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Probability — CBSE Class 10 Maths Important Questions

13 CBSE Class 10 Maths practice questions on Probability, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 1 October 2026

Last updated 25 October 2026

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Key concepts
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Probability — CBSE Class 10 Maths Important Questions

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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. Questions use dice, coins, a well-shuffled deck of 525252 cards, and drawing balls or numbered cards from a bag. 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. For cards, practise finding the probability of a king, a red card, a black queen or a face card.

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.

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Important questions with answers

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

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

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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.

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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.

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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.

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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.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): When two coins are tossed together, the probability of getting one head and one tail is 13.\dfrac{1}{3}.1/3.

Reason (R): When two coins are tossed, the equally likely outcomes are HH, HT, THHH,\ HT,\ THHH, HT, TH and TTTTTT.

  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

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Answer: (d) A is false but R is true.

'One head and one tail' covers two of the four equally likely outcomes (HTHTHT and THTHTH), so its probability is 24=12\dfrac{2}{4}=\dfrac{1}{2}2/4=1/2, not 13\dfrac{1}{3}1/3. The value 13\dfrac{1}{3}1/3 comes from wrongly treating 'two heads', 'two tails' and 'one of each' as equally likely.

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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.

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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.

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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.

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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.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Three coins are tossed together. Find the probability of getting (i) exactly two heads, (ii) at least two heads, (iii) at most one head.

Show model answer

The 888 equally likely outcomes are HHH,HHT,HTH,THH,HTT,THT,TTH,TTTHHH, HHT, HTH, THH, HTT, THT, TTH, TTTHHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

(i) Exactly two heads: HHT,HTH,THH⇒3HHT, HTH, THH\Rightarrow 3HHT, HTH, THH⇒ 3 outcomes. P=38.P=\dfrac{3}{8}.P=3/8.

(ii) At least two heads: these three and HHH⇒4HHH\Rightarrow 4HHH⇒ 4 outcomes. P=48=12.P=\dfrac{4}{8}=\dfrac{1}{2}.P=4/8=1/2.

(iii) At most one head: HTT,THT,TTH,TTT⇒4HTT, THT, TTH, TTT\Rightarrow 4HTT, THT, TTH, TTT⇒ 4 outcomes. P=48=12P=\dfrac{4}{8}=\dfrac{1}{2}P=4/8=1/2 (the complement of (ii): 1−12=121-\dfrac{1}{2}=\dfrac{1}{2}1-1/2=1/2).

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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.

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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.

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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.

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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.

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(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.

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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.

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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

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Sin, cos, tan, identities — all the Class 10 trigonometry formulas explained the way a real tutor would teach them, not memorised.

Frequently asked questions

  • What are the important questions in Probability Class 10?
    Dice problems using the 36 outcomes of two dice, playing-card problems on a 52-card deck (sometimes with cards removed), balls or numbered cards from a box, coins, and complementary events. This page has 13 questions with answers, including a spinning-wheel case study.
  • How do you solve probability questions on a deck of 52 cards?
    Know the deck: 4 suits of 13, 26 red and 26 black, 12 face cards and 4 aces. The probability of a red king is 252=126\tfrac{2}{52}=\tfrac1{26}2/52=1/26. If cards are removed, reduce both the total and the favourable count before dividing.
  • What is the probability that the sum on two dice is a perfect square?
    The possible perfect-square sums are 444 and 999. A sum of 444 comes from (1,3),(2,2),(3,1)(1,3),(2,2),(3,1)(1,3),(2,2),(3,1) and a sum of 999 from (3,6),(4,5),(5,4),(6,3)(3,6),(4,5),(5,4),(6,3)(3,6),(4,5),(5,4),(6,3), giving 777 outcomes out of 363636, so the probability is 736\tfrac{7}{36}7/36.
  • Which numbers cannot be a probability?
    Any value below 000 or above 111 — such as −0.2-0.2-0.2, 1.51.51.5 or 76\tfrac{7}{6}7/6 — because 0≤P(E)≤10\le P(E)\le10≤ P(E)≤1. A probability of 000 means an impossible event and 111 a sure event. Percentages work too: 115%115\%115% is not a valid probability.
  • Where can I find Probability Class 10 notes?
    Use the Key concepts & formulas section of this page: the classical definition P(E)=favourabletotalP(E)=\tfrac{\text{favourable}}{\text{total}}P(E)=favourable/total, complementary events, the 363636 outcomes of two dice and the make-up of a 525252-card deck. Exercise 14.1 itself is solved on Classmate AI's NCERT Solutions page for Chapter 14.
  • Where can I get Probability Class 10 extra questions?
    This page adds a bag of 555 red, 888 white and 777 black balls, cards numbered 111 to 252525, three coins tossed together, a deck with the spade face cards removed, a box of 121212 balls where adding 666 black ones doubles the chance of black, and a spinning-wheel case study, all with answers.
  • Where can I practise Probability Class 10 MCQs?
    Four MCQs and an assertion–reason item open the page: a prime number on one die, a value that cannot be a probability, a red king from a deck, a perfect-square sum on two dice, and one head with one tail on two coins. After a free sign-up, Create Test builds longer timed Probability quizzes.

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