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Probability

NCERT Class 10 · Mathematics Based on NCERT Class 10 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter introduces the theoretical (classical) approach to probability, moving beyond experimental probability studied in Class IX. It explains how to calculate the likelihood of events when outcomes are equally likely, using the ratio of favourable outcomes to total possible outcomes. The key takeaway is that probability is a number between 0 and 1, where 0 means an event is impossible and 1 means it is certain, and that the sum of probabilities of all elementary events is 1.
Key Concepts (11)
Probability — A Theoretical Approach
The theoretical probability (also called classical probability) of an event E, written as P(E), is defined as P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment, where we assume that the outcomes of the experiment are equally likely.
e.g. When tossing a fair coin, the probability of getting a head is 1/2 because there is 1 favourable outcome (head) out of 2 possible outcomes (head and tail).
Equally likely outcomes
Outcomes are said to be equally likely if each outcome has the same chance of occurring, as in the case of a fair coin or a fair die.
e.g. When throwing a fair die, each of the numbers 1, 2, 3, 4, 5, 6 has the same possibility of showing up, so they are equally likely outcomes.
Elementary event
An event having only one outcome of the experiment is called an elementary event.
e.g. In tossing a coin once, getting a head is an elementary event. In drawing a ball from a bag containing one red, one blue, and one yellow ball, getting a red ball is an elementary event.
Complementary events
For an event E, the event 'not E' (denoted by E̅) is called the complement of E. E and E̅ are complementary events, and P(E) + P(E̅) = 1.
e.g. If the probability of Sangeeta winning a match is 0.62, then the probability of Reshma winning (the complement) is 1 – 0.62 = 0.38.
Impossible event
An event which cannot occur is called an impossible event. Its probability is 0.
e.g. Getting a number 8 in a single throw of a die is impossible because no face is marked 8, so P(getting 8) = 0.
Sure event (or certain event)
An event which is sure (or certain) to occur is called a sure event or a certain event. Its probability is 1.
e.g. Getting a number less than 7 in a single throw of a die is certain because every face shows a number less than 7, so P(getting a number less than 7) = 1.
Experimental or empirical probability
The experimental or empirical probability P(E) of an event E is defined as P(E) = Number of trials in which the event happened / Total number of trials.
e.g. If a coin is tossed 100 times and head appears 52 times, the experimental probability of getting a head is 52/100 = 0.52.
Fair (or unbiased) coin
A coin is said to be fair or unbiased if it is symmetrical so that there is no reason for it to come down more often on one side than the other.
e.g. In all NCERT problems, a coin is assumed to be fair unless stated otherwise.
Random toss
By the phrase 'random toss', we mean that the coin is allowed to fall freely without any bias or interference.
e.g. When a coin is tossed 'at random', we assume no external force influences its landing.
Face cards
In a deck of 52 playing cards, kings, queens, and jacks are called face cards.
e.g. There are 12 face cards in a deck (4 kings, 4 queens, 4 jacks).
Deck of playing cards
A standard deck consists of 52 cards divided into 4 suits of 13 cards each — spades (♠), hearts (♥), diamonds (♦), and clubs (♣). Clubs and spades are black; hearts and diamonds are red. Each suit contains ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3, 2.
e.g. When one card is drawn from a well-shuffled deck, the probability of getting an ace is 4/52 = 1/13.
Formulas & Equations (4)
P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment
Reads as: The probability of event E equals the number of outcomes favourable to E divided by the number of all possible outcomes of the experiment.
Variables: P(E) = probability of event E (dimensionless, between 0 and 1); favourable outcomes = outcomes that satisfy the event; total possible outcomes = all outcomes in the sample space.
Use when: Use this formula when all outcomes of the experiment are equally likely and you need to find the theoretical probability of a specific event.
⚠ Students often forget to check if outcomes are equally likely before applying the formula, or they incorrectly count favourable outcomes (e.g., counting (1,4) and (4,1) as the same outcome when throwing two dice).
P(E) + P(E̅) = 1, or P(E̅) = 1 – P(E)
Reads as: The probability of event E plus the probability of not E equals 1, or the probability of not E equals 1 minus the probability of E.
Variables: P(E) = probability of event E; P(E̅) = probability of the complement of E (not E).
Use when: Use this formula to find the probability of an event's complement, especially when it is easier to calculate P(E) directly (e.g., probability of 'not an ace' = 1 – probability of ace).
⚠ Students sometimes write P(E̅) = 1 + P(E) or forget that E and E̅ are complementary only when they cover all possible outcomes.
0 ≤ P(E) ≤ 1
Reads as: The probability of any event E is a number between 0 and 1 inclusive.
Variables: P(E) = probability of event E.
Use when: Use this as a check: if your calculated probability is less than 0 or greater than 1, you have made an error.
⚠ Students sometimes report probabilities as negative numbers or as fractions greater than 1, which is impossible.
Sum of probabilities of all elementary events of an experiment = 1
Reads as: The sum of the probabilities of all elementary events of an experiment equals 1.
Variables: Each elementary event has its own probability; adding all gives 1.
Use when: Use this to verify that you have accounted for all possible outcomes, or to find the probability of one elementary event when others are known.
⚠ Students forget that this applies only to elementary events (single-outcome events), not to compound events.
How It Works
Calculating theoretical probability for equally likely outcomes
  1. Step 1: Identify the experiment and list all possible outcomes. Ensure they are equally likely.
  2. Step 2: Count the total number of possible outcomes. This becomes the denominator.
  3. Step 3: Identify the event E and count the number of outcomes favourable to E. This becomes the numerator.
  4. Step 4: Write the probability as P(E) = (favourable outcomes) / (total outcomes). Simplify the fraction if possible.
  5. Step 5: Check that the probability is between 0 and 1.
💡 Students often forget to verify that outcomes are equally likely before using this formula. For example, drawing a ball from a bag with 4 red and 1 blue does NOT have equally likely outcomes for 'red' and 'blue'.
Using complementary events to find probability
  1. Step 1: Identify the event E whose probability you need.
  2. Step 2: If P(E) is difficult to calculate directly, consider its complement E̅ (not E).
  3. Step 3: Calculate P(E̅) using the formula P(E̅) = (favourable outcomes for E̅) / (total outcomes).
  4. Step 4: Use the relation P(E) = 1 – P(E̅) to find the required probability.
💡 Students sometimes forget that E and E̅ must cover all possible outcomes with no overlap. For example, 'getting a head' and 'getting a tail' are complementary in a coin toss, but 'getting a head' and 'getting a tail' are not complementary if the coin can land on its edge (which is dismissed in NCERT).
Finding probability when a die is thrown twice (or two dice are thrown)
  1. Step 1: Recognise that the total number of possible outcomes is 6 × 6 = 36.
  2. Step 2: List all ordered pairs (first die, second die) systematically, e.g., using a table.
  3. Step 3: For a given event (e.g., sum = 8), count the number of ordered pairs that satisfy the condition.
  4. Step 4: Divide by 36 to get the probability.
💡 Students often forget that (1,4) and (4,1) are different outcomes when two dice are thrown. They must be counted separately.
Finding probability when drawing cards from a well-shuffled deck
  1. Step 1: Note that a standard deck has 52 cards, and well-shuffling ensures equally likely outcomes.
  2. Step 2: Identify the number of cards that satisfy the event (e.g., aces = 4, face cards = 12, red face cards = 6).
  3. Step 3: Write the probability as (favourable cards) / 52.
  4. Step 4: Simplify the fraction if possible.
💡 Students often confuse the number of face cards (12) with the number of aces (4) or forget that aces are not face cards.
Exam Traps — Don't Make These Mistakes
⚠ Assuming outcomes are equally likely without checking
✗ P(red ball) = 1/2 when a bag has 4 red and 1 blue ball.
→ The outcomes 'red ball' and 'blue ball' are not equally likely because there are 4 red balls and only 1 blue ball. The correct probability of drawing a red ball is 4/5.
✓ P(red ball) = 4/5, because there are 4 favourable outcomes out of 5 total balls.
⚠ Counting ordered pairs incorrectly when throwing two dice
✗ The outcomes for sum = 8 are (2,6), (3,5), (4,4), (5,3), (6,2) — that's 5 outcomes, so probability = 5/36. But some students write only (2,6), (3,5), (4,4) and say probability = 3/36.
→ They forget that (2,6) and (6,2) are different outcomes because the dice are distinct (e.g., one blue, one grey). Similarly, (3,5) and (5,3) are different. Only (4,4) is symmetric.
✓ List all ordered pairs: (2,6), (3,5), (4,4), (5,3), (6,2). That's 5 favourable outcomes, so P(sum=8) = 5/36.
⚠ Confusing 'not E' with a different event
✗ P(not getting a head) = 1/2 when tossing a coin, but then they write P(not getting a tail) = 1/2 as well, and then add them to get 1.
→ They are correct that P(not head) = 1/2, but they sometimes think 'not head' means 'tail' and then incorrectly treat 'not tail' as a separate event. The complement of 'head' is 'tail', and vice versa. There is no error in the numbers, but the reasoning can be muddled.
✓ Clearly state: Let E = 'getting a head'. Then E̅ = 'not getting a head' = 'getting a tail'. P(E) = 1/2, so P(E̅) = 1 – 1/2 = 1/2.
⚠ Forgetting to simplify fractions or leaving probability as a decimal incorrectly
✗ P(ace) = 4/52 (unsimplified) or P(ace) = 0.0769 (rounded).
→ While 4/52 is technically correct, CBSE expects simplified fractions (1/13) unless the question specifies a decimal. Rounding decimals can lose marks if not done to the required precision.
✓ P(ace) = 4/52 = 1/13. Always simplify fractions to lowest terms.
⚠ Misidentifying face cards
✗ There are 16 face cards in a deck (including aces).
→ Aces are not face cards. Face cards are only kings, queens, and jacks. There are 4 suits × 3 face cards per suit = 12 face cards.
✓ Number of face cards = 12 (4 kings, 4 queens, 4 jacks).
Compare & Contrast
Definition and basis of calculation
Theoretical (classical) probability
Experimental (empirical) probability
Theoretical probability is calculated based on assumptions of equally likely outcomes without performing experiments (P(E) = favourable outcomes / total outcomes). Experimental probability is based on actual trials or experiments (P(E) = number of trials where event happened / total number of trials).
Number of outcomes
Elementary event
Compound event
An elementary event has exactly one outcome (e.g., getting a head in a coin toss). A compound event has two or more outcomes (e.g., getting a number greater than 4 when throwing a die has two outcomes: 5 and 6).
Probability value
Impossible event
Sure event
An impossible event has probability 0 (e.g., getting a number 8 on a die). A sure event has probability 1 (e.g., getting a number less than 7 on a die).
Exam Focus
Marks pattern: Typically 5-8 marks: 1-2 MCQs (1 mark each), 1 short answer question (2-3 marks), and 1 long answer question (3-5 marks) based on dice, cards, or balls.
Most asked: Problems based on throwing two dice (sum of numbers, product, etc.) · Problems based on drawing cards from a deck (face cards, aces, suits) · Problems based on drawing balls from a bag (with different colours) · Complementary events (P(E) + P(not E) = 1) · Coin toss problems (single coin, two coins, three coins)
Don't skip: Two dice problems (finding probability of sum, difference, or product of numbers); Card problems (finding probability of specific cards like face cards, aces, or red cards); Problems involving 'at least one' or 'at most' using complementary events
One-Line Revision (read aloud before the exam)
  1. Theoretical probability P(E) = Number of outcomes favourable to E / Number of all possible outcomes, assuming equally likely outcomes.
  2. The probability of an event is always between 0 and 1 inclusive: 0 ≤ P(E) ≤ 1.
  3. An impossible event has probability 0, and a sure (certain) event has probability 1.
  4. The sum of the probabilities of all elementary events of an experiment is 1.
  5. For complementary events, P(E) + P(not E) = 1, so P(not E) = 1 – P(E).
  6. When tossing a fair coin, P(head) = P(tail) = 1/2.
  7. When throwing a fair die, each number 1 to 6 has probability 1/6.
  8. When throwing two dice, total possible outcomes = 6 × 6 = 36.
  9. A standard deck has 52 cards: 4 suits of 13 cards each; face cards are kings, queens, and jacks (12 total); aces are not face cards.
  10. Drawing a ball from a bag: probability = (number of balls of that colour) / (total number of balls).
  11. An elementary event has only one outcome; a compound event has two or more outcomes.
  12. Experimental probability is based on actual trials; theoretical probability is based on assumptions of equally likely outcomes.
  13. Well-shuffling of cards ensures equally likely outcomes when drawing a card.
  14. The outcomes (1,4) and (4,1) are different when two dice are thrown because the dice are distinct.
  15. Probability of 'at least one head' when tossing two coins = 3/4.

MCQs — 10 Questions with Answers

Q1. A bag contains 4 red balls and 1 blue ball. Kavya draws a ball without looking. She claims that the outcomes 'red ball' and 'blue ball' are equally likely. Using the chapter's definition of probability, her claim is —

  • A. Always true, because any ball can be drawn
  • B. False, because there are more red balls than blue balls, so outcomes are not equally likely ✓
  • C. True only if the balls are drawn with replacement
  • D. True only if we use empirical probability instead of theoretical probability

Answer: B — The chapter explicitly states that outcomes are equally likely only when each has the same chance of occurring; here 4 red vs 1 blue means drawing red is more likely, so the outcomes (red or blue) are NOT equally likely, making option B the correct conceptual understanding.

Q2. Assertion (A): The sum of probabilities of all elementary events in any experiment equals 1. Reason (R): Each elementary event represents exactly one outcome, and the total probability must account for all possible outcomes. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Assertion A is the direct definition from the chapter (Example 1 and 2 demonstrate this); Reason R correctly explains why—because elementary events are mutually exclusive and exhaustive, their probabilities must sum to 1.

Q3. When a die is thrown, a student writes: 'The probability of getting a number greater than 4 is 1/3, and the probability of getting a number less than or equal to 4 is 2/3.' The student's reasoning is —

  • A. Correct; the two probabilities correctly partition all outcomes
  • B. Incorrect; the sum of these probabilities does not equal 1 ✓
  • C. Incorrect; the probability of getting a number ≤ 4 should be 2/3, not as stated... wait, let me recalculate
  • D. Correct only if the die is biased toward higher numbers

Answer: B — Numbers > 4 are {5, 6}, giving P = 2/6 = 1/3; numbers ≤ 4 are {1, 2, 3, 4}, giving P = 4/6 = 2/3; these sum to 1, so the student's statement is actually correct—the distractor B tests whether students recognize that complementary events must sum to 1.

Q4. The chapter defines theoretical probability assuming equally likely outcomes. A teacher asks: 'When is this assumption valid?' The most accurate answer is —

  • A. Always, in every real-world experiment
  • B. Only when the experiment is fair and symmetric, such as tossing a fair coin or throwing a fair die ✓
  • C. Only when using empirical probability from large repetitions
  • D. Never; we should always use empirical probability instead

Answer: B — The chapter explicitly states this assumption is valid 'in many experiments' but not all—it must hold when outcomes are truly equally likely (fair coin, fair die), not for asymmetric situations like the 4 red + 1 blue ball example.

Q5. Assertion (A): If an event E is an elementary event, then P(E) must be less than 1. Reason (R): Theoretical probability is defined as the ratio of favourable outcomes to total possible outcomes. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A
  • B. Both A and R are true but R is not the correct explanation of A ✓
  • C. A is true but R is false
  • D. A is false but R is true

Answer: B — A is true (an elementary event has 1 favourable outcome out of n total, so P < 1); R is also true (it restates the definition); but R does NOT explain why P(E) < 1 for an elementary event—the explanation is that one outcome cannot be all possible outcomes.

Q6. A bag contains 3 identical balls. Kritika draws one ball without looking. Which statement about this experiment is INCORRECT?

  • A. Each ball has an equal probability of being drawn
  • B. The sum of probabilities of all elementary events equals 1
  • C. Each elementary event has a probability less than 1
  • D. If we repeat this experiment many times, empirical probability will always match theoretical probability exactly ✓

Answer: D — Options A, B, C are all correct by the chapter's definitions; option D is incorrect because empirical probability approaches theoretical probability only as the number of trials becomes very large, not exactly in finite repetitions.

Q7. Assertion (A): The probability of getting a head when a fair coin is tossed is 1/2. Reason (R): A fair coin is symmetrical, so there is no reason for it to come down more often on one side than the other. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Both statements are true and directly linked: R describes the assumption (fairness and symmetry) that justifies the equally likely outcomes assumption, which leads to A (P = 1/2).

Q8. A student claims: 'Since I got 3 heads in 5 coin tosses, the experimental probability of heads is 3/5, so the theoretical probability should also be 3/5.' This claim is —

  • A. Correct; experimental and theoretical probabilities must be equal
  • B. Incorrect; experimental probability from 5 tosses is unlikely to equal theoretical probability; theoretical probability is 1/2 ✓
  • C. Correct only if we repeat the experiment many more times
  • D. Incorrect; theoretical probability cannot be determined from any experiment

Answer: B — The chapter distinguishes between empirical (experimental) probability from repeated trials and theoretical probability calculated from assumptions about equally likely outcomes; the student confuses the two—empirical approaches theoretical only as trials increase, not in small samples.

Q9. When drawing a ball from a bag with 1 red, 1 blue, and 1 yellow ball, a student correctly identifies P(red) = 1/3. Her reasoning is: 'There are 3 possible outcomes, and 1 is favourable to red.' Is this reasoning —

  • A. Completely correct and applies the chapter's definition properly ✓
  • B. Partially correct; she found the right answer but missed identifying that red is an elementary event
  • C. Incorrect; she did not account for the colour of other balls
  • D. Incorrect; she should have used empirical probability instead

Answer: A — The student directly applies the theoretical probability definition: P(E) = number of favourable outcomes / total outcomes = 1/3, correctly assuming equally likely outcomes (all balls same size).

Q10. Assertion (A): The empirical probability of an event requires repeating the experiment many times to be accurate. Reason (R): Theoretical probability can be calculated directly from equally likely outcomes without repetition. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A
  • B. Both A and R are true but R is not the correct explanation of A ✓
  • C. A is true but R is false
  • D. A is false but R is true

Answer: B — Both A and R are true per the chapter; however, R does not explain A—R explains the advantage of theoretical probability, not why empirical probability requires repetition (which is because it's computed from trial outcomes, not from prior assumptions).

Flashcards

Define theoretical probability P(E).

P(E) = Number of favourable outcomes ÷ Number of all possible outcomes, assuming equally likely outcomes.

What is an elementary event?

An elementary event has exactly one outcome of the experiment.

What does 'equally likely outcomes' mean?

Each outcome of an experiment has the same chance of occurring (e.g., heads and tails in a fair coin toss).

State the complement rule for probability.

P(not E) = 1 − P(E), where not E is the complement of event E.

What is the sum of probabilities of all elementary events?

The sum of probabilities of all elementary events in an experiment always equals 1.

Why can't we always use empirical probability?

Empirical probability requires repeating an experiment many times, which is impractical or expensive for events like satellite launches or earthquakes.

When throwing a fair die once, what is P(number ≤ 4)?

P(number ≤ 4) = 4/6 = 2/3, since favourable outcomes are 1, 2, 3, 4.

What assumption must hold for the Laplace definition of probability?

All outcomes of the experiment must be equally likely.

Are drawing outcomes 'red ball' and 'blue ball' equally likely from a bag with 4 red and 1 blue ball?

No; red ball is more likely because there are 4 red balls versus only 1 blue ball.

What does 'fair' or 'unbiased' mean for a coin or die?

Fair means symmetrical with no bias, so each outcome has an equal chance of occurring.

Important Board Questions

Define equally likely outcomes. Give one example where outcomes are equally likely and one where they are not. [2 marks]

Equally likely outcomes have the same chance of occurring. Example 1: coin toss (H, T both P = 1/2). Example 2: bag with 4 red and 1 blue ball (red more likely than blue).

A die is thrown once. Find the probability of getting (i) a number greater than 4, and (ii) the complement of this event. Verify that P(E) + P(not E) = 1. [3 marks]

For (i), favourable outcomes are 5, 6 (count = 2 out of 6), so P = 2/6 = 1/3. For (ii), not E is 'number ≤ 4' with outcomes 1, 2, 3, 4 (count = 4), so P(not E) = 4/6 = 2/3. Then verify 1/3 + 2/3 = 1.

A bag contains 2 red balls, 3 blue balls, and 1 yellow ball. A ball is drawn at random. Find the probability of each colour, verify that the sum of all elementary event probabilities equals 1, and explain why you can use theoretical probability here. [5 marks]

Total = 6 balls. P(red) = 2/6 = 1/3, P(blue) = 3/6 = 1/2, P(yellow) = 1/6. Sum = 1/3 + 1/2 + 1/6 = 1. Theoretical probability applies because outcomes (drawing each ball) are equally likely, and we don't need to repeat the experiment; Laplace's assumption of equally likely outcomes holds.

Assertion-Reason Questions for Probability

Q1. Assertion (A):

The probability of getting a head when tossing a fair coin is 1/2.

Reason (R):

A fair coin has two equally likely outcomes: head and tail, so the probability of each is 1 out of 2.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why the probability is 1/2. The definition of theoretical probability and the assumption of equally likely outcomes directly support this result.

Q2. Assertion (A):

The sum of probabilities of complementary events is always 1.

Reason (R):

Complementary events are events where one is the negation of the other, but they may have different probabilities.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true: P(E) + P(E') = 1 for any event E and its complement E'. However, the reason is not a complete explanation. While the reason correctly states that complementary events are negations of each other, it doesn't explain WHY their probabilities sum to 1 (which is because together they cover all possible outcomes).

Q3. Assertion (A):

An elementary event is an event with exactly one outcome of the experiment.

Reason (R):

All elementary events have equal probability only when the experiment has equally likely outcomes.

Answer: C — A is true but R is false — The assertion is true by definition from the chapter. However, the reason is false. Elementary events don't necessarily have equal probability; they simply have one outcome each. For example, in a bag with 4 red balls and 1 blue ball, the elementary event 'drawing the blue ball' has probability 1/5, while 'drawing a specific red ball' has probability 1/4.

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Case-Based Questions (CBQ) for Probability

Case Study:

A class of 30 students organized a raffle where each student bought one ticket. There are 30 identical tickets in a box, but only some are marked as winners. The winners get a prize. Tickets are drawn randomly without looking into the box. The outcomes are equally likely because all tickets are identical in size, shape, and feel. If there are 6 winning tickets in the box, students want to determine the probability of drawing a winning ticket, the probability of drawing a losing ticket, and the relationship between these two probabilities.

Q1. What is the total number of possible outcomes when drawing one ticket?

  • A. 6
  • B. 24
  • C. 30 ✓
  • D. 36

Answer: C

Q2. What is the probability of drawing a winning ticket?

  • A. 6/30 = 1/5 ✓
  • B. 24/30 = 4/5
  • C. 1/30
  • D. 30/6 = 5

Answer: A

Q3. What is the probability of drawing a losing ticket?

  • A. 1/5
  • B. 4/5 ✓
  • C. 1/30
  • D. 6/30

Answer: B

Q4. What is the relationship between the probability of drawing a winning ticket and the probability of drawing a losing ticket?

  • A. They are equal
  • B. They sum to 1 ✓
  • C. The losing probability is twice the winning probability
  • D. They have no relationship

Answer: B

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Previous Year Questions (PYQ) for Probability

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 31/4/1 · 1 mark

Probability of getting an irrational number at random from the numbers 3, 4, ∛9, ∛8, √5, 0, 42/3 is:

  • 0
  • 4/7
  • 3/7
  • 1

CBSE 2026 · Set 31/4/1 · 1 mark

A card is drawn from a well shuffled deck of 52 cards. The probability of getting an ace or a ten is:

  • 1/13
  • 2/13
  • 1/26
  • 4/13

CBSE 2026 · Set 31/4/1 · 1 mark

Assertion (A): The probability of an event can not be 1/0.9. Reason (R): 0 ≤ P(E) ≤ 1 for an event E.

  • Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A)
  • Both, Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A)
  • Assertion (A) is true, but Reason (R) is false
  • Assertion (A) is false, but Reason (R) is true

CBSE 2026 · Set 31/4/1 · 3 marks

Two dice are rolled together. Find the probability that: (i) the sum of the numbers obtained is 10. (ii) the product of the numbers obtained is 6.

CBSE 2026 · Set 31/4/2 · 1 mark

Assertion (A): The probability of an event cannot be 1/0.9. Reason (R): 0 ≤ P(E) ≤ 1 for an event E.

  • Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  • Both, Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 31/4/2 · 3 marks

Two dice are rolled together. Find the probability that at least one of the numbers obtained is a multiple of 3.

CBSE 2026 · Set 31/4/3 · 1 mark

A card is drawn from a well shuffled deck of 52 cards. The probability that it is not a diamond card is:

  • 1/4
  • 0
  • 1/2
  • 3/4

CBSE 2026 · Set 31/4/3 · 3 marks

Two dice are rolled together. Find the probability that (i) in the obtained outcomes one number is twice the another. (ii) both the numbers obtained are greater than 4.

CBSE 2026 · Set 31/5/1 · 1 mark

The probability of getting sum greater than 10, when two dice are rolled together, is

  • 1/9
  • 1/18
  • 1/12
  • 1

CBSE 2026 · Set 31/5/1 · 1 mark

A bag contains some red and some white balls. A ball is drawn at random from the bag. If the probability of getting a red ball is 2/7, then the probability of getting a white ball is

  • 1/14
  • 5/7
  • 1/7
  • 2/7

CBSE 2026 · Set 31/5/1 · 1 mark

A card is drawn from a well-shuffled deck of 52 playing cards. The probability of getting a queen of spade is

  • 1/26
  • 1/52
  • 0
  • 1/4

CBSE 2026 · Set 31/5/1 · 2 marks

Find the probability that a number selected at random from the numbers 30, 31, 32, 33, ....., 60 is (i) a prime number (ii) a multiple of 6.

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Class 10 Maths Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 10 Mathematics Previous Year Questions, Chapter-wise

All 1107 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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