Classmate AI
Learn · Solve · Practice
Chapter 14CBSE Class 10 Maths100% Free

Probability — CBSE Class 10 Maths Important Questions

13 hand-picked CBSE Class 10 Maths important questions for Probability, each with a full model answer — the formats and topics most likely to appear in your board exam.

13
Questions
6
Question types
32
Total marks
₹0
With answers

Probability — CBSE Class 10 Maths Important Questions

Get the Sample Space Right First

Your Personal AI Tutor

Teaches Class 8–10 CBSE/ICSE Maths & Science by asking the right questions — not handing over answers.

Start your Freemium plan
Quick answer

For an event EEE with equally likely outcomes, P(E)=number of outcomes favourable to Etotal number of outcomes.P(E)=\dfrac{\text{number of outcomes favourable to }E}{\text{total number of outcomes}}.P(E)=number of outcomes favourable to E/total number of outcomes. Always 0≤P(E)≤10\le P(E)\le 10≤ P(E)≤ 1, and P(not E)=1−P(E)P(\text{not }E)=1-P(E)P(not E)=1-P(E). A standard deck has 525252 cards; two dice give 363636 equally likely outcomes.

Syllabus reference (2026–27): NCERT (National Council of Educational Research and Training) · CBSE (Central Board of Secondary Education)

About Probability

This chapter uses the classical (theoretical) definition of probability. Typical board questions involve dice, coins, a well-shuffled deck of 525252 cards, and drawing balls or numbered cards from a bag. It is a high-scoring chapter (about 4-6 marks, usually including a case-study). The keys are counting the total and favourable outcomes correctly, and remembering that P(E)+P(not E)=1P(E)+P(\text{not }E)=1P(E)+P(not E)=1. Cards questions are especially common — for example finding the probability of drawing a king, a red card, a black queen, or a face card from a well-shuffled deck of 52 playing cards.

Theoretical (classical) probabilityComplement of an event: $P(\text{not }E)=1-P(E)$Probability with dice (two dice give 36 outcomes)Probability with a deck of 52 playing cardsDrawing balls or numbered cards from a bag

Key concepts & formulas

Classical definition

If all outcomes are equally likely, P(E)=favourable outcomestotal outcomes.P(E)=\dfrac{\text{favourable outcomes}}{\text{total outcomes}}.P(E)=favourable outcomes/total outcomes. Probability is always between 000 (impossible event) and 111 (sure event).

Complementary events

EEE and 'not EEE' are complementary: P(E)+P(not E)=1P(E)+P(\text{not }E)=1P(E)+P(not E)=1, so P(not E)=1−P(E)P(\text{not }E)=1-P(E)P(not E)=1-P(E). This shortcut is often faster than direct counting.

Two dice

Throwing two dice gives 6×6=366\times 6=366× 6=36 equally likely ordered pairs. Build a table of sums or products to count favourable outcomes; doublets (same number on both) number 666.

A deck of 52 cards

525252 cards =4=4=4 suits (spades and clubs are black; hearts and diamonds are red) of 131313 each. Face cards are king, queen, jack (121212 in all); there are 444 aces. Adjust the totals when some cards are removed.

Free download

Get all 13 Probability questions as a PDF

The full question bank with model answers — perfect for offline revision and last-minute practice.

Free · No spam · Unsubscribe anytime

Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

A die is thrown once. The probability of getting a prime number is:

  1. (a)

    12\dfrac{1}{2}1/2

  2. (b)

    13\dfrac{1}{3}1/3

  3. (c)

    23\dfrac{2}{3}2/3

  4. (d)

    16\dfrac{1}{6}1/6

Show model answer

Answer: (a) 12\dfrac{1}{2}1/2

Prime numbers on a die are 2,3,52,3,52,3,5 (three outcomes). P=36=12.P=\dfrac{3}{6}=\dfrac{1}{2}.P=3/6=1/2.

Still stuck? Ask the AI tutor to explain this step by step →
Q2MCQEasy1 mark

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

  1. (a)

    −1.5-1.5-1.5

  2. (b)

    0.70.70.7

  3. (c)

    23\dfrac{2}{3}2/3

  4. (d)

    15\dfrac{1}{5}1/5

Show model answer

Answer: (a) −1.5-1.5-1.5

The probability of any event always satisfies 0≤P(E)≤10\le P(E)\le 10≤ P(E)≤ 1, so a negative value like −1.5-1.5-1.5 is impossible.

Still stuck? Ask the AI tutor to explain this step by step →
Q3MCQModerate1 mark

One card is drawn at random from a well-shuffled deck of 525252 cards. The probability of getting a red king is:

  1. (a)

    126\dfrac{1}{26}1/26

  2. (b)

    113\dfrac{1}{13}1/13

  3. (c)

    213\dfrac{2}{13}2/13

  4. (d)

    152\dfrac{1}{52}1/52

Show model answer

Answer: (a) 126\dfrac{1}{26}1/26

There are 222 red kings (king of hearts and king of diamonds). P=252=126.P=\dfrac{2}{52}=\dfrac{1}{26}.P=2/52=1/26.

Still stuck? Ask the AI tutor to explain this step by step →
Q4MCQHOTS1 mark

Two dice are thrown together. The probability that the sum of the numbers appearing on them is a perfect square is:

  1. (a)

    736\dfrac{7}{36}7/36

  2. (b)

    16\dfrac{1}{6}1/6

  3. (c)

    536\dfrac{5}{36}5/36

  4. (d)

    14\dfrac{1}{4}1/4

Show model answer

Answer: (a) 736\dfrac{7}{36}7/36

Possible sums range from 222 to 121212; the perfect squares are 444 and 999. Sum =4=4=4: (1,3),(2,2),(3,1)=3(1,3),(2,2),(3,1)=3(1,3),(2,2),(3,1)=3 ways. Sum =9=9=9: (3,6),(4,5),(5,4),(6,3)=4(3,6),(4,5),(5,4),(6,3)=4(3,6),(4,5),(5,4),(6,3)=4 ways. Total =7.=7.=7. P=736.P=\dfrac{7}{36}.P=7/36.

Still stuck? Ask the AI tutor to explain this step by step →

Want every Probability question solved live, at your pace?

Practise free with the AI tutor →

Assertion–Reason questions (1 mark)

Q5Assertion–ReasonEasy1 mark

Assertion (A): When a die is thrown once, the probability of getting a number greater than 444 is 13.\dfrac{1}{3}.1/3.

Reason (R): The probability of an event =number of favourable outcomestotal number of outcomes.=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}.=number of favourable outcomes/total number of outcomes.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

Show model answer

Answer: (a) Both A and R are true and R is the correct explanation of A.

Numbers greater than 444 are 555 and 666, so P=26=13P=\dfrac{2}{6}=\dfrac{1}{3}P=2/6=1/3, obtained exactly by the formula in R.

Still stuck? Ask the AI tutor to explain this step by step →

Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

A bag contains 555 red, 888 white and 777 black balls. A ball is drawn at random. Find the probability that it is (i) red, (ii) not black.

Show model answer

Total balls =5+8+7=20.=5+8+7=20.=5+8+7=20.

(i) P(red)=520=14.P(\text{red})=\dfrac{5}{20}=\dfrac{1}{4}.P(red)=5/20=1/4.

(ii) Black balls =7=7=7, so P(not black)=1−720=1320.P(\text{not black})=1-\dfrac{7}{20}=\dfrac{13}{20}.P(not black)=1-7/20=13/20.

Still stuck? Ask the AI tutor to explain this step by step →
Q7Very ShortModerate2 marks

Cards numbered 111 to 252525 are placed in a box and mixed thoroughly. One card is drawn at random. Find the probability that the number on the drawn card is divisible by 333 or 555.

Show model answer

Divisible by 333: 3,6,9,12,15,18,21,24⇒83,6,9,12,15,18,21,24\Rightarrow 83,6,9,12,15,18,21,24 8 numbers. Divisible by 555: 5,10,15,20,25⇒55,10,15,20,25\Rightarrow 55,10,15,20,25 5 numbers. Divisible by both (151515): 111 number.

Favourable =8+5−1=12.=8+5-1=12.=8+5-1=12. P=1225.P=\dfrac{12}{25}.P=12/25.

Still stuck? Ask the AI tutor to explain this step by step →

Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Two dice are thrown simultaneously. Find the probability that (i) the sum of the numbers is 777, (ii) the sum is a prime number, (iii) the same number appears on both dice.

Show model answer

Total outcomes =36.=36.=36.

(i) Sum =7=7=7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)=6(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)=6(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)=6 ways. P=636=16.P=\dfrac{6}{36}=\dfrac{1}{6}.P=6/36=1/6.

(ii) Prime sums are 2,3,5,7,112,3,5,7,112,3,5,7,11 with 1+2+4+6+2=151+2+4+6+2=151+2+4+6+2=15 ways. P=1536=512.P=\dfrac{15}{36}=\dfrac{5}{12}.P=15/36=5/12.

(iii) Doublets: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)=6(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)=6(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)=6 ways. P=636=16.P=\dfrac{6}{36}=\dfrac{1}{6}.P=6/36=1/6.

Still stuck? Ask the AI tutor to explain this step by step →
Q9Short AnswerModerate3 marks

From a well-shuffled deck of 525252 cards, all the three face cards of spades (king, queen and jack of spades) are removed. One card is then drawn at random from the remaining cards. Find the probability that the drawn card is (i) a heart, (ii) a queen, (iii) a black card.

Show model answer

Remaining cards =52−3=49.=52-3=49.=52-3=49.

(i) Hearts are untouched =13.=13.=13. P(heart)=1349.P(\text{heart})=\dfrac{13}{49}.P(heart)=13/49.

(ii) Only the queen of spades was removed, so 333 queens remain. P(queen)=349.P(\text{queen})=\dfrac{3}{49}.P(queen)=3/49.

(iii) Black cards were 262626; removing 333 spades leaves 23.23.23. P(black)=2349.P(\text{black})=\dfrac{23}{49}.P(black)=23/49.

Still stuck? Ask the AI tutor to explain this step by step →
Q10Short AnswerHOTS3 marks

A number xxx is selected at random from the numbers 1,2,3,41, 2, 3, 41, 2, 3, 4 and a number yyy is selected at random from the numbers 1,4,9,161, 4, 9, 161, 4, 9, 16. Find the probability that the product xyxyxy is less than 161616.

Show model answer

Total outcomes =4×4=16.=4\times 4=16.=4× 4=16. Count the pairs with xy<16xy<16xy<16:

x=1x=1x=1: y=1,4,9y=1,4,9y=1,4,9 give 1,4,91,4,91,4,9 (all <16<16<16) ⇒3.\Rightarrow 3.3.
x=2x=2x=2: y=1,4y=1,4y=1,4 give 2,82,82,8 ⇒2.\Rightarrow 2.2.
x=3x=3x=3: y=1,4y=1,4y=1,4 give 3,123,123,12 ⇒2.\Rightarrow 2.2.
x=4x=4x=4: y=1y=1y=1 gives 444 ⇒1.\Rightarrow 1.1.

Favourable =3+2+2+1=8.=3+2+2+1=8.=3+2+2+1=8. P(xy<16)=816=12.P(xy<16)=\dfrac{8}{16}=\dfrac{1}{2}.P(xy<16)=8/16=1/2.

Still stuck? Ask the AI tutor to explain this step by step →

Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

Two dice are thrown together. Find the probability that (i) the product of the numbers is 121212, (ii) the sum of the numbers is 888, (iii) a doublet appears, (iv) the sum is divisible by 333, (v) both numbers are odd.

Show model answer

Total outcomes =36.=36.=36.

(i) Product 121212: (2,6),(6,2),(3,4),(4,3)=4(2,6),(6,2),(3,4),(4,3)=4(2,6),(6,2),(3,4),(4,3)=4 ways. P=436=19.P=\dfrac{4}{36}=\dfrac{1}{9}.P=4/36=1/9.

(ii) Sum 888: (2,6),(6,2),(3,5),(5,3),(4,4)=5(2,6),(6,2),(3,5),(5,3),(4,4)=5(2,6),(6,2),(3,5),(5,3),(4,4)=5 ways. P=536.P=\dfrac{5}{36}.P=5/36.

(iii) Doublets =6=6=6 ways. P=636=16.P=\dfrac{6}{36}=\dfrac{1}{6}.P=6/36=1/6.

(iv) Sum divisible by 333 (sums 3,6,9,123,6,9,123,6,9,12): 2+5+4+1=122+5+4+1=122+5+4+1=12 ways. P=1236=13.P=\dfrac{12}{36}=\dfrac{1}{3}.P=12/36=1/3.

(v) Both odd (1,3,51,3,51,3,5 on each): 3×3=93\times 3=93× 3=9 ways. P=936=14.P=\dfrac{9}{36}=\dfrac{1}{4}.P=9/36=1/4.

Still stuck? Ask the AI tutor to explain this step by step →
Q12Long AnswerHOTS5 marks

A box contains 121212 balls, out of which xxx are black. (i) If one ball is drawn at random, write the probability that it is black. (ii) If 666 more black balls are put in the box, the probability of drawing a black ball is now double what it was in (i). Find xxx. (iii) Using this value of xxx, find the probability of drawing a black ball after the 666 balls are added.

Show model answer

(i) P(black)=x12.P(\text{black})=\dfrac{x}{12}.P(black)=x/12.

(ii) After adding 666 black balls, total =18=18=18 and black =x+6=x+6=x+6, so P=x+618.P=\dfrac{x+6}{18}.P=x+6/18. Given this is double the earlier value:
x+618=2×x12=x6.\dfrac{x+6}{18}=2\times\dfrac{x}{12}=\dfrac{x}{6}.x+6/18=2×x/12=x/6. Cross-multiplying: 6(x+6)=18x⇒6x+36=18x⇒12x=36⇒x=3.6(x+6)=18x\Rightarrow 6x+36=18x\Rightarrow 12x=36\Rightarrow x=3.6(x+6)=18x 6x+36=18x 12x=36 x=3.

(iii) New probability =x+618=918=12.=\dfrac{x+6}{18}=\dfrac{9}{18}=\dfrac{1}{2}.=x+6/18=9/18=1/2.

Still stuck? Ask the AI tutor to explain this step by step →

Case-based questions (4 marks)

Q13Case-basedModerate4 marks

At a school fair, a game uses a fair spinning wheel divided into 888 equal sectors numbered 1,2,3,…,81, 2, 3, \ldots, 81, 2, 3, , 8. A player wins a prize depending on where the pointer stops. Assume each number is equally likely.

(i) Find the probability that the pointer stops at an odd number.

(ii) Find the probability that it stops at a number that is a multiple of 333.

(iii) Find the probability that it stops at a prime number.

Show model answer

Total outcomes =8.=8.=8.

(i) Odd numbers: 1,3,5,7⇒4.1,3,5,7\Rightarrow 4.1,3,5,7 4. P(odd)=48=12.P(\text{odd})=\dfrac{4}{8}=\dfrac{1}{2}.P(odd)=4/8=1/2.

(ii) Multiples of 333: 3,6⇒2.3,6\Rightarrow 2.3,6 2. P=28=14.P=\dfrac{2}{8}=\dfrac{1}{4}.P=2/8=1/4.

(iii) Primes: 2,3,5,7⇒4.2,3,5,7\Rightarrow 4.2,3,5,7 4. P(prime)=48=12.P(\text{prime})=\dfrac{4}{8}=\dfrac{1}{2}.P(prime)=4/8=1/2.

Still stuck? Ask the AI tutor to explain this step by step →

All CBSE Class 10 Maths Chapters

Real NumbersPolynomialsPair of Linear Equations in Two VariablesQuadratic EquationsArithmetic ProgressionsTrianglesCoordinate GeometryIntroduction to TrigonometrySome Applications of TrigonometryCirclesAreas Related to CirclesSurface Areas and VolumesStatisticsProbability

Related study guide

Quadratic Equations Class 10: Score Full 6 Marks
Factorisation, formula, completing the square — the three methods, when to use which, and the small mistakes that cost board marks.
Trigonometry Class 10 CBSE: Understand Every Formula in 1 Hour
Sin, cos, tan, identities — all the Class 10 trigonometry formulas explained the way a real tutor would teach them, not memorised.

Frequently asked questions

  • Are these Probability important questions free?
    Yes. All 13 CBSE Class 10 Maths important questions for Probability are free, with full model answers and no login required.
  • Do these Probability questions follow the latest CBSE syllabus?
    Yes — they are aligned to the NCERT 2026–27 syllabus for CBSE Class 10 Maths, so nothing here is outside the current course.
  • How should I practise the Probability important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Probability?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the CBSE paper is covered.

Stuck on Probability? Let the AI tutor help

Free to start · Step-by-step Socratic help · CBSE Class 10 Maths

Practise Probability free →
💬 Share on WhatsApp
Practise Probability free with the AI tutor →