NCERT Solutions Class 10 Maths Chapter 15 Probability Exercise 15.1

NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 help students understand the basic concepts of Probability in a clear and exam-oriented manner. This exercise introduces students to the fundamental idea of probability and how to calculate it in simple situations.

Prepared according to the latest CBSE Class 10 Maths syllabus, Exercise 15.1 focuses on finding the probability of an event using the classical definition. Regular practice of this exercise helps students strengthen conceptual clarity and improve accuracy in board examinations.

NCERT Solutions Class 10 Maths Chapter 15 Probability Exercise 15.1

Probability
Medium
Q.
Complete 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_____.
Probability
Medium
Q.
Which 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.
Probability
Medium
Q.
Why 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?
Probability
Medium
Q.
It 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?
Probability
Medium
Q.
Gopi 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 (see the following figure). What is the probability that the fish taken out is a male fish?
Probability
Medium
Q.
A 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.
Probability
Medium
Q.
(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?
Probability
Medium
Q.
A child has a die whose six faces show the letters as given below:
A B C D E F
The die is thrown once. What is the probability of getting (i) A?  (ii) D?
Probability
Medium
Q.
A lot consists of 144 ball pens of which 20 are defective and the others 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?
Probability
Medium
Q.
Two dice, one blue and one grey, are thrown at the same time.
(i) Complete the following table:
Event:
‘Sum of two dice’
2
3
4
5
6
7
8
9
10
11
12
Probability
1 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaIXaaabaGaaG4maiaaiAdaaaaaaa@3D0A@
 
 
 
 
 
 5 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaI1aaabaGaaG4maiaaiAdaaaaaaa@3D0E@
 
 
 
 1 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaIXaaabaGaaG4maiaaiAdaaaaaaa@3D0A
(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.
Probability
Medium
Q.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i)  If two coins are tossed simultaneously there are three possible outcomes — two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes — an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.
Probability
Medium
Q.
Two 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?
Probability
Medium
Q.
A 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.
Probability
Medium
Q.
A 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.
Probability
Medium
Q.
A 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.

NCERT Solutions Class 10 Maths Chapter 15 Probability Exercise 15.1

The solutions are explained step-by-step so students can clearly understand how to identify favorable outcomes, total outcomes, and apply the probability formula correctly.

Q1. Complete the following statements

(i) Probability of an event E + Probability of the event ‘not E’ = 1

P(E)+P(not E)=1P(E)+P(\text{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 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.

0≤P(E)≤10 \le P(E) \le 1

Q2. Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.
Solution: Not equally likely, because it depends on many factors (car condition, battery, etc.).

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Solution: Not equally likely, because it depends on the player’s skill.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.
Solution: Equally likely (assuming random guess), because two outcomes are possible.

(iv) A baby is born. It is a boy or a girl.
Solution: Equally likely (in basic probability assumption), because both outcomes are considered equally possible.

Q3. Why 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?

Solution:
Because tossing a coin has two equally likely outcomes: Head (H) or Tail (T).
So, both teams get equal chance (1/2 each), hence it is fair.

Q4. Which of the following cannot be the probability of an event?

(A)

23\frac{2}{3}

(B) -1.5 (C) 15% (D) 0.7

Solution:
Probability always lies between 0 and 1.

0≤P(E)≤10 \le P(E) \le 1

So (B) -1.5 cannot be a probability because it is negative.

Q5. If

P(E)=0.05P(E)=0.05

 

, what is the probability of ‘not E’?

Solution:

P(E)+P(not E)=1P(E)+P(\text{not }E)=1

0.05+P(not E)=10.05 + P(\text{not }E)=1

P(not E)=1−0.05=0.95P(\text{not }E)=1-0.05=0.95

✅ Answer: 0.95

Q6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking. Find probability that she takes out:

(i) an orange flavoured candy
(ii) a lemon flavoured candy

Solution:
(i) Orange candy is not present, so probability = 0 (impossible event).
(ii) Only lemon candies are present, so probability = 1 (sure event).

Q7. In a group of 3 students, probability of 2 students not having the same birthday is 0.992. Find probability that 2 students have the same birthday.

Solution:
Let event

EE

: “2 students have the same birthday”
Then

not E\text{not }E

: “2 students do not have the same birthday”

Given:

P(not E)=0.992P(\text{not }E)=0.992

P(E)=1−P(not E)=1−0.992=0.008P(E)=1-P(\text{not }E)=1-0.992=0.008

✅ Answer: 0.008

Q8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random. Find probability that it is:

(i) red
(ii) not red

Solution:
Total balls

=3+5=8=3+5=8

(i)

P(red)=38P(\text{red})=\frac{3}{8}

(ii) Not red means black: 5 balls

P(not red)=58P(\text{not red})=\frac{5}{8}

(or

1−38=581-\frac{3}{8}=\frac{5}{8}

)

Q9. A box contains 5 red, 8 white and 4 green marbles. One marble is taken out. Find probability that it is:

(i) red
(ii) white
(iii) not green

Solution:
Total marbles

=5+8+4=17=5+8+4=17

(i)

P(red)=517P(\text{red})=\frac{5}{17}

(ii)

P(white)=817P(\text{white})=\frac{8}{17}

(iii) Not green means red or white =

17−4=1317-4=13

P(not green)=1317P(\text{not green})=\frac{13}{17}

Q10. Piggy bank has 100 (50p), 50 (Re1), 20 (Rs2), 10 (Rs5). One coin falls out randomly. Find probability that coin:

(i) is a 50p coin
(ii) is not a Rs 5 coin

Solution:
Total coins

=100+50+20+10=180=100+50+20+10=180

(i)

P(50p)=100180=59P(50p)=\frac{100}{180}=\frac{5}{9}

(ii) Rs5 coins = 10

P(Rs5)=10180=118P(\text{Rs5})=\frac{10}{180}=\frac{1}{18}

P(not Rs5)=1−118=1718P(\text{not Rs5})=1-\frac{1}{18}=\frac{17}{18}

Q11. Tank has 5 male fish and 8 female fish. One fish is taken out randomly. Find probability it is a male fish.

Solution:
Total fish

=5+8=13=5+8=13

P(male)=513P(\text{male})=\frac{5}{13}

Q12. Spinning arrow points to one of 1 to 8 (equally likely). Find probability it points at:

(i) 8
(ii) an odd number
(iii) a number greater than 2
(iv) a number less than 9

Solution: Total outcomes = 8

(i) Only 8 is favourable

P(8)=18P(8)=\frac{1}{8}

(ii) Odd numbers: 1,3,5,7 = 4

P(odd)=48=12P(\text{odd})=\frac{4}{8}=\frac{1}{2}

(iii) Greater than 2: 3,4,5,6,7,8 = 6

P(>2)=68=34P(>2)=\frac{6}{8}=\frac{3}{4}

(iv) Less than 9: all 1–8 = 8

P(<9)=88=1P(<9)=\frac{8}{8}=1

Q13. A die is thrown once. Find probability of getting:

(i) a prime number
(ii) a number lying between 2 and 6
(iii) an odd number

Solution: Total outcomes = 6 (1–6)

(i) Prime numbers on die: 2,3,5 = 3

P(prime)=36=12P(\text{prime})=\frac{3}{6}=\frac{1}{2}

(ii) Numbers between 2 and 6: 3,4,5 = 3

P(2 and 6 between)=36=12P(2\text{ and }6\text{ between})=\frac{3}{6}=\frac{1}{2}

(iii) Odd numbers: 1,3,5 = 3

P(odd)=36=12P(\text{odd})=\frac{3}{6}=\frac{1}{2}

Q14. One card drawn from well-shuffled deck (52 cards). Find probability of getting:

(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) jack of hearts
(v) a spade
(vi) queen of diamonds

Solution: Total cards = 52
Hearts = 13, Diamonds = 13 (red)
Face cards = J,Q,K in each suit =

3×4=123\times4=12

(i) Red kings = King of hearts + King of diamonds = 2

P=252=126P=\frac{2}{52}=\frac{1}{26}

(ii) Face cards = 12

P=1252=313P=\frac{12}{52}=\frac{3}{13}

(iii) Red face cards = (J,Q,K of hearts) 3 + (J,Q,K of diamonds) 3 = 6

P=652=326P=\frac{6}{52}=\frac{3}{26}

(iv) Jack of hearts = 1 card

P=152P=\frac{1}{52}

(v) Spades = 13

P=1352=14P=\frac{13}{52}=\frac{1}{4}

(vi) Queen of diamonds = 1 card

P=152P=\frac{1}{52}

Q15. Five cards (10, J, Q, K, A of diamonds) are shuffled face down. One card picked.

(i) Probability card is queen
(ii) If queen is drawn and put aside, probability second card is: (a) an ace (b) a queen

Solution: Total cards = 5

(i) One queen

P(Q)=15P(Q)=\frac{1}{5}

(ii) Queen removed ⇒ remaining cards = 4 (10, J, K, A)

(a) One ace

P(A)=14P(A)=\frac{1}{4}

(b) Queen is not there now

P(Q)=0P(Q)=0

Q16. 12 defective pens mixed with 132 good pens. One pen taken randomly. Find probability it is good.

Solution:
Total pens

=12+132=144=12+132=144

P(good)=132144=1112P(\text{good})=\frac{132}{144}=\frac{11}{12}

Q17.

(i) Lot of 20 bulbs has 4 defective. One drawn. Probability defective?
(ii) If drawn bulb is not defective and not replaced, probability next bulb is not defective?

Solution:
Total bulbs = 20, defective = 4, good = 16

(i)

P(defective)=420=15P(\text{defective})=\frac{4}{20}=\frac{1}{5}

(ii) First bulb is good and removed ⇒ remaining bulbs = 19, remaining good = 15

P(not defective)=1519P(\text{not defective})=\frac{15}{19}

Q18. Box has 90 discs numbered 1 to 90. One drawn. Find probability it bears:

(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5

Solution: Total discs = 90

(i) Two-digit numbers from 10 to 90 = 81

P=8190=910P=\frac{81}{90}=\frac{9}{10}

(ii) Perfect squares ≤ 90: 1,4,9,16,25,36,49,64,81 = 9

P=990=110P=\frac{9}{90}=\frac{1}{10}

(iii) Divisible by 5: 5,10,15,...,90 (count = 18)

P=1890=15P=\frac{18}{90}=\frac{1}{5}

Q19. Die faces: A, B, C, D, E, A. Thrown once. Find probability of getting:

(i) A
(ii) D

Solution: Total outcomes = 6
A appears 2 times, D appears 1 time

(i)

P(A)=26=13P(A)=\frac{2}{6}=\frac{1}{3}

(ii)

P(D)=16P(D)=\frac{1}{6}

Q20. Die dropped randomly on rectangular region (3 m × 2 m) with a circle of diameter 1 m inside. Find probability it lands inside circle.

Solution:
Area of rectangle

=3×2=6 m2=3\times2=6 \,m^2


Circle diameter = 1 m ⇒ radius

r=0.5r=0.5

m
Area of circle

=πr2=π(0.5)2=π4=\pi r^2=\pi(0.5)^2=\frac{\pi}{4}

P=Area of circleArea of rectangle=π/46=π24P=\frac{\text{Area of circle}}{\text{Area of rectangle}} =\frac{\pi/4}{6}=\frac{\pi}{24}

✅ Answer:

π24\frac{\pi}{24}

Q21. Lot has 144 ball pens, 20 defective, rest good. Nuri buys if good. One pen given randomly. Find probability:

(i) She will buy it
(ii) She will not buy it

Solution:
Good pens

=144−20=124=144-20=124

(i)

P(buy)=124144=3136P(\text{buy})=\frac{124}{144}=\frac{31}{36}

(ii)

P(not buy)=20144=536P(\text{not buy})=\frac{20}{144}=\frac{5}{36}

Q22. Sum of 2 dice: complete probability table and judge argument

(i) Complete table:

Sum 2 3 4 5 6 7 8 9 10 11 12
Probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

(ii) Student says each has probability 1/11. Do you agree?
Solution: No. Because sums are not equally likely (7 comes in 6 ways, 2 comes in 1 way, etc.). So probabilities are not equal.

Q23. Tossing a 1-rupee coin 3 times. Hanif wins if all same (HHH or TTT). Find probability Hanif loses.

Solution:
Total outcomes

=23=8=2^3=8


Winning outcomes: HHH, TTT = 2

P(win)=28=14P(\text{win})=\frac{2}{8}=\frac{1}{4}

P(lose)=1−14=34P(\text{lose})=1-\frac{1}{4}=\frac{3}{4}

✅ Answer:

34\frac{3}{4}

Q24. A die thrown twice. Find probability that:

(i) 5 will not come up either time
(ii) 5 will come up at least once

Solution:
On one throw, probability of not getting 5 =

56\frac{5}{6}

(i)

P(no 5 in both)=56×56=2536P(\text{no 5 in both})=\frac{5}{6}\times\frac{5}{6}=\frac{25}{36}

(ii)

P(at least one 5)=1−2536=1136P(\text{at least one 5})=1-\frac{25}{36}=\frac{11}{36}

Q25. Which arguments are correct? Give reasons.

(i) Two coins tossed: outcomes are “two heads, two tails, or one of each”, so each has probability 1/3.
Solution: Incorrect.
Actual outcomes: HH, HT, TH, TT (4 equally likely)

  • P(HH)=14P(HH)=\frac{1}{4}

  • P(TT)=14P(TT)=\frac{1}{4}

  • “one of each” = HT or TH = 2 outcomes

P(one of each)=24=12P(\text{one of each})=\frac{2}{4}=\frac{1}{2}

So probabilities are not 1/3 each.

(ii) One die thrown: outcomes are odd or even, so probability odd = 1/2.
Solution: Correct.
Odd: 1,3,5 (3 outcomes)
Even: 2,4,6 (3 outcomes)

P(odd)=36=12P(\text{odd})=\frac{3}{6}=\frac{1}{2}

FAQs – Class 10 Maths Chapter 15 Exercise 15.1

Q1. What is the focus of Exercise 15.1?
Exercise 15.1 focuses on calculating the probability of simple events using the classical definition of probability.

Q2. What is the basic formula of probability?
Probability of an event (P) =
Number of favourable outcomes / Total number of possible outcomes

Q3. Why is Exercise 15.1 important for board exams?
Probability is a scoring chapter in CBSE Class 10 exams. Questions from Exercise 15.1 are usually straightforward and concept-based.

Q4. How can students prepare effectively for Exercise 15.1?
Students should clearly understand the concept of outcomes, practice coin and dice-based questions, and revise the basic probability formula thoroughly.