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Class 11 Maths — questions with answers and worked explanations.
These questions are drawn from the Pariksha Sutra question bank for Probability, part of the Class 11 Maths syllabus. Each one shows the correct answer and, where a method helps, the working behind it.
Read the question, decide your answer before looking, then check the explanation — that is what turns practice into marks. If a question catches you out, the explanation is the part worth re-reading.
A random experiment must have an uncertain result each trial while the complete sample space is known. Tossing a coin yields either heads or tails, both possible outcomes are known but the actual result cannot be predicted. Adding 2 and 3 always gives 5, so the outcome is certain. Heating water to 100 °C at sea level always occurs under those conditions, not random. Measuring the boiling point of pure water may vary slightly due to experimental error, but the process itself is deterministic. Hence the random experiment is tossing a coin and noting the face shown.
A single sample point represents one specific outcome of the experiment.
Since it cannot be broken down into further outcomes, it is the most basic event.
Such an event is defined as a simple (elementary) event. Simple (elementary) event
By definition a probability is a non‑negative real number, so P(A)≥0.
The total probability of the sample space is 1, and any event cannot be more likely than the whole space, giving P(A)≤1.
Hence every event satisfies 0 ≤ P(A) ≤ 1, which is the correct answer.
Total possible outcomes when two dice are thrown = 6 × 6 = 36.
Favourable outcomes giving sum 7 are (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 ways.
Probability = favourable/total = 6/36 = 1/6.
Hence the answer is 1/6.
A sample space lists all possible outcomes of an experiment.
Any collection of one or more of these outcomes is called a subset of the sample space.
Such a subset represents a condition whose occurrence we can talk about, i.e., an event.
Hence the correct term is Event.
A but not B means outcomes that belong to A while excluding any that are also in B.
Hence we take the part of A and remove the part common with B, which is the complement of B within A.
This is expressed as the intersection of A with the complement of B, i.e., A ∩ B′.
Thus the correct choice is A intersection B'.
Total cards = 52; each suit (hearts, diamonds, clubs, spades) has 13 cards.
Number of favorable outcomes (hearts) = 13.
Probability = favorable/total = 13/52 = 1/4.
Hence the answer is 1/4.
Probability of an event and its complement sum to 1.
Here the complement of “rain today’’ is “does not rain today”.
So P(not rain)=1−P(rain)=1−0.7=0.3.
Hence the answer is 0.3.
Each toss of a fair coin has 2 possible results (H or T).
For three independent tosses the total number of ordered triples is 2×2×2.
Multiplying gives 8 distinct outcomes such as HHT, TTH, etc.
Hence the sample space contains 8 outcomes.
The sample space for a single die is {1,2,3,4,5,6}.
An event is the set of outcomes that satisfy the condition.
‘Number greater than 6’ has no outcomes in the sample space, so the event set is empty.
An event with no possible outcomes is called an impossible event. Impossible event.
By axiom 1, the probability of any event lies between 0 and 1.
Axiom 2 states that the probability of the sample space S (the sure event) equals 1.
Thus P(S)=1, not 0, 0.5 or infinite.
The correct answer is 1.
Total outcomes when two dice are thrown = 6 × 6 = 36.
Favourable outcomes for a doublet are (1,1),(2,2)…(6,6) → 6 cases.
Probability = favourable/total = 6/36 = 1/6.
Hence the answer is 1/6.
A die has six faces numbered 1 through 6, so each face is a possible outcome of a single roll.
The sample space S must list every elementary outcome that can occur.
Thus S = {1, 2, 3, 4, 5, 6} is the complete set of outcomes.
All possible outcomes on a die are {1,2,3,4,5,6}.
Every outcome is less than 7, so the event occurs for every trial.
Since its probability is 1, it is a sure (certain) event.
Probability is defined as favorable outcomes divided by total outcomes, which lies between 0 and 1 inclusive.
An impossible event has no favorable outcomes, so its count is zero.
Zero divided by any non‑zero total gives 0, thus the probability of the empty set is 0.
Total balls = 5 (red) + 3 (black) = 8.
Probability of drawing a red ball = number of favourable outcomes / total outcomes = 5 / 8.
Thus the required probability is 5/8.
Two coins each have two outcomes: H or T.
The combined outcome is an ordered pair (coin 1, coin 2).
List all possible pairs: (H,H), (H,T), (T,H), (T,T).
Thus the sample space contains 4 elements, so the answer is 4.
Even numbers on a die are {2,4,6}; this is event A.
The complement A′ consists of all outcomes not in A, i.e., the odd numbers.
The odd faces of a die are {1,3,5}.
Hence A′ = {1, 3, 5}.
For mutually exclusive events A and B, they cannot occur together, so P(A∩B)=0.
The additivity axiom states P(A∪B)=P(A)+P(B)−P(A∩B).
Substituting P(A∩B)=0 gives P(A∪B)=P(A)+P(B).
Thus the correct answer is P(A) + P(B).
Count total letters in “ASSASSINATION”: 13.
Identify vowels A, A, I, A, I, O → 6 vowels.
Probability = number of favorable outcomes / total outcomes = 6/13.
Thus the answer is 6/13.
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