Probability | ISC Class 11 Maths Notes

This note covers random experiments, outcomes, sample spaces, types of events, operations on events, mutually exclusive and exhaustive events, axiomatic probability, equally likely outcomes, the addition theorem and complementary probabilities.

What are random experiments, outcomes and sample spaces?

Probability measures the chance that a specified event, or set of outcomes, occurs.

A random experiment is a procedure with known possible outcomes whose particular result cannot be predicted with certainty before it is performed. An outcome is a possible result. Performing the experiment once is called a trial.

The sample space, denoted by S, is the set of all possible outcomes of an experiment. A set is a collection of distinct elements; an element of a sample space is also called a sample point. Curly brackets enclose the elements of a set.

How does the description determine the outcomes?

For a coin, let H mean head and T mean tail. When a coin is tossed twice, write the first result before the second. Thus HT means head followed by tail, whereas TH means tail followed by head.

The sample space is S = {HH, HT, TH, TT}. These are four different outcomes, even though HT and TH both contain one head. Recording the sequence preserves the distinction between the first and second tosses.

Worked example 1. A coin is tossed twice. Write the sample space and identify the outcomes having exactly one head.

Answer: S = {HH, HT, TH, TT}. Exactly one head occurs in HT and TH, so the required set is {HT, TH}, containing 2 outcomes.

A standard die has six faces numbered 1, 2, 3, 4, 5 and 6. For one throw, S = {1, 2, 3, 4, 5, 6}. For two throws, an ordered pair records the first score and then the second.

There are 36 possible ordered pairs for two die throws. The first position has six possibilities and each permits six possibilities in the second position. Specifying outcomes is separate from assigning their probabilities: listing outcomes alone does not establish that they have equal chances.

What is an event, and when does it occur?

Definition: An event is any subset of a sample space. A subset is a set whose elements all belong to the original set.

Let E denote an event associated with sample space S. The notation E ⊆ S means that E is a subset of S. An event occurs when the actual outcome belongs to its set, rather than when every outcome in that set happens.

For a die throw, the event “a number less than 4 appears” is E = {1, 2, 3}. If the result is 1, 2 or 3, E occurs. If the result is 4, 5 or 6, E does not occur.

How can different descriptions identify the same event?

For two coin tosses, the following sets connect verbal conditions to outcomes. The letters A, B, C and D name the events in this table. The symbol ∅ denotes the empty set, a set containing no elements.

At least means the stated number or more; at most means the stated number or fewer. “Exactly” allows only the stated number. Thus “at most one head” includes no heads, represented by TT, as well as exactly one head.

Events B and C are equal because their elements are identical, despite their different descriptions. Comparing the sets is therefore more reliable than comparing the wording alone. The empty set and the whole sample space are themselves events, because each is a subset of S.

How do simple, compound, sure and impossible events differ?

A simple event, also called an elementary event, contains one sample point. A compound event contains more than one sample point. Classification depends on the number of outcomes in the event, not on the length of its verbal description.

For two coin tosses, {HH}, {HT}, {TH} and {TT} are the four simple events. Notice the distinction between HH, which is an outcome, and {HH}, which is the event containing that outcome.

Which events are certain or impossible?

The sure event is S itself, because every possible result belongs to it. The impossible event is ∅, because no possible result belongs to it. Both classifications refer to a particular experiment and its sample space.

Worked example 2. A standard die numbered 1 to 6 is rolled. Classify the events “a multiple of 7 appears” and “an odd or even number appears”.

Answer: No face is a multiple of 7, so the first event is ∅ and is impossible. Every face is odd or even, so the second event is S = {1, 2, 3, 4, 5, 6} and is sure.

For three coin tosses, S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. The event of exactly one head is {HTT, THT, TTH}, a compound event with three sample points.

The event of at most one head is {TTT, THT, HTT, TTH}. The event of at least one head contains every outcome except TTT. Neither expression means “exactly one head”, so these descriptions must not be interchanged.

If a sample space contains n distinct outcomes, where n denotes their number, it has n simple events. Each such event contains a different outcome. An event can also carry more than one applicable description: the sure event for a die throw contains several outcomes and is compound.

How do “not”, “or”, “and” and “but not” describe events?

Let A and B be events in the same sample space S. Their complement, union, intersection and difference translate verbal conditions into sets. Establishing these sets first helps prevent using a probability rule for the wrong event.

The prime mark in A′ denotes a complement. The symbols ∪ and ∩ denote union and intersection respectively. The minus sign between sets denotes set difference, so A − B = A ∩ B′. Its meaning differs from subtraction between numerical probabilities.

How are these operations applied to a die?

A prime number is a whole number greater than 1 divisible, without a remainder, only by 1 and itself. Thus 2, 3 and 5 are the prime numbers on a standard die.

Worked example 3. A standard die numbered 1 to 6 is rolled. Let A = {2, 3, 5}, the prime-number event, and B = {1, 3, 5}, the odd-number event. Find their union, intersection, difference A − B and complement A′.

Answer: A ∪ B = {1, 2, 3, 5}; A ∩ B = {3, 5}; A − B = {2}; A′ = {1, 4, 6}.

The word or is inclusive here: outcomes common to both events remain in the union. The outcomes 3 and 5 are therefore included in “prime or odd”. They also make up the intersection because each satisfies both conditions.

For two die throws, let A mean that the first score is six and B mean that the sum is at least 11. Then B = {(5, 6), (6, 5), (6, 6)}, while A ∩ B = {(6, 5), (6, 6)}.

The outcome (5, 6) belongs to B but not to A, since the first score is five. This illustrates why “and” requires both conditions to hold for the same outcome, rather than allowing separate outcomes to satisfy separate conditions.

How are mutually exclusive and exhaustive events distinguished?

Two events are mutually exclusive when they cannot occur together in one trial. Their sets are disjoint, meaning that they have no common element. In symbols, A ∩ B = ∅.

Events are exhaustive when their union is the whole sample space. At least one of them necessarily occurs whenever the experiment is performed. Exhaustiveness concerns coverage of the sample space; mutual exclusiveness concerns overlap between events.

Can exhaustive events overlap?

For a standard die, take A = {1, 2, 3}, B = {3, 4} and C = {5, 6}, where C names a third event. Their union is S = {1, 2, 3, 4, 5, 6}, so they are exhaustive.

However, A and B both contain 3. They are not mutually exclusive. By contrast, the odd-number and even-number events are both mutually exclusive and exhaustive: no face is both odd and even, and every face is one or the other.

Worked example 4. A coin is tossed three times. Let A mean no head, B exactly one head, and C at least two heads. Determine whether the events are mutually exclusive and exhaustive.

Answer: A = {TTT}, B = {HTT, THT, TTH}, and C = {HHT, HTH, THH, HHH}. Every pair has an empty intersection, and their union contains all 8 outcomes. The events are mutually exclusive and exhaustive.

For more than two events, pairwise disjoint means that every pair has an empty intersection. Checking just one pair is insufficient. In the worked example, none of the three head-count categories can hold together with either of the others.

Note: Distinct simple events are mutually exclusive because each contains a different single outcome. To establish that a collection is exhaustive, check that no sample point has been left out.

Complementary events A and A′ satisfy both requirements. They cannot occur together, because A′ contains no outcome from A. They are exhaustive because every outcome in S lies either inside A or outside A.

What rules define axiomatic probability?

Probability assigns a number to an event to express its chance of occurrence. Write P(E) for the probability of event E. The assignment P acts on events, so its inputs are subsets of the sample space.

The power set of S is the set of all subsets of S. It supplies the events to which probabilities are assigned. An axiom is a basic rule accepted as the starting point for further reasoning.

What are the three axioms?

1. Non-negativity: P(E) ≥ 0 for every event E. The symbol ≥ means “greater than or equal to”.
2. Normalisation: P(S) = 1. The probability of the sure event is one.
3. Additivity for disjoint events: if E and F are mutually exclusive events, P(E ∪ F) = P(E) + P(F).

Every probability lies between 0 and 1, including these endpoints. For a finite sample space, meaning one containing a definite number of outcomes, the probabilities assigned to all its individual outcomes add to 1. An event's probability is the sum of the probabilities of the outcomes belonging to it.

Result: The impossible event has probability zero

The events E and ∅ are disjoint and their union is E. Additivity therefore gives P(E) = P(E) + P(∅). Subtracting P(E) from both sides gives P(∅) = 0.

Worked example 5. Check two assignments to six distinct outcomes: the first gives each outcome probability 1/6; the second assigns 0.1, 0.2, 0.3, 0.4, 0.5 and 0.6 respectively.

Answer: The first assignment is valid: each probability is non-negative and their sum is 6 × 1/6 = 1. The second is invalid: its probabilities sum to 2.1 rather than 1.

A coin model assigning P(H) = 1/4 and P(T) = 3/4 also satisfies the axioms. Here P(H) and P(T) are shorthand for the probabilities of the single-outcome events {H} and {T}. The axioms do not require equal probabilities for different outcomes.

When can probability be calculated by counting outcomes?

Equally likely outcomes have the same probability. If a finite sample space has n outcomes and each has probability p, where p denotes their common probability, adding all outcome probabilities gives np = 1. Consequently, p = 1/n.

Let n(S) denote the number of outcomes in S, and n(E) the number in event E. Outcomes belonging to E are called favourable outcomes for E. “Favourable” describes membership of the event, not whether the result is desirable.

Result: Probability for equally likely outcomes

P(E) = n(E)/n(S), provided the sample space is finite and its outcomes are equally likely. The numerator counts favourable outcomes and the denominator counts all possible outcomes in that same sample space.

Worked example 6. A bag contains 9 discs of similar shape and size: 4 red, 3 blue and 2 yellow. One disc is drawn at random, with every disc equally likely to be selected. Find the probabilities of the three colours.

Answer: P(red) = 4/9, P(blue) = 3/9 = 1/3 and P(yellow) = 2/9. These add to 1 because the three colour events are mutually exclusive and exhaustive.

The equally likely outcomes here are the individual discs. The three colour names do not have equal probabilities: there are different numbers of discs of each colour. Replacing the nine individual outcomes with three equally weighted colours would change the given model.

What happens when probabilities are unequal?

Worked example 7. For two coin tosses, suppose the assigned probabilities are P(HH) = 1/4, P(HT) = 1/7, P(TH) = 2/7 and P(TT) = 9/28. Find the probability that both tosses give the same result.

Answer: The required event is E = {HH, TT}. Add its outcome probabilities: P(E) = 1/4 + 9/28 = 7/28 + 9/28 = 4/7. Counting two outcomes out of four would incorrectly give 1/2.

These four assignments sum to 1, but they are unequal. The event-sum method works with their actual probabilities. The counting formula is a special case of that method, obtained when each elementary probability is the same.

How does the addition theorem calculate “A or B”?

Theorem: Addition of probabilities for two events

For events A and B in the same sample space, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The intersection term removes the duplicate contribution from outcomes that satisfy both events.

Adding P(A) and P(B) counts every common outcome twice. The union must count each outcome once. Subtracting the probability of the intersection once therefore corrects the total, while keeping the common outcomes within the union.

How can the theorem be proved?

1. Write A ∪ B = A ∪ (B − A). The events A and B − A are mutually exclusive.
2. Apply additivity to obtain P(A ∪ B) = P(A) + P(B − A).
3. Write B = (A ∩ B) ∪ (B − A). These two parts of B are also mutually exclusive.
4. Therefore P(B − A) = P(B) − P(A ∩ B). Substituting this expression proves the addition theorem.

A rectangle labelled S encloses two overlapping circles labelled A and B. The left-only region A − B and right-only region B − A are blue; the central intersection A ∩ B is unshaded.

See Fig. 14.1 in your NCERT textbook

Worked example 8. A coin is tossed three times, with all 8 sequences equally likely. Let A = {HHT, HTH, THH} and B = {HTH, THH, HHH}. Calculate P(A ∪ B).

Answer: P(A) = 3/8 and P(B) = 3/8. Their intersection is {HTH, THH}, with probability 2/8. Thus P(A ∪ B) = 3/8 + 3/8 − 2/8 = 4/8 = 1/2.

When A and B are mutually exclusive, their intersection is empty and its probability is zero. The theorem then reduces to P(A ∪ B) = P(A) + P(B). The simpler formula therefore needs the disjointness condition.

The same theorem can find a missing intersection by rearrangement: P(A ∩ B) = P(A) + P(B) − P(A ∪ B). This calculates the probability of “and” when the two individual probabilities and their union probability are known.

How do complements simplify “not” and “neither” probabilities?

Result: Probability of a complementary event

For any event A, P(A′) = 1 − P(A). The sets A and A′ are mutually exclusive and exhaustive, so their probabilities add to P(S) = 1. Subtracting P(A) gives the result.

A complement must be taken relative to the specified sample space. It contains every outcome outside A, including all alternatives allowed by the experiment. Finding the complement can be simpler than listing a long collection of favourable outcomes directly.

Worked example 9. One card is drawn with equal likelihood from 52 cards, of which 4 are aces. Find the probability that the selected card is not an ace.

Answer: Let A be the event of drawing an ace. Then P(A) = 4/52 = 1/13. The event of not drawing an ace is A′, so P(A′) = 1 − 1/13 = 12/13.

How do “neither” and “not both” differ?

De Morgan's laws describe complements of unions and intersections: (A ∪ B)′ = A′ ∩ B′, while (A ∩ B)′ = A′ ∪ B′. The outer prime means the complement of the whole event inside the brackets.

“Neither A nor B” means that both events fail, so it is the complement of their union. “Not both A and B” also allows exactly one to occur, so it is the complement of their intersection.

The last expression adds the probabilities of A − B and B − A, which are mutually exclusive. It removes the intersection from each individual event. Unlike the inclusive union, exactly one event excludes every outcome common to both.

How can a complete probability calculation be organised?

Start by specifying the experiment and naming the required event. Then decide whether the information consists of equally likely outcomes, unequal outcome probabilities, or probabilities of whole events. This determines how the event's probability can be calculated.

What sequence keeps the reasoning clear?

1. State the sample space, or the complete set of probabilities supplied in the question.
2. Translate the required condition into an event, preserving words such as “exactly”, “at least”, “and” and “or”.
3. Select the appropriate event-sum, counting, addition or complement rule, and state any equal-likelihood or disjointness condition used.
4. Substitute the given values, simplify the result and check that it lies between zero and one.

Worked example 10. The probabilities that Anil and Ashima qualify an examination are 0.05 and 0.10 respectively; the probability that both qualify is 0.02. Find the probabilities that neither qualifies, at least one fails to qualify, and exactly one qualifies.

Answer: Let E mean Anil qualifies and F mean Ashima qualifies. P(E ∪ F) = 0.05 + 0.10 − 0.02 = 0.13. Neither qualifies with probability 1 − 0.13 = 0.87. At least one fails with probability 1 − 0.02 = 0.98. Exactly one qualifies with probability 0.05 + 0.10 − 2 × 0.02 = 0.11.

This calculation distinguishes three different events. “Neither qualifies” excludes every outcome in the union. “At least one fails” excludes only the outcomes where both qualify. “Exactly one qualifies” retains the two separate possibilities in which just one student qualifies.

Check the interpretation before judging the arithmetic. All three results lie between zero and one, but this check alone would not expose an answer obtained for the wrong event. Writing the required event in words alongside its formula makes that distinction visible.

When both individual event probabilities and their union are supplied, use the rearranged addition theorem to obtain the overlap. When a complement is requested, identify the whole event being negated before subtracting its probability from one.

Glossary

• Random experiment — A procedure with known possible outcomes whose particular result cannot be predicted with certainty beforehand.
• Sample space — The set containing all possible outcomes of the random experiment being considered.
• Sample point — An individual outcome belonging to the sample space of an experiment.
• Event — A subset of the sample space, occurring when the actual outcome belongs to that subset.
• Simple event — An event containing exactly one sample point, also called an elementary event.
• Compound event — An event containing more than one sample point of the given sample space.
• Sure event — The whole sample space, containing every possible outcome and having probability one.
• Impossible event — The empty set, containing no possible outcome and having probability zero.
• Complementary event — The event containing all sample points outside a specified event but within its sample space.
• Mutually exclusive events — Events that cannot occur together because their sets have no common outcome.
• Exhaustive events — Events whose union contains the whole sample space, ensuring that at least one occurs.
• Equally likely outcomes — Outcomes assigned the same probability, allowing probability calculation through a ratio of outcome counts.

Common errors and misconceptions

• Misconception: HT and TH are the same outcome. Correct: They record different orders: head then tail, and tail then head.
• Misconception: “At most one head” means exactly one head. Correct: It includes no head as well as exactly one head.
• Misconception: “A or B” excludes their common outcomes. Correct: The union includes outcomes in either event or in both.
• Misconception: Exhaustive events must be mutually exclusive. Correct: Their union must cover the sample space, but they may overlap.
• Misconception: Favourable outcomes divided by total outcomes works for unequal outcome probabilities. Correct: That formula requires equally likely outcomes; otherwise add the actual probabilities.
• Misconception: P(A ∪ B) equals P(A) + P(B) for any two events. Correct: In general, subtract P(A ∩ B) to correct double counting.
• Misconception: “Neither event” and “not both events” are interchangeable. Correct: Neither complements the union; not both complements the intersection.
• Misconception: An assignment is valid whenever every probability lies between zero and one. Correct: The probabilities of all sample points must also sum to one.

Exam-style questions with model answers

Q1. A coin is tossed twice. Using H for head and T for tail, write the sample space and the event of exactly one head. [2 marks]

1. The sample space is S = {HH, HT, TH, TT}, with the first letter recording the first toss.
2. The event of exactly one head is {HT, TH}; each of these outcomes has one head and one tail.

Q2. A die numbered 1 to 6 is rolled. Let A = {2, 3, 5} and B = {1, 3, 5}. Find A ∪ B, A ∩ B and A − B, explaining their meanings. [3 marks]

1. The union A ∪ B = {1, 2, 3, 5}. This includes every number in A or B, counting common numbers once.
2. The intersection A ∩ B = {3, 5}. These numbers belong to both events and therefore satisfy “A and B”.
3. The difference A − B = {2}. It contains the number in A that is not in B, representing “A but not B”.

Q3. A coin is tossed three times. Let A mean no head, B exactly one head and C at least two heads. Using H for head and T for tail, show that the events are mutually exclusive and exhaustive. [4 marks]

1. There are eight outcomes: S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Each three-letter sequence records the tosses in order.
2. The event sets are A = {TTT}, B = {HTT, THT, TTH}, and C = {HHT, HTH, THH, HHH}.
3. Every pair has an empty intersection: A ∩ B = A ∩ C = B ∩ C = ∅. Hence the events are mutually exclusive.
4. The union A ∪ B ∪ C contains all eight outcomes in S. Hence the events are exhaustive as well.

Q4. A bag contains 9 similar discs: 4 red, 3 blue and 2 yellow. One disc is selected, with each disc equally likely. Find the probabilities of red, yellow, blue, not blue, and red or blue. [5 marks]

1. There are nine equally likely individual outcomes. Four discs are red, so P(red) = 4/9, using favourable outcomes divided by total outcomes.
2. Two of the nine discs are yellow. Therefore P(yellow) = 2/9, with the same total of nine possible individual selections.
3. Three of the nine discs are blue. Thus P(blue) = 3/9 = 1/3 after simplifying the fraction.
4. “Not blue” is the complement of blue. Its probability is 1 − 1/3 = 2/3, including all red and yellow discs.
5. Red and blue are mutually exclusive for a single selected disc. Therefore P(red or blue) = 4/9 + 3/9 = 7/9.

Q5. The probabilities that Anil and Ashima qualify an examination are 0.05 and 0.10 respectively. The probability that both qualify is 0.02. Find the probabilities that at least one qualifies, neither qualifies, at least one does not qualify, and exactly one qualifies, defining your events. [5 marks]

1. Let E be the event that Anil qualifies, and F the event that Ashima qualifies. Thus P(E) = 0.05, P(F) = 0.10 and P(E ∩ F) = 0.02.
2. At least one qualifies when E ∪ F occurs. The addition theorem gives P(E ∪ F) = 0.05 + 0.10 − 0.02 = 0.13.
3. Neither qualifies in the complement of E ∪ F. Therefore the required probability is 1 − P(E ∪ F) = 1 − 0.13 = 0.87.
4. At least one does not qualify in the complement of E ∩ F. Its probability is 1 − 0.02 = 0.98.
5. Exactly one qualifies in either E − F or F − E. These events are disjoint, giving 0.05 + 0.10 − 2 × 0.02 = 0.11.

Q6. A coin is tossed twice. Using H for head and T for tail, the outcome probabilities are P(HH) = 1/4, P(HT) = 1/7, P(TH) = 2/7 and P(TT) = 9/28. Check that the assignment is valid and find the probability of matching results. [3 marks]

1. Each assigned probability lies between zero and one, so none violates the required bounds for the probability of an outcome.
2. The total is 1/4 + 1/7 + 2/7 + 9/28 = (7 + 4 + 8 + 9)/28 = 1. Therefore the assignment is valid.
3. Matching results form the event {HH, TT}. Its probability is 1/4 + 9/28 = 4/7. Equal-outcome counting is inapplicable because the four probabilities differ.

Q7. For events A and B in the same sample space, prove P(A ∪ B) = P(A) + P(B) − P(A ∩ B), using additivity for mutually exclusive events. [4 marks]

1. Separate the union as A ∪ B = A ∪ (B − A). The two events on the right have no common outcome.
2. Apply additivity to this disjoint union: P(A ∪ B) = P(A) + P(B − A).
3. Also B = (A ∩ B) ∪ (B − A), with disjoint parts. Thus P(B − A) = P(B) − P(A ∩ B).
4. Substitute this expression into the equation for the union to obtain P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Q8. A card is drawn with equal likelihood from a 52-card deck containing 13 diamonds, 4 aces and 26 black cards. Find the probabilities of a diamond, not an ace, a black card, not a diamond and not a black card. [5 marks]

1. All 52 cards are equally likely outcomes, and 13 are diamonds. Hence the probability of a diamond is 13/52 = 1/4.
2. The probability of an ace is 4/52 = 1/13. Its complement, not an ace, has probability 1 − 1/13 = 12/13.
3. There are 26 black cards among the 52 possible selections. The probability of selecting a black card is therefore 26/52 = 1/2.
4. Not a diamond is the complement of selecting a diamond. Subtracting its probability from one gives 1 − 1/4 = 3/4.
5. Not a black card is the complement of selecting a black card. The required probability is 1 − 1/2 = 1/2.

Key takeaways

• A sample space lists all possible outcomes; an event is a subset that occurs when the actual result belongs to it.
• Simple events contain one outcome, compound events contain several, the sure event is S, and the impossible event is empty.
• Union represents inclusive “or”, intersection represents “and”, and a complement contains the outcomes outside the specified event.
• Mutually exclusive events have no common outcome, while exhaustive events together cover the whole sample space.
• Probability assignments must be non-negative, give the whole sample space probability one, and satisfy additivity for mutually exclusive events.
• Counting favourable outcomes and dividing by total outcomes requires a finite sample space with equally likely outcomes.
• The addition theorem subtracts the intersection once, correcting the double counting caused by adding two overlapping event probabilities.
• Neither event occurring complements their union; at least one failing to occur complements their intersection.

Test yourself

For a coin tossed twice, why are HT and TH different outcomes, where H means head and T means tail?

The letters record order: HT means head then tail, while TH means tail then head.

For a die numbered 1 to 6, what set represents the event of a multiple of 7?

The empty set ∅ represents it, since none of the six faces is a multiple of 7.

What does A ∪ B include when A and B are events?

It includes outcomes in A, in B, or in both events, with each distinct outcome included once.

For S = {1, 2, 3, 4, 5, 6}, are A = {1, 2, 3}, B = {3, 4} and C = {5, 6} exhaustive and mutually exclusive?

They are exhaustive because their union is S. They are not mutually exclusive because A and B share 3.

Can a coin model assign head probability 1/4 and tail probability 3/4?

Yes. Both probabilities are non-negative and their sum is one, so the assignment satisfies the outcome conditions.

When may P(A ∪ B) be found by adding P(A) and P(B) using the disjoint-event rule?

When A and B are mutually exclusive, their intersection is empty and contributes probability zero.

If an ace has probability 1/13, what is the probability of not drawing an ace?

The complementary probability is one minus the ace probability: 1 − 1/13 = 12/13.

Why does “not both A and B” differ from “neither A nor B”?

Not both allows exactly one event to occur; neither requires both events to fail.