Class 10 Mathematics - NCERT Mathematics
NCERT Solutions for Class 10 Maths Chapter 14: Probability
Chapter 14 covers the theoretical (classical) approach to probability. Students learn to calculate the probability of an event using the formula P(E) = favourable outcomes / total outcomes. The chapter covers problems involving coins, dice, playing cards, coloured balls, numbered tokens, and real-life scenarios. This chapter is part of the Statistics and Probability unit which carries 11 marks in the CBSE Class 10 board exam 2027. It is one of the most scoring chapters as problems follow a predictable pattern.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 14 Probability has 1 exercise: Exercise 14.1 (25 questions). The older Exercise 14.2 (5 questions on compound events) has been removed from the CBSE syllabus and will NOT be asked in the 2027 board exam. However, solutions for Exercise 14.2 are provided below for state board students (UP Board, Bihar Board, MP Board, Rajasthan Board) who may still include these problems.
Table of Contents
14.1Exercise 14.1 - Probability of Events
Q1Complete the following statements: (i) Probability of an event E + Probability of the event 'not E' = ____. (ii) The probability of an event that cannot happen is ____. Such an event is called ____. (iii) The probability of an event that is certain to happen is ____. Such an event is called ____. (iv) The sum of the probabilities of all the elementary events of an experiment is ____. (v) The probability of an event is greater than or equal to ____ and less than or equal to ____.
(i) P(E) + P(not E) = 1
(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure (or certain) event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.
Q2Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car. The car starts or does not start. (ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. (iii) A trial is made to answer a true-false question. The answer is right or wrong. (iv) A baby is born. It is a boy or a girl.
(i) Not equally likely. Whether a car starts depends on many factors like fuel, battery, engine condition. The two outcomes are not equally likely.
(ii) Not equally likely. Whether a player scores depends on skill, distance, fatigue. The outcomes are not equally likely.
(iii) Equally likely. In a true-false question, if you guess randomly, the probability of right and wrong is the same (1/2 each). The two outcomes are equally likely.
(iv) Equally likely. A baby is equally likely to be a boy or a girl (approximately). The two outcomes are equally likely.
Q3Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Tossing a coin is considered fair because the two outcomes (Head and Tail) are equally likely. Each outcome has a probability of 1/2. Since neither team has an advantage, neither outcome favours one team over the other. This makes it an unbiased method of making a decision. The coin toss gives each team an equal chance of winning the toss.
Q4Which of the following cannot be the probability of an event? (A) 2/3 (B) -1.5 (C) 15% (D) 0.7
The probability of an event must satisfy: 0 <= P(E) <= 1.
(A) 2/3 = 0.667 - Valid (between 0 and 1)
(B) -1.5 - Not valid (negative, less than 0)
(C) 15% = 0.15 - Valid (between 0 and 1)
(D) 0.7 - Valid (between 0 and 1)
Answer: (B) -1.5 cannot be the probability of an event because probability can never be negative.
Q5If P(E) = 0.05, what is the probability of 'not E'?
P(not E) = 1 - P(E) = 1 - 0.05 = 0.95
Q6A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange flavoured candy? (ii) a lemon flavoured candy?
The bag contains only lemon flavoured candies.
(i) P(orange flavoured):
There are no orange candies in the bag. This is an impossible event.
P(orange) = 0/total = 0
(ii) P(lemon flavoured):
All candies are lemon flavoured. This is a sure event.
P(lemon) = total/total = 1
Q7It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Let E = event that 2 students have the same birthday.
Then 'not E' = event that 2 students do not have the same birthday.
P(not E) = 0.992
P(E) = 1 - P(not E) = 1 - 0.992 = 0.008
Q8A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?
Total balls = 3 + 5 = 8
(i) P(red) = 3/8
(ii) P(not red) = 1 - P(red) = 1 - 3/8 = 5/8
Alternatively: Not red means black. P(black) = 5/8.
Q9A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?
Total marbles = 5 + 8 + 4 = 17
(i) P(red) = 5/17
(ii) P(white) = 8/17
(iii) P(not green) = 1 - P(green) = 1 - 4/17 = 13/17
Alternatively: Not green = red + white = 5 + 8 = 13. P = 13/17.
Q10A piggy bank contains hundred 50p coins, fifty Re 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50p coin? (ii) will not be a Rs 5 coin?
Total coins = 100 + 50 + 20 + 10 = 180
(i) P(50p coin) = 100/180 = 5/9
(ii) P(not Rs 5 coin) = 1 - P(Rs 5 coin) = 1 - 10/180 = 1 - 1/18 = 17/18
Q11Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Total fish = 5 + 8 = 13
P(male fish) = 5/13 = 5/13
Q12A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. What is the probability that it will point at (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
Total outcomes = 8, Sample space = {1, 2, 3, 4, 5, 6, 7, 8}
(i) P(8) = 1/8
(ii) P(odd number):
Odd numbers = {1, 3, 5, 7} = 4 numbers
P = 4/8 = 1/2
(iii) P(number greater than 2):
Numbers > 2 = {3, 4, 5, 6, 7, 8} = 6 numbers
P = 6/8 = 3/4
(iv) P(number less than 9):
All 8 numbers are less than 9. This is a sure event.
P = 8/8 = 1
Q13A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.
Sample space when a die is thrown = {1, 2, 3, 4, 5, 6}. Total outcomes = 6.
(i) P(prime number):
Prime numbers on a die = {2, 3, 5} = 3 numbers
P = 3/6 = 1/2
(ii) P(number lying between 2 and 6):
Numbers between 2 and 6 (exclusive) = {3, 4, 5} = 3 numbers
P = 3/6 = 1/2
(iii) P(odd number):
Odd numbers = {1, 3, 5} = 3 numbers
P = 3/6 = 1/2
Q14One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds.
A standard deck: 52 cards, 4 suits (hearts, diamonds = red; clubs, spades = black), 13 cards per suit, 12 face cards (J, Q, K in each suit).
(i) P(king of red colour):
Red kings = King of hearts + King of diamonds = 2
P = 2/52 = 1/26
(ii) P(face card):
Face cards = 12 (3 per suit x 4 suits)
P = 12/52 = 3/13
(iii) P(red face card):
Red face cards = 6 (3 from hearts + 3 from diamonds)
P = 6/52 = 3/26
(iv) P(jack of hearts):
Only 1 jack of hearts in the deck.
P = 1/52 = 1/52
(v) P(a spade):
13 spades in a deck.
P = 13/52 = 1/4
(vi) P(queen of diamonds):
Only 1 queen of diamonds.
P = 1/52 = 1/52
Q15Five cards - the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
5 cards: 10, J, Q, K, A of diamonds.
(i) P(queen):
1 queen out of 5 cards.
P = 1/5 = 1/5
(ii) After queen is removed, 4 cards remain: 10, J, K, A
(a) P(ace):
1 ace out of 4 remaining cards.
P = 1/4 = 1/4
(b) P(queen):
Queen has already been removed. No queen left.
P = 0/4 = 0 (impossible event)
Q1612 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Total pens = 12 + 132 = 144
Good pens = 132
P(good pen) = 132/144 = 11/12
Q17(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
(i) Total bulbs = 20, Defective = 4
P(defective) = 4/20 = 1/5
(ii) One good bulb has been removed (not replaced).
Remaining bulbs = 19, Remaining defective = 4, Remaining good = 15
P(not defective) = 15/19 = 15/19
Q18A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
Total discs = 90.
(i) P(two-digit number):
Two-digit numbers from 1 to 90: 10, 11, 12, ..., 90
Count = 90 - 10 + 1 = 81
P = 81/90 = 9/10
(ii) P(perfect square):
Perfect squares from 1 to 90: 1, 4, 9, 16, 25, 36, 49, 64, 81
Count = 9 (since 10^2 = 100 > 90)
P = 9/90 = 1/10
(iii) P(divisible by 5):
Multiples of 5 from 1 to 90: 5, 10, 15, ..., 90
Count = 90/5 = 18
P = 18/90 = 1/5
Q19A child has a die whose six faces show the letters as given below: A B C D E A The die is thrown once. What is the probability of getting (i) A? (ii) D?
The 6 faces show: A, B, C, D, E, A. Total outcomes = 6.
(i) P(A):
A appears on 2 faces.
P = 2/6 = 1/3
(ii) P(D):
D appears on 1 face.
P = 1/6 = 1/6
Q20Suppose you drop a die at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with diameter 7 cm? (The rectangle has dimensions 14 cm by 7 cm.)
Area of rectangle = 14 x 7 = 98 cm^2
Diameter of circle = 7 cm, so radius = 3.5 cm
Area of circle = (pi) x r^2 = (22/7) x 3.5 x 3.5 = (22/7) x 12.25 = 38.5 cm^2
P(landing inside circle) = Area of circle / Area of rectangle
= 38.5 / 98
= 77/196
= 11/28
So the probability is 11/28 or approximately 0.393.
Q21A lot consists of 144 ball pens of which 20 are defective and the rest are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) she will buy it? (ii) she will not buy it?
Total pens = 144, Defective = 20, Good = 144 - 20 = 124
(i) P(she will buy) = P(good pen) = 124/144 = 31/36
(ii) P(she will not buy) = P(defective pen) = 20/144 = 5/36
Verification: 31/36 + 5/36 = 36/36 = 1
Q22Refer to Example 13. (i) Complete the following table: Event: Sum on 2 dice = 2,3,4,5,6,7,8,9,10,11,12 Probability: 1/36, ... (ii) A student argues that 'there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.
When two dice are thrown, total outcomes = 6 x 6 = 36.
(i) Probability table:
| Sum | Ways to get it | Probability |
|---|---|---|
| 2 | (1,1) = 1 way | 1/36 |
| 3 | (1,2),(2,1) = 2 ways | 2/36 = 1/18 |
| 4 | (1,3),(2,2),(3,1) = 3 ways | 3/36 = 1/12 |
| 5 | (1,4),(2,3),(3,2),(4,1) = 4 ways | 4/36 = 1/9 |
| 6 | (1,5),(2,4),(3,3),(4,2),(5,1) = 5 ways | 5/36 |
| 7 | (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6 ways | 6/36 = 1/6 |
| 8 | (2,6),(3,5),(4,4),(5,3),(6,2) = 5 ways | 5/36 |
| 9 | (3,6),(4,5),(5,4),(6,3) = 4 ways | 4/36 = 1/9 |
| 10 | (4,6),(5,5),(6,4) = 3 ways | 3/36 = 1/12 |
| 11 | (5,6),(6,5) = 2 ways | 2/36 = 1/18 |
| 12 | (6,6) = 1 way | 1/36 |
Total: 1+2+3+4+5+6+5+4+3+2+1 = 36 = 36/36 = 1
(ii) No, the student's argument is wrong. The 11 outcomes (sums 2 to 12) are not equally likely. For example, sum 7 can be obtained in 6 ways (probability 1/6), while sum 2 can only be obtained in 1 way (probability 1/36). The classical probability formula requires equally likely outcomes, and since the sums have different numbers of ways, each sum does NOT have probability 1/11.
Q23A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
When a coin is tossed 3 times, total outcomes = 2^3 = 8.
Sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Hanif wins if all tosses are the same: {HHH, TTT} = 2 outcomes
Hanif loses in all other cases = 8 - 2 = 6 outcomes
P(Hanif loses) = 6/8 = 3/4
Q24A die is thrown once. Determine the probability of getting (i) a number greater than or equal to 3 (ii) a number less than or equal to 1 (iii) a number more than 6 (iv) a number less than 6.
Sample space = {1, 2, 3, 4, 5, 6}. Total = 6.
(i) P(>= 3):
Numbers >= 3: {3, 4, 5, 6} = 4 numbers
P = 4/6 = 2/3
(ii) P(<= 1):
Numbers <= 1: {1} = 1 number
P = 1/6 = 1/6
(iii) P(more than 6):
No number on a die is more than 6. Impossible event.
P = 0/6 = 0
(iv) P(less than 6):
Numbers < 6: {1, 2, 3, 4, 5} = 5 numbers
P = 5/6 = 5/6
Q25One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability of getting: (i) A queen of clubs or a king of hearts (ii) A queen or a jack
Total cards = 52.
(i) P(queen of clubs or king of hearts):
Queen of clubs = 1 card, King of hearts = 1 card.
These are different cards, so no overlap.
Favourable outcomes = 1 + 1 = 2
P = 2/52 = 1/26
(ii) P(queen or jack):
Queens = 4, Jacks = 4. No overlap (a card cannot be both).
Favourable = 4 + 4 = 8
P = 8/52 = 2/13
old-14.2Old Exercise 14.2 - Problems on Compound Events (For State Boards)
This exercise has been removed from the CBSE 2026-27 syllabus. It is provided here for UP Board, Bihar Board, MP Board, and other state board students.
Q1Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?
Days = {Tue, Wed, Thu, Fri, Sat} = 5 days.
Total outcomes = 5 x 5 = 25 (each person independently chooses one of 5 days).
(i) P(same day):
Favourable: (Tue,Tue), (Wed,Wed), (Thu,Thu), (Fri,Fri), (Sat,Sat) = 5
P = 5/25 = 1/5
(ii) P(consecutive days):
Consecutive pairs: (Tue,Wed), (Wed,Tue), (Wed,Thu), (Thu,Wed), (Thu,Fri), (Fri,Thu), (Fri,Sat), (Sat,Fri) = 8
P = 8/25 = 8/25
(iii) P(different days):
P = 1 - P(same day) = 1 - 1/5 = 4/5
Q2A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score is noted. Complete the following table which gives a few values of the total score on the two throws: (What is the probability that the total score is (i) even? (ii) 6? (iii) at least 6?)
The die faces: {1, 2, 2, 3, 3, 6}. When thrown twice, total outcomes = 6 x 6 = 36.
Possible sums and their counts:
| + | 1 | 2 | 2 | 3 | 3 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 3 | 4 | 4 | 7 |
| 2 | 3 | 4 | 4 | 5 | 5 | 8 |
| 2 | 3 | 4 | 4 | 5 | 5 | 8 |
| 3 | 4 | 5 | 5 | 6 | 6 | 9 |
| 3 | 4 | 5 | 5 | 6 | 6 | 9 |
| 6 | 7 | 8 | 8 | 9 | 9 | 12 |
(i) P(even total):
Even sums: 2(1), 4(8), 6(4), 8(4), 12(1) = 1+8+4+4+1 = 18
P = 18/36 = 1/2
(ii) P(total = 6):
Sum 6 appears 4 times in the table.
P = 4/36 = 1/9
(iii) P(total at least 6):
Sums >= 6: 6(4), 7(2), 8(4), 9(4), 12(1) = 4+2+4+4+1 = 15
P = 15/36 = 5/12
Q3A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Let number of blue balls = x.
Total balls = 5 + x.
P(red) = 5/(5 + x)
P(blue) = x/(5 + x)
Given: P(blue) = 2 x P(red)
x/(5 + x) = 2 x 5/(5 + x)
x/(5 + x) = 10/(5 + x)
Since denominators are equal:
x = 10
Number of blue balls = 10
Verification: Total = 15. P(red) = 5/15 = 1/3. P(blue) = 10/15 = 2/3 = 2 x (1/3).
Q4A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Initially: Total = 12, Black = x
P(black initially) = x/12
After adding 6 black balls: Total = 18, Black = x + 6
P(black now) = (x + 6)/18
Given: (x + 6)/18 = 2 x (x/12)
(x + 6)/18 = 2x/12 = x/6
Cross-multiplying:
6(x + 6) = 18x
6x + 36 = 18x
36 = 12x
x = 3
x = 3 (initially 3 black balls)
Verification: P(initial) = 3/12 = 1/4. P(new) = 9/18 = 1/2 = 2 x (1/4).
Q5A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue marbles in the jar.
Total marbles = 24
P(green) = 2/3
Number of green marbles = P(green) x Total = (2/3) x 24 = 16
Number of blue marbles = 24 - 16 = 8
Verification: P(green) = 16/24 = 2/3. P(blue) = 8/24 = 1/3.
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A card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability of getting (i) a face card (ii) a card which is neither a king nor a red card.
Total cards = 52. (i) Face cards = 12 (J, Q, K in each of 4 suits). P(face card) = 12/52 = 3/13. (ii) Kings = 4, Red cards = 26. Red cards already include 2 red kings, so cards that are king or red = 26 red cards + 2 black kings = 28. Cards neither king nor red = 52 - 28 = 24. P = 24/52 = 6/13.
Two dice are thrown simultaneously. Find the probability of getting (i) a doublet (ii) a sum of 8.
Total outcomes = 36 when two dice are thrown. (i) Doublets: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 outcomes. P(doublet) = 6/36 = 1/6. (ii) Sum of 8: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 outcomes. P(sum = 8) = 5/36.
A bag contains 5 red balls, 4 green balls and 3 yellow balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is (i) not yellow (ii) either red or yellow.
Total balls = 5 + 4 + 3 = 12. (i) P(not yellow) = 1 - P(yellow) = 1 - 3/12 = 1 - 1/4 = 3/4. (ii) Red or yellow balls = 5 + 3 = 8. P(red or yellow) = 8/12 = 2/3.
From a deck of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining cards, a card is drawn at random. Find the probability that the drawn card is (i) a black king (ii) a card of red colour (iii) a jack.
Removed: Red J, Q, K = 6 cards (3 from hearts + 3 from diamonds). Remaining = 52 - 6 = 46 cards. (i) Black kings remain = 2 (king of clubs, king of spades). P = 2/46 = 1/23. (ii) Red cards remaining = 26 - 6 = 20 (hearts and diamonds minus face cards removed). P = 20/46 = 10/23. (iii) Jacks remaining = 4 - 2 = 2 (only black jacks). P = 2/46 = 1/23.
A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.
First 50 natural numbers: 1, 2, 3, ..., 50. Total = 50. Multiples of both 3 and 4 means multiples of LCM(3,4) = 12. Multiples of 12 from 1 to 50: 12, 24, 36, 48 = 4 numbers. P(multiple of 3 and 4) = 4/50 = 2/25.
A bag contains 4 red, 3 blue, and 2 yellow balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is (i) red (ii) yellow (iii) not blue.
Total balls = 4 + 3 + 2 = 9. (i) P(red) = 4/9. (ii) P(yellow) = 2/9. (iii) P(not blue) = 1 - P(blue) = 1 - 3/9 = 1 - 1/3 = 2/3. Alternatively, not blue = 4 + 2 = 6, P = 6/9 = 2/3.
Two dice are thrown at the same time. Find the probability of getting (i) same number on both dice (ii) different numbers on both dice.
Total outcomes = 36. (i) Same number (doublet): (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 outcomes. P(same) = 6/36 = 1/6. (ii) P(different) = 1 - P(same) = 1 - 1/6 = 5/6. Alternatively, for each value on die 1, die 2 has 5 choices that differ, giving 6 x 5 = 30 outcomes. P = 30/36 = 5/6.
A box contains cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is (i) a prime number (ii) a composite number (iii) a number divisible by 3.
Total cards = 20. (i) Primes from 1 to 20: 2, 3, 5, 7, 11, 13, 17, 19 = 8 numbers. P(prime) = 8/20 = 2/5. (ii) Composite numbers from 1 to 20: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20 = 11 numbers. (Note: 1 is neither prime nor composite.) P(composite) = 11/20. (iii) Divisible by 3: 3, 6, 9, 12, 15, 18 = 6 numbers. P(div by 3) = 6/20 = 3/10.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) the jack of hearts (iv) a red face card.
Total = 52. (i) Red kings = 2 (heart and diamond). P = 2/52 = 1/26. (ii) Face cards = 12 (J, Q, K from each of 4 suits). P = 12/52 = 3/13. (iii) Jack of hearts = 1 unique card. P = 1/52. (iv) Red face cards = 6 (3 from hearts + 3 from diamonds: JH, QH, KH, JD, QD, KD). P = 6/52 = 3/26.
A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.
Let black balls = x. Total = 15 + x. P(white) = 15/(15+x). P(black) = x/(15+x). Given P(black) = 3 x P(white): x/(15+x) = 3 x 15/(15+x) = 45/(15+x). Since denominators are equal: x = 45. Number of black balls = 45. Verification: P(white) = 15/60 = 1/4. P(black) = 45/60 = 3/4 = 3 x 1/4.
A die is thrown once. Find the probability of getting a number (i) less than 3 (ii) greater than or equal to 4.
Sample space = {1,2,3,4,5,6}, Total = 6. (i) Numbers less than 3: {1, 2} = 2 outcomes. P = 2/6 = 1/3. (ii) Numbers >= 4: {4, 5, 6} = 3 outcomes. P = 3/6 = 1/2.
A box contains 20 balls bearing numbers 1, 2, 3, ..., 20. A ball is drawn at random from the box. What is the probability that the number on the ball is (i) an odd number (ii) divisible by 2 or 3 (iii) a prime number?
Total = 20. (i) Odd numbers: 1,3,5,7,9,11,13,15,17,19 = 10. P = 10/20 = 1/2. (ii) Divisible by 2: 2,4,6,...,20 = 10. Divisible by 3: 3,6,9,12,15,18 = 6. Divisible by both (6): 6,12,18 = 3. By inclusion-exclusion: 10 + 6 - 3 = 13. P = 13/20. (iii) Primes: 2,3,5,7,11,13,17,19 = 8. P = 8/20 = 2/5.
Two different dice are thrown together. Find the probability that the numbers obtained (i) have a sum less than 7 (ii) have a product less than 16 (iii) is a doublet of odd numbers.
Total outcomes = 36. (i) Sum < 7: Sum 2(1), 3(2), 4(3), 5(4), 6(5) = 1+2+3+4+5 = 15 outcomes. P = 15/36 = 5/12. (ii) Product < 16: We list pairs with product >= 16 and subtract. Products >= 16: (3,6)=18, (6,3)=18, (4,4)=16, (4,5)=20, (5,4)=20, (4,6)=24, (6,4)=24, (5,5)=25, (5,6)=30, (6,5)=30, (6,6)=36 = 11 outcomes. P(product < 16) = (36-11)/36 = 25/36. (iii) Doublet of odd numbers: (1,1), (3,3), (5,5) = 3 outcomes. P = 3/36 = 1/12.
Cards marked with numbers 3, 4, 5, ..., 50 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the drawn card is (i) divisible by 7 (ii) a perfect square.
Numbers: 3, 4, 5, ..., 50. Total = 50 - 3 + 1 = 48 cards. (i) Divisible by 7: 7, 14, 21, 28, 35, 42, 49 = 7 numbers. P = 7/48. (ii) Perfect squares from 3 to 50: 4, 9, 16, 25, 36, 49 = 6 numbers. P = 6/48 = 1/8.
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Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): If P(E) = 0.3, then P(not E) = 0.7.
Reason (R): P(E) + P(not E) = 1 for any event E.
Answer: (A)
Both are true. P(not E) = 1 - 0.3 = 0.7. The reason (complement rule) correctly explains why P(not E) = 0.7.
AR2
Assertion (A): When two coins are tossed simultaneously, the probability of getting at least one head is 3/4.
Reason (R): Total outcomes when two coins are tossed = 4 (HH, HT, TH, TT).
Answer: (A)
Both are true. At least one head: HH, HT, TH = 3 outcomes out of 4. P = 3/4. The reason correctly states the total outcomes, which is used to calculate the assertion.
AR3
Assertion (A): The probability of getting a sum of 13 when two dice are thrown is 0.
Reason (R): The maximum sum possible with two dice is 12.
Answer: (A)
Both are true. Maximum sum = 6 + 6 = 12. Sum of 13 is impossible. P = 0. The reason correctly explains why it is impossible.
AR4
Assertion (A): If a die is thrown once, the probability of getting a number greater than 6 is 1.
Reason (R): A standard die has numbers 1 to 6.
Answer: (D)
Assertion is FALSE - no number on a standard die is greater than 6, so P(>6) = 0, not 1. Reason is TRUE - a die has numbers 1 to 6. So answer is D (Assertion false, Reason true).
AR5
Assertion (A): A card is drawn from a deck of 52 cards. The probability of getting a card which is neither a heart nor a king is 9/13.
Reason (R): Hearts = 13 cards, Kings = 4 cards. Cards that are heart or king = 13 + 4 - 1 = 16.
Answer: (A)
Hearts = 13, Kings = 4, but King of hearts is counted in both, so hearts or kings = 13 + 4 - 1 = 16. Neither = 52 - 16 = 36. P = 36/52 = 9/13. Both assertion and reason are true, and reason explains assertion.
AR6
Assertion (A): The probability of selecting a vowel from the English alphabet is 5/26.
Reason (R): There are 5 vowels (A, E, I, O, U) in the English alphabet of 26 letters.
Answer: (A)
Both are true. Vowels = 5 out of 26 letters. P = 5/26. The reason correctly states the count used in the calculation.
Case Study Questions
At a school fun fair, a lucky draw box contains 30 tokens numbered 1 to 30. A student picks one token at random to win prizes.
(i) What is the probability that the token has a number which is a multiple of 5? (1 mark)
Multiples of 5 from 1 to 30: 5, 10, 15, 20, 25, 30 = 6 numbers.
P = 6/30 = 1/5
(ii) What is the probability that the number on the token is a perfect square? (1 mark)
Perfect squares from 1 to 30: 1, 4, 9, 16, 25 = 5 numbers.
P = 5/30 = 1/6
(iii) What is the probability that the token number is a prime number greater than 10? (2 marks)
Primes from 1 to 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
Primes > 10: 11, 13, 17, 19, 23, 29 = 6 numbers.
P = 6/30 = 1/5
(iv) Find the probability that the token number is neither divisible by 2 nor by 5. (2 marks)
Divisible by 2: 2,4,6,...,30 = 15 numbers.
Divisible by 5: 5,10,15,20,25,30 = 6 numbers.
Divisible by both (10): 10,20,30 = 3 numbers.
Divisible by 2 or 5 = 15 + 6 - 3 = 18.
Neither = 30 - 18 = 12.
P = 12/30 = 2/5
Rahul and Priya are playing a card game. They use a standard deck of 52 playing cards. The deck has 4 suits - hearts and diamonds (red), clubs and spades (black). Each suit has 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King. The 12 face cards are Jacks, Queens, and Kings. One card is drawn at random.
(i) What is the probability of drawing a black card? (1 mark)
Black cards = clubs + spades = 13 + 13 = 26.
P = 26/52 = 1/2
(ii) What is the probability of drawing an ace? (1 mark)
Aces = 4 (one in each suit).
P = 4/52 = 1/13
(iii) What is the probability of drawing a card that is neither a king nor a queen? (2 marks)
Kings = 4, Queens = 4. Total king or queen = 8.
Neither king nor queen = 52 - 8 = 44.
P = 44/52 = 11/13
(iv) If all the face cards are removed, what is the probability of drawing a card of hearts from the remaining cards? (2 marks)
Face cards removed = 12 (3 from each suit).
Remaining = 52 - 12 = 40 cards.
Hearts remaining = 13 - 3 = 10 (Ace through 10 of hearts).
P = 10/40 = 1/4
MCQ Practice
MCQ1. The probability of an event that is certain to occur is:
(A) 0
(B) 1
(C) 1/2
(D) 3/4
Answer: (B) - A certain (sure) event always occurs. Its probability is 1.
MCQ2. A die is thrown once. The probability of getting an even number is:
(A) 1/2
(B) 1/3
(C) 1/6
(D) 2/3
Answer: (A) - Even numbers on a die: {2, 4, 6} = 3 out of 6. P = 3/6 = 1/2.
MCQ3. A card is selected from a deck of 52 cards. The probability of it being a red face card is:
(A) 3/26
(B) 3/13
(C) 2/13
(D) 1/2
Answer: (A) - Red face cards: J, Q, K of hearts and J, Q, K of diamonds = 6 cards. P = 6/52 = 3/26.
MCQ4. If P(E) = 0.05, what is the probability of 'not E'?
(A) 0.05
(B) 0.5
(C) 0.95
(D) 0
Answer: (C) - P(not E) = 1 - P(E) = 1 - 0.05 = 0.95.
MCQ5. Someone is asked to take a number from 1 to 100. The probability that it is a prime is:
(A) 1/5
(B) 6/25
(C) 1/4
(D) 13/50
Answer: (C) - Primes from 1 to 100: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97 = 25 primes. P = 25/100 = 1/4.
MCQ6. A bag contains 3 red, 5 blue and 7 green balls. A ball is picked at random. The probability that it is not blue is:
(A) 1/3
(B) 2/3
(C) 5/15
(D) 10/15
Answer: (B) - Total = 15. P(not blue) = 1 - 5/15 = 1 - 1/3 = 2/3.
MCQ7. The probability that a non-leap year has 53 Sundays is:
(A) 1/7
(B) 2/7
(C) 3/7
(D) 5/7
Answer: (A) - A non-leap year has 365 days = 52 weeks + 1 day. The extra day can be any one of the 7 days. For 53 Sundays, the extra day must be Sunday. P = 1/7.
MCQ8. Two coins are tossed simultaneously. The probability of getting at most one head is:
(A) 1/4
(B) 1/2
(C) 3/4
(D) 1
Answer: (C) - Outcomes: HH, HT, TH, TT = 4. At most one head means 0 or 1 heads: HT, TH, TT = 3. P = 3/4.
MCQ9. Which of the following cannot be the probability of an event?
(A) 1/3
(B) 0.1
(C) 3%
(D) 17/16
Answer: (D) - P(E) must satisfy 0 <= P(E) <= 1. 17/16 = 1.0625 > 1, so it cannot be a probability.
MCQ10. If a card is drawn from a deck of 52 cards, the probability of getting a card which is not a heart is:
(A) 1/4
(B) 3/4
(C) 12/13
(D) 1/13
Answer: (B) - Hearts = 13. P(heart) = 13/52 = 1/4. P(not heart) = 1 - 1/4 = 3/4.
MCQ11. A box has 5 blue, 4 red and 3 green balls. If a ball is selected at random, what is the probability of it being a blue ball?
(A) 5/12
(B) 4/12
(C) 3/12
(D) 1/3
Answer: (A) - Total = 5+4+3 = 12. P(blue) = 5/12.
MCQ12. Two dice are thrown at the same time and the product of numbers appearing on them is noted. The probability that the product is less than 9 is:
(A) 1/3
(B) 2/9
(C) 4/9
(D) 2/3
Answer: (C) - Total = 36. Products < 9: (1,1)=1, (1,2)=2, (2,1)=2, (1,3)=3, (3,1)=3, (1,4)=4, (4,1)=4, (2,2)=4, (1,5)=5, (5,1)=5, (1,6)=6, (6,1)=6, (2,3)=6, (3,2)=6, (2,4)=8, (4,2)=8 = 16 outcomes. P = 16/36 = 4/9.
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Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
If P(E) = 0.6, find P(not E).
P(not E) = 1 - P(E) = 1 - 0.6 = 0.4.
A die is thrown once. What is the probability of getting a composite number?
Composite numbers on a die: {4, 6} = 2 numbers. P = 2/6 = 1/3.
What is the probability of a sure event?
The probability of a sure (certain) event is 1.
In a single throw of a die, what is the probability of getting a number greater than 4?
Numbers > 4: {5, 6} = 2. P = 2/6 = 1/3.
If the probability of winning a game is 0.3, what is the probability of losing it?
P(losing) = 1 - P(winning) = 1 - 0.3 = 0.7.
2-3 Mark Questions
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing (i) an ace (ii) a red card (iii) a face card.
(i) Aces = 4. P = 4/52 = 1/13. (ii) Red cards = 26. P = 26/52 = 1/2. (iii) Face cards = 12. P = 12/52 = 3/13.
A box contains 5 red, 8 white and 4 green marbles. One marble is drawn at random. What is the probability that the marble drawn is (i) red? (ii) white? (iii) not green?
Total = 17. (i) P(red) = 5/17. (ii) P(white) = 8/17. (iii) P(not green) = 1 - 4/17 = 13/17.
A bag contains 6 red, 5 blue and 4 green balls. A ball is drawn at random. Find the probability that the ball drawn is (i) not blue (ii) either red or green.
Total = 15. (i) P(not blue) = 1 - 5/15 = 10/15 = 2/3. (ii) Red or green = 6+4 = 10. P = 10/15 = 2/3.
A box contains 90 discs numbered from 1 to 90. If one disc is drawn at random, find the probability that it bears (i) a number divisible by 5 (ii) a perfect square.
Total = 90. (i) Multiples of 5: 5,10,...,90 = 18. P = 18/90 = 1/5. (ii) Perfect squares: 1,4,9,16,25,36,49,64,81 = 9. P = 9/90 = 1/10.
3-4 Mark Questions
All the red face cards are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards. Find the probability that the card drawn is (i) a red card (ii) a face card (iii) a card of clubs.
Red face cards removed = 6. Remaining = 46 cards. (i) Red cards left = 26-6 = 20. P = 20/46 = 10/23. (ii) Face cards left = 12-6 = 6. P = 6/46 = 3/23. (iii) Clubs = 13 (untouched). P = 13/46.
From a deck of 52 playing cards, Jacks and Kings are removed. From the remaining, a card is drawn at random. Find the probability that (i) it is a face card (ii) it is a red card (iii) it is a card of spades.
Removed: 4 Jacks + 4 Kings = 8. Remaining = 44. (i) Face cards left = only 4 Queens. P = 4/44 = 1/11. (ii) Red removed = 2 red Jacks + 2 red Kings = 4. Red left = 26-4 = 22. P = 22/44 = 1/2. (iii) Spades left = 13-2 (J and K of spades removed) = 11. P = 11/44 = 1/4.
Two dice are thrown simultaneously. Find the probability of getting (i) a sum greater than 10 (ii) a sum equal to 7 (iii) a doublet.
Total = 36. (i) Sum > 10: sum 11: (5,6),(6,5)=2; sum 12: (6,6)=1. Total=3. P = 3/36 = 1/12. (ii) Sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6. P = 6/36 = 1/6. (iii) Doublets: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) = 6. P = 6/36 = 1/6.
Key Concepts
Important Formulas
Study Plan - Probability
Basics of Probability and Simple Events
Read NCERT theory on classical probability. Understand sample space, favourable outcomes, equally likely outcomes, impossible and sure events. Solve Exercise 14.1 Q1-Q8.
Dice and Spinner Problems
Solve Exercise 14.1 Q9-Q13 and Q19, Q24. Master die problems (single die, two dice). Practice listing all possible outcomes systematically. Learn the sum table for two dice.
Playing Card Problems
Memorise deck structure (52 cards, 4 suits, 13 per suit, 12 face cards). Solve Q14, Q15, Q25. Practice 'cards removed' type problems where some cards are taken out before drawing.
Numbered Tokens, Balls, and Real-Life Problems
Solve Q16-Q23. Focus on complement-based problems (P(not E) = 1 - P(E)), defective items, numbered discs (divisibility, perfect squares, primes), and geometric probability (Q20).
PYQs, MCQs, AR, Case Studies, and Revision
Solve all 14 PYQs. Practice 12 MCQs and 6 AR questions. Attempt both case studies under timed conditions. Review common mistakes and tips. Make a cheat sheet of card deck facts.
Common Mistakes to Avoid
- ✗Confusing 'between 2 and 6' (exclusive: 3,4,5) with 'from 2 to 6' (inclusive: 2,3,4,5,6) - always check whether the boundary values are included
- ✗In card problems, forgetting that face cards are only Jack, Queen, King (not Ace). There are 12 face cards, not 16.
- ✗Using the wrong total when cards are removed - if 6 cards are removed from a deck, the new total is 46, not 52
- ✗Forgetting that 1 is neither prime nor composite when counting primes from a set of numbers
- ✗In two-dice problems, treating (3,5) and (5,3) as the same outcome - they are different outcomes because the dice are distinguishable
- ✗Not simplifying fractions in the final answer - always reduce 12/52 to 3/13, 6/36 to 1/6, etc.
- ✗In complement problems, calculating P(E) and P(not E) separately and getting a sum that is not 1 - always verify P(E) + P(not E) = 1
- ✗In numbered disc/token problems, miscounting due to off-by-one errors - for example, numbers from 10 to 90 inclusive is 81 numbers (90 - 10 + 1), not 80
Tips for Board Exam
- ✓Memorise the deck of cards structure: 52 cards, 4 suits (2 red: hearts/diamonds, 2 black: clubs/spades), 13 per suit (A,2-10,J,Q,K), 12 face cards (J,Q,K x 4), 4 aces. This saves time in card problems.
- ✓For two-dice problems, draw the 6x6 grid once and use it for all parts. The sum table helps: sum 7 is most likely (6/36), sums 2 and 12 are least likely (1/36 each).
- ✓Always use the complement rule P(not E) = 1 - P(E) when the event is easier to count in reverse. For example, P(at least one head) = 1 - P(no heads).
- ✓For 'numbers from a to b' problems: count = b - a + 1 (inclusive). Multiples of k from 1 to n: count = floor(n/k). Perfect squares from 1 to n: count = floor(sqrt(n)).
- ✓Write the answer as a simplified fraction. Avoid decimals unless asked. Show the total outcomes and favourable outcomes clearly for full marks.
- ✓In board exams, card problems carry 3-4 marks. Read carefully whether cards have been removed or added before drawing. The total changes and so do the favourable outcomes.
- ✓When asked 'probability of A or B', add the individual favourable counts only if A and B are mutually exclusive (no overlap). If there is overlap, subtract it: P(A or B) = P(A) + P(B) - P(A and B).
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Start Probability - Choose Your ModeAbout Probability - Class 10 Mathematics
Chapter 14 Probability is the final chapter of CBSE Class 10 Mathematics and is part of the Statistics and Probability unit which carries 11 marks in the board exam 2027. This chapter typically contributes 4-5 marks and is considered one of the easiest and most scoring chapters because every problem follows a single formula: P(E) = number of favourable outcomes divided by total number of equally likely outcomes.
What This Chapter Covers
The chapter introduces the classical or theoretical approach to probability. Unlike the experimental approach (studied in Class 9, where you actually toss coins and count results), the theoretical approach calculates probability using logic and counting. The core idea is simple: if all outcomes of an experiment are equally likely, then the probability of an event is the ratio of favourable outcomes to total outcomes.
The Probability Formula
P(E) = Number of outcomes favourable to event E / Total number of possible outcomes
This formula works only when all outcomes are equally likely. For a fair coin, heads and tails are equally likely. For a fair die, all six faces are equally likely. For a well-shuffled deck, all 52 cards are equally likely to be drawn.
Key Properties of Probability
The probability of any event lies between 0 and 1 inclusive. P(E) = 0 means the event is impossible (it can never happen, like rolling a 7 on a standard die). P(E) = 1 means the event is certain (it will always happen, like getting a number less than 7 on a standard die). The sum of probabilities of all elementary events equals 1.
The complement rule states that P(not E) = 1 - P(E). This is extremely useful when counting the outcomes where E does not happen is easier than counting where E happens. For example, P(at least one head in 3 coin tosses) = 1 - P(no heads) = 1 - 1/8 = 7/8.
Probability with Coins
A fair coin has two equally likely outcomes: Head (H) and Tail (T). For n coin tosses, total outcomes = 2^n. For 1 toss: 2 outcomes. For 2 tosses: 4 outcomes (HH, HT, TH, TT). For 3 tosses: 8 outcomes. The key pattern is that each additional toss doubles the total outcomes.
Common questions involve finding probabilities like P(all heads), P(at least one head), P(exactly 2 heads). For these, list the sample space systematically and count favourable outcomes.
Probability with Dice
A standard die has six equally likely outcomes: {1, 2, 3, 4, 5, 6}. Common events include: prime numbers {2,3,5} with P = 1/2, even numbers {2,4,6} with P = 1/2, numbers greater than 4 {5,6} with P = 1/3.
When two dice are thrown together, total outcomes = 36 (6 x 6). The most commonly asked concept is the sum of two dice. Sum 7 is the most likely (6 ways out of 36, P = 1/6), while sums 2 and 12 are the least likely (1 way each, P = 1/36). Students should memorise or be able to quickly construct the 6x6 outcome table.
Probability with Playing Cards
Playing card problems are the most frequently tested type in CBSE board exams. A standard deck has 52 cards divided into 4 suits. Hearts and diamonds are red (26 cards total). Clubs and spades are black (26 cards total). Each suit has 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.
Face cards are Jack, Queen, and King only (not Ace). There are 12 face cards total (3 per suit). Aces are not face cards - this is a common source of errors.
Common probabilities: P(any specific card) = 1/52, P(any suit) = 1/4, P(face card) = 3/13, P(red card) = 1/2, P(ace) = 1/13, P(king) = 1/13.
A frequently asked variation involves removing some cards and then drawing from the remaining. For example, if all face cards are removed, only 40 cards remain (10 per suit), and probabilities change accordingly.
Probability with Numbered Tokens and Balls
Problems involving numbered tokens (1 to n) test number theory concepts within probability. Common events include: multiples of k (count = floor(n/k)), perfect squares (count = floor(sqrt(n))), prime numbers, two-digit numbers, and divisibility conditions.
Ball-in-bag problems test basic probability. Given balls of different colours, students calculate the probability of drawing a specific colour or not drawing a specific colour. The complement rule is often needed.
Geometric Probability
Exercise 14.1 Q20 introduces geometric probability where the probability is calculated as the ratio of areas. If a dart is thrown randomly at a rectangular board containing a circle, the probability of landing inside the circle equals the area of the circle divided by the area of the rectangle.
CBSE 2026-27 Syllabus Details
For the CBSE 2026-27 session, Chapter 14 has 1 exercise: Exercise 14.1 with 25 questions. The older Exercise 14.2 (5 questions on compound events and finding unknowns) has been removed from the CBSE syllabus. Padhaao provides Exercise 14.2 solutions for state board students who may still need them.
How This Chapter Is Tested in Board Exams
Probability questions in CBSE board exams typically include one MCQ worth 1 mark (basic concept like complement, impossible event, or simple die/coin), one short answer worth 2-3 marks (ball-in-bag, die throw, or simple card problem), and sometimes a long answer worth 3-4 marks (playing card problem with multiple parts or numbered token with divisibility).
The most frequently tested problem types are: drawing a card from a deck (especially with cards removed), balls drawn from a bag (complement questions), dice thrown once or twice, numbered discs/tokens (multiples, primes, squares), and finding unknown quantities given a probability condition.
Connection to Other Chapters
Probability connects to Chapter 13 (Statistics) as both are in the same unit. The frequency-based (experimental) approach to probability from Class 9 connects to the theoretical approach in Class 10. In higher classes (11 and 12), students study conditional probability, Bayes theorem, and probability distributions, building directly on the foundation laid in this chapter.
The counting techniques used in probability (listing outcomes, finding multiples, primes, perfect squares) connect to number theory concepts from Chapter 1 (Real Numbers). The area-based probability in Q20 connects to mensuration concepts from Chapters 11 and 12.
Frequently Asked Questions
How many exercises are in NCERT Class 10 Maths Chapter 14 for CBSE 2026-27?
For CBSE 2026-27, Chapter 14 Probability has 1 exercise: Exercise 14.1 with 25 questions. The older Exercise 14.2 (5 questions on compound events) has been removed from the CBSE syllabus. Padhaao provides Exercise 14.2 solutions for state board students.
What is the formula for probability of an event?
P(E) = Number of outcomes favourable to E / Total number of possible outcomes. This formula assumes all outcomes are equally likely. The probability always lies between 0 and 1.
How many marks does Probability carry in the CBSE Class 10 board exam?
Probability is part of the Statistics and Probability unit which carries 11 marks total (shared with Chapter 13 Statistics). Chapter 14 alone typically contributes 4-5 marks: 1 MCQ (1 mark) + 1 short answer on dice/cards/balls (2-3 marks).
How many cards are in a standard deck and what are the suits?
A standard deck has 52 cards divided into 4 suits: Hearts and Diamonds (red), Clubs and Spades (black). Each suit has 13 cards: Ace, 2 through 10, Jack, Queen, King. There are 12 face cards (J, Q, K in each suit) and 4 aces. Aces are NOT face cards.
What is the complement of an event in probability?
The complement of event E (written as 'not E' or E') includes all outcomes that are not in E. P(not E) = 1 - P(E). This rule is extremely useful when it is easier to count outcomes where the event does not happen.
What is the probability of getting a sum of 7 when two dice are thrown?
When two dice are thrown, total outcomes = 36. Sum of 7 can be obtained in 6 ways: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). P(sum = 7) = 6/36 = 1/6. Sum 7 is the most likely sum when two dice are thrown.
Is Exercise 14.2 deleted from CBSE 2026-27 syllabus?
Yes, Exercise 14.2 (5 questions on compound probability) has been removed from CBSE 2026-27. Only Exercise 14.1 (25 questions) is in the syllabus. State board students may still need Exercise 14.2.
What type of probability questions are most commonly asked in CBSE board exams?
The most common types are: (1) Drawing a card from a deck of 52 (or modified deck) - asked almost every year as a 3-mark question. (2) Balls drawn from a bag with complement. (3) Die thrown once or twice. (4) Numbered tokens with divisibility, primes, or perfect squares. (5) MCQ on basic probability concepts.