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Probability — ICSE Class 10 Maths Important Questions

13 ICSE 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

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Key concepts
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Probability — ICSE Class 10 Maths Important Questions

Probability, One Outcome at a Time

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Quick answer

Paper-favourite ICSE Probability questions use P(E)=number of favourable outcomestotal number of outcomesP(E)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}P(E)=number of favourable outcomes/total number of outcomes for a single event with equally likely outcomes: drawing a card from a well-shuffled pack, throwing a die, tossing coins, and picking a coloured ball from a bag. Complementary probability P(not E)=1−P(E)P(\text{not }E)=1-P(E)P(not E)=1-P(E) and impossible/sure events appear almost every year.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Probability

Within the ICSE Class 10 Maths chapter Probability you find the probability of a single event with equally likely outcomes using P(E)=favourable outcomestotal outcomesP(E)=\dfrac{\text{favourable outcomes}}{\text{total outcomes}}P(E)=favourable outcomes/total outcomes. Standard experiments include drawing a card from a pack of 525252, throwing a die, tossing coins and drawing a coloured ball or marble from a bag, along with the complement rule and the ideas of sure and impossible events.

Definition $P(E)=\dfrac{\text{favourable}}{\text{total}}$Playing cards ($52$-card pack)Dice and coinsColoured balls / marbles from a bagComplementary, sure and impossible events

Key concepts & formulas

Probability of an event

For equally likely outcomes, P(E)=number of favourable outcomestotal number of outcomesP(E)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}P(E)=number of favourable outcomes/total number of outcomes, and 0≤P(E)≤10\le P(E)\le10≤ P(E)≤1.

Complement rule

P(not E)=1−P(E)P(\text{not }E)=1-P(E)P(not E)=1-P(E). The probability of a sure event is 111 and of an impossible event is 000.

A pack of cards

A pack has 525252 cards: 262626 red (hearts, diamonds) and 262626 black (spades, clubs); 444 suits of 131313; face cards are Jack, Queen, King (121212 in all); 444 aces.

Dice and coins

One die has 666 equally likely faces {1,2,3,4,5,6}\{1,2,3,4,5,6\}\1,2,3,4,5,6\. A fair coin has outcomes {H,T}\{H,T\}\H,T\; two coins give {HH,HT,TH,TT}\{HH,HT,TH,TT\}\HH,HT,TH,TT\ (4 outcomes).

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Important questions with answers

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Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

A die is thrown once. The probability of getting an even number is:

  1. (a)

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

  2. (b)

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

  3. (c)

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

  4. (d)

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

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

Even numbers are {2,4,6}\{2,4,6\}\2,4,6\, so P=36=12.P=\dfrac{3}{6}=\dfrac{1}{2}.P=3/6=1/2.

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Q2MCQEasy1 mark

One card is drawn from a well-shuffled pack of 525252 cards. The probability that it is a king is:

  1. (a)

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

  2. (b)

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

  3. (c)

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

  4. (d)

    413\dfrac{4}{13}4/13

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Answer: (b) 113\dfrac{1}{13}1/13.

There are 444 kings, so P=452=113.P=\dfrac{4}{52}=\dfrac{1}{13}.P=4/52=1/13.

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Q3MCQModerate1 mark

A bag contains 555 red and 333 green balls. The probability of drawing a green ball is:

  1. (a)

    58\dfrac{5}{8}5/8

  2. (b)

    38\dfrac{3}{8}3/8

  3. (c)

    35\dfrac{3}{5}3/5

  4. (d)

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

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Answer: (b) 38\dfrac{3}{8}3/8.

Total balls =5+3=8=5+3=8=5+3=8, favourable (green) =3=3=3, so P=38.P=\dfrac{3}{8}.P=3/8.

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Q4MCQHOTS1 mark

A card is drawn from a pack of 525252. The probability that it is neither a heart nor a king is:

  1. (a)

    913\dfrac{9}{13}9/13

  2. (b)

    413\dfrac{4}{13}4/13

  3. (c)

    313\dfrac{3}{13}3/13

  4. (d)

    1013\dfrac{10}{13}10/13

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Answer: (a) 913\dfrac{9}{13}9/13.

Hearts =13=13=13; kings not already counted (the king of hearts is in the hearts) =3=3=3. Cards that are a heart or a king =13+3=16=13+3=16=13+3=16. So neither =52−16=36=52-16=36=52-16=36, giving P=3652=913.P=\dfrac{36}{52}=\dfrac{9}{13}.P=36/52=9/13.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The probability of getting a number greater than 666 on a single throw of a die is 000.

Reason (R): An event that cannot occur is an impossible event and has probability 000.

  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: (a) A die shows only 111 to 666, so a number greater than 666 is impossible and has probability 000. R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Two fair coins are tossed together. Find the probability of getting (i) exactly one head, (ii) at least one head.

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The sample space is {HH,HT,TH,TT}\{HH, HT, TH, TT\}\HH, HT, TH, TT\, so total outcomes =4=4=4.

(i) Exactly one head: {HT,TH}=2\{HT, TH\}=2\HT, TH\=2 outcomes, so P=24=12.P=\dfrac{2}{4}=\dfrac{1}{2}.P=2/4=1/2.

(ii) At least one head: {HH,HT,TH}=3\{HH, HT, TH\}=3\HH, HT, TH\=3 outcomes, so P=34.P=\dfrac{3}{4}.P=3/4.

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Q7Very ShortModerate2 marks

A card is drawn at random from a well-shuffled pack of 525252 cards. Find the probability that it is (i) a red face card, (ii) an ace.

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Total outcomes =52=52=52.

(i) Red face cards are the Jack, Queen, King of hearts and of diamonds =2×3=6=2\times3=6=2×3=6. So P=652=326.P=\dfrac{6}{52}=\dfrac{3}{26}.P=6/52=3/26.

(ii) There are 444 aces, so P=452=113.P=\dfrac{4}{52}=\dfrac{1}{13}.P=4/52=1/13.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

A bag contains 666 red, 444 white and 555 blue balls. A ball is drawn at random. Find the probability that it is (i) white, (ii) not blue, (iii) red or white.

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Total balls =6+4+5=15.=6+4+5=15.=6+4+5=15.

(i) White: favourable =4=4=4, so P=415.P=\dfrac{4}{15}.P=4/15.

(ii) Not blue: not-blue balls =6+4=10=6+4=10=6+4=10, so P=1015=23.P=\dfrac{10}{15}=\dfrac{2}{3}.P=10/15=2/3. (Check: P(blue)=515=13P(\text{blue})=\dfrac{5}{15}=\dfrac13P(blue)=5/15=13, and 1−13=231-\dfrac13=\dfrac231-13=23.)

(iii) Red or white: favourable =6+4=10=6+4=10=6+4=10, so P=1015=23.P=\dfrac{10}{15}=\dfrac{2}{3}.P=10/15=2/3.

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Q9Short AnswerModerate3 marks

A die is thrown once. Find the probability of getting (i) a prime number, (ii) a number divisible by 333, (iii) a number less than 555.

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Total outcomes ={1,2,3,4,5,6}=6.=\{1,2,3,4,5,6\}=6.=\1,2,3,4,5,6\=6.

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

(ii) Numbers divisible by 333 are {3,6}=2\{3,6\}=2\3,6\=2, so P=26=13.P=\dfrac{2}{6}=\dfrac{1}{3}.P=2/6=1/3.

(iii) Numbers less than 555 are {1,2,3,4}=4\{1,2,3,4\}=4\1,2,3,4\=4, so P=46=23.P=\dfrac{4}{6}=\dfrac{2}{3}.P=4/6=2/3.

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Q10Short AnswerHOTS3 marks

A bag contains xxx white and 666 red balls. If the probability of drawing a white ball is 25\dfrac{2}{5}2/5, find the value of xxx and the number of balls in the bag.

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Total balls =x+6=x+6=x+6. Probability of a white ball:

P(white)=xx+6=25.P(\text{white})=\dfrac{x}{x+6}=\dfrac{2}{5}.P(white)=x/x+6=2/5.

Cross-multiplying: 5x=2(x+6)=2x+12.5x=2(x+6)=2x+12.5x=2(x+6)=2x+12.

5x−2x=12⇒3x=12⇒x=4.5x-2x=12\Rightarrow 3x=12\Rightarrow x=4.5x-2x=12 3x=12 x=4.

So there are 444 white balls, and the total number of balls =4+6=10.=4+6=10.=4+6=10.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

From a well-shuffled pack of 525252 cards, one card is drawn at random. Find the probability that the card is (i) a spade, (ii) a face card, (iii) a black ace, (iv) a card bearing a number between 222 and 888 (both inclusive) of hearts, (v) not a diamond.

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Total outcomes =52.=52.=52.

(i) Spade: there are 131313 spades, so P=1352=14.P=\dfrac{13}{52}=\dfrac{1}{4}.P=13/52=1/4.

(ii) Face card: Jacks, Queens, Kings =3×4=12=3\times4=12=3×4=12, so P=1252=313.P=\dfrac{12}{52}=\dfrac{3}{13}.P=12/52=3/13.

(iii) Black ace: aces of spades and clubs =2=2=2, so P=252=126.P=\dfrac{2}{52}=\dfrac{1}{26}.P=2/52=1/26.

(iv) Numbers 222 to 888 inclusive of hearts ={2,3,4,5,6,7,8}=7=\{2,3,4,5,6,7,8\}=7=\2,3,4,5,6,7,8\=7 cards, so P=752.P=\dfrac{7}{52}.P=7/52.

(v) Not a diamond: diamonds =13=13=13, so non-diamonds =52−13=39=52-13=39=52-13=39, giving P=3952=34.P=\dfrac{39}{52}=\dfrac{3}{4}.P=39/52=3/4.

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Q12Long AnswerHOTS5 marks

A box contains cards numbered 111 to 303030. One card is drawn at random. Find the probability that the number on the card is (i) a multiple of 555, (ii) a perfect square, (iii) a prime number greater than 101010, (iv) an even number or a multiple of 333.

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Total outcomes =30.=30.=30.

(i) Multiples of 555: {5,10,15,20,25,30}=6\{5,10,15,20,25,30\}=6\5,10,15,20,25,30\=6, so P=630=15.P=\dfrac{6}{30}=\dfrac{1}{5}.P=6/30=1/5.

(ii) Perfect squares: {1,4,9,16,25}=5\{1,4,9,16,25\}=5\1,4,9,16,25\=5, so P=530=16.P=\dfrac{5}{30}=\dfrac{1}{6}.P=5/30=1/6.

(iii) Primes greater than 101010 (up to 303030): {11,13,17,19,23,29}=6\{11,13,17,19,23,29\}=6\11,13,17,19,23,29\=6, so P=630=15.P=\dfrac{6}{30}=\dfrac{1}{5}.P=6/30=1/5.

(iv) Even numbers =15=15=15 (i.e. 2,4,…,302,4,\dots,302,4,,30); multiples of 333 =10=10=10 (i.e. 3,6,…,303,6,\dots,303,6,,30); numbers that are both even and a multiple of 333 (multiples of 666) ={6,12,18,24,30}=5=\{6,12,18,24,30\}=5=\6,12,18,24,30\=5. By counting the union, favourable =15+10−5=20=15+10-5=20=15+10-5=20, so

P=2030=23.P=\dfrac{20}{30}=\dfrac{2}{3}.P=20/30=2/3.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

In a school lucky draw, a bag holds 888 identical tickets numbered 111 to 888. A student draws one ticket at random. Answer the following.

(i) Write the total number of possible outcomes.

(ii) Find the probability of drawing an odd-numbered ticket.

(iii) Find the probability of drawing a ticket whose number is a multiple of 333.

(iv) Find the probability that the number is greater than 888, and state what kind of event this is.

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(i) The tickets are numbered 111 to 888, so the total number of possible outcomes is 888.

(ii) Odd numbers are {1,3,5,7}=4\{1,3,5,7\}=4\1,3,5,7\=4 outcomes, so P(odd)=48=12.P(\text{odd})=\dfrac{4}{8}=\dfrac{1}{2}.P(odd)=4/8=1/2.

(iii) Multiples of 333 are {3,6}=2\{3,6\}=2\3,6\=2 outcomes, so P=28=14.P=\dfrac{2}{8}=\dfrac{1}{4}.P=2/8=1/4.

(iv) No ticket has a number greater than 888, so favourable outcomes =0=0=0 and P=08=0P=\dfrac{0}{8}=0P=0/8=0. This is an impossible event.

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All ICSE Class 10 Maths Chapters

GST (Goods and Services Tax)Banking (Recurring Deposit Accounts)Shares and DividendsLinear Inequations (In One Variable)Quadratic EquationsSolving (Simple) Problems Based on Quadratic EquationsRatio and ProportionRemainder and Factor TheoremsMatricesArithmetic ProgressionGeometric ProgressionReflectionSection and Mid-Point FormulaEquation of a LineSimilarity (With Applications to Maps and Models)Loci (Locus and Its Constructions)CirclesTangents and Intersecting ChordsConstructions (Circles)Cylinder, Cone and Sphere (Surface Area and Volume)Trigonometrical IdentitiesHeights and DistancesGraphical Representation (Histograms and Ogives)Measures of Central Tendency (Mean, Median, Quartiles and Mode)Probability

Frequently asked questions

  • Do these Probability questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE 2026–27 syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

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