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Maths · Class 11 · Chapter 14

Probability

Class 10 probability was about counting favourable outcomes. Here the same ideas are rebuilt on the language of sets: outcomes form a sample space, events are subsets, and "or", "and" and "not" become union, intersection and complement. Conditional probability and Bayes' theorem follow in Class 12.

In this chapter: random experiments, outcomes and sample space, events and their types, the algebra of events, mutually exclusive and exhaustive events, the axiomatic definition of probability, equally likely outcomes, and the rules for P(A or B), P(not A) and P(A but not B).

Random experiments and sample space

An experiment is random if it has more than one possible outcome and the outcome cannot be predicted in advance. Tossing a coin, rolling a die and drawing a card are the standard examples.

Each possible result is an outcome, and the set of all outcomes is the sample space S. Each element of S is a sample point.

  • One coin tossed twice: S = {HH, HT, TH, TT}, 4 outcomes.
  • One die rolled: S = {1, 2, 3, 4, 5, 6}.
  • Two dice rolled: S = {(x, y) : x, y = 1, 2, ..., 6}, 36 outcomes.
  • A coin tossed n times: 2n outcomes.

For experiments done in stages, a tree diagram lists the sample space without missing anything.

Sample space of three coin tosses as a treewww.iitmedicoguide.com1st toss2nd toss3rd tossoutcomeHHHHTHHTHHHTHTHTTTHHTHHTTHTHHTTHTTTTTTat least two heads4 of the 8 outcomeswww.iitmedicoguide.com
Three tosses of a coin give 2 × 2 × 2 = 8 equally likely outcomes. The four highlighted ones have at least two heads, so that event has probability 4/8 = 1/2.

Events

An event is any subset E of the sample space. The event E occurs if the outcome of the experiment is an element of E. For a die, "an even number shows" is the event E = {2, 4, 6}; if a 4 turns up, E has occurred.

TypeMeaningExample (one die)
Impossible eventThe empty set φ"a 7 shows"
Sure eventS itself"a number less than 7 shows"
Simple eventExactly one sample point{5}
Compound eventMore than one sample point{1, 3, 5}

Algebra of events

In wordsIn set notation
not A (complementary event)A′ = S − A
A or B (at least one of them)A ∪ B
A and B (both)A ∩ B
A but not BA − B = A ∩ B′
  • A and B are mutually exclusive if A ∩ B = φ: they cannot occur together. Simple events of an experiment are always mutually exclusive.
  • E1, E2, ..., En are exhaustive if E1 ∪ E2 ∪ ... ∪ En = S: at least one of them must occur.
  • If they are also pairwise disjoint, they are mutually exclusive and exhaustive, which means exactly one of them occurs.

For a die, A = {1, 3, 5} and B = {2, 4, 6} are mutually exclusive and exhaustive. A = {1, 2, 3} and B = {3, 4, 5, 6} are exhaustive but not mutually exclusive, because 3 belongs to both.

Axiomatic approach to probability

Probability is a function P that assigns a real number P(E) to each event E of S, satisfying three axioms:

  1. P(E) ≥ 0 for every event E.
  2. P(S) = 1.
  3. If E and F are mutually exclusive, P(E ∪ F) = P(E) + P(F).

From these it follows that P(φ) = 0, that 0 ≤ P(E) ≤ 1, and that the probability of an event is the sum of the probabilities of the sample points in it. If S has n outcomes that are all equally likely, each has probability 1/n and

P(E) = n(E)/n(S) = (number of outcomes favourable to E)/(total number of outcomes)

This formula holds only when the outcomes are equally likely. The sums of two dice, 2 to 12, are not equally likely, so you must count from the 36 ordered pairs.

Worked example: Two dice are rolled. Find the probability that (a) the sum is 8, (b) a doublet appears.
Solution: n(S) = 36. (a) Sum 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), so P = 5/36. (b) Doublets: (1, 1), (2, 2), ..., (6, 6), so P = 6/36 = 1/6.

Rules for combining events

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)P(A ∪ B) = P(A) + P(B)if A and B are mutually exclusiveP(A′) = 1 − P(A)P(A but not B) = P(A ∩ B′) = P(A) − P(A ∩ B)P(neither A nor B) = P(A′ ∩ B′) = 1 − P(A ∪ B)

The first rule is the probability version of n(A ∪ B) = n(A) + n(B) − n(A ∩ B) from the Sets chapter; the overlap would otherwise be counted twice. The complement rule is the standard trick for "at least one" questions.

Worked example: (a) P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Find P(not A), P(not B) and P(A or B). (b) One card is drawn from a well-shuffled pack of 52. Find the probability that it is a king or a heart.
Solution: (a) P(not A) = 0.58, P(not B) = 0.52, and P(A or B) = 0.42 + 0.48 − 0.16 = 0.74.
(b) P(king) = 4/52, P(heart) = 13/52, and P(king of hearts) = 1/52. So P(king or heart) = (4 + 13 − 1)/52 = 16/52 = 4/13.

Using combinations

When items are chosen together, count with nCr. If 2 people are chosen at random from 4 men and 6 women, P(both are women) = 6C2/10C2 = 15/45 = 1/3, and P(at least one man) = 1 − 1/3 = 2/3.

Common mistakes: (1) Using n(E)/n(S) when the outcomes are not equally likely. (2) Listing (3, 5) and (5, 3) as one outcome for two dice; they are different. (3) Adding probabilities of events that overlap without subtracting P(A ∩ B). (4) Confusing mutually exclusive with exhaustive. (5) Accepting an answer greater than 1 or less than 0 without rechecking. (6) Solving "at least one" questions by listing cases when 1 − P(none) is much shorter.

Exam focus

  • Writing sample spaces for coins, dice and cards, and counting favourable outcomes correctly.
  • Translating "or", "and", "not" and "but not" into set operations.
  • Addition rule and complement rule, including P(neither A nor B).
  • Deciding whether given events are mutually exclusive, exhaustive, or both.
  • Probability problems that need nCr for counting selections.

Practice questions

Two coins are tossed. The probability of getting at least one head is:

  1. 1/4
  2. 1/2
  3. 3/4
  4. 1
Show answer
C. 1 − P(TT) = 1 − 1/4 = 3/4.

Two dice are rolled. The probability that the sum is 8 is:

  1. 5/36
  2. 1/6
  3. 1/9
  4. 7/36
Show answer
A. Five favourable pairs out of 36.

If P(A) = 0.42, P(B) = 0.48 and P(A ∩ B) = 0.16, then P(A ∪ B) is:

  1. 0.90
  2. 0.74
  3. 0.58
  4. 0.26
Show answer
B. 0.42 + 0.48 − 0.16 = 0.74.

A card is drawn from a pack of 52. The probability that it is a king or a heart is:

  1. 17/52
  2. 1/4
  3. 1/13
  4. 4/13
Show answer
D. (4 + 13 − 1)/52 = 16/52 = 4/13; the king of hearts is counted once.

Which of the following cannot be the probability of an event?

  1. 0.3
  2. 0
  3. 1.2
  4. 1
Show answer
C. A probability always lies between 0 and 1.

A and B are mutually exclusive with P(A) = 0.3 and P(B) = 0.4. Then P(neither A nor B) is:

  1. 0.12
  2. 0.3
  3. 0.7
  4. 0.42
Show answer
B. P(A ∪ B) = 0.7, so P(A′ ∩ B′) = 1 − 0.7 = 0.3.

Two people are chosen at random from 4 men and 6 women. The probability that both are women is:

  1. 1/3
  2. 2/5
  3. 3/5
  4. 1/5
Show answer
A. 6C2/10C2 = 15/45.

A coin is tossed three times. The probability of exactly two heads is:

  1. 1/2
  2. 1/8
  3. 7/8
  4. 3/8
Show answer
D. HHT, HTH, THH: 3 of the 8 outcomes.
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