYQ papers →Ready to drill Mathematics?
Unlock all MCQs, chapter tests, mocks & PYQs for ₹199/year.
Get UniDrill Pro