NCERT Solutions Class 10 Maths Chapter 15 Probability Exercise 15.2

NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.2 help students understand and solve questions based on probability of events and complementary events. This exercise strengthens conceptual clarity and improves logical thinking, which is essential for scoring well in board exams.

Prepared according to the latest CBSE Class 10 Maths syllabus, Exercise 15.2 focuses on applying the probability formula in slightly advanced situations, including finding the probability of “not happening” of an event. Regular practice of this exercise helps students improve accuracy and avoid common mistakes in calculations.

NCERT Solutions Class 10 Maths Chapter 15 Probability Exercise 15.2

Probability

Medium

Q.

If P(E) = 0.05, what is the probability of ‘not E’?

View Solution

Probability

Medium

Q.

A 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?

View Solution

Probability

Medium

Q.

A 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?

View Solution

Probability

Medium

Q.

A 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?

View Solution

Probability

Medium

Q.

A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 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 50 p coin? (ii) will not be a ₹ 5 coin?

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Probability

Medium

Q.

A 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 (see the following figure), 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?

View Solution

Probability

Medium

Q.

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) a red face card
(iv)  the jack of hearts
(v)   a spade
(vi)  the queen of diamond

View Solution

Probability

Medium

Q.

Five 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?

View Solution

Probability

Medium

Q.

12 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.

View Solution

Probability

Medium

Q.

A 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.

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Probability

Medium

Q.

Suppose you drop a die at random on the rectangular region shown in the following figure. What is the probability that it will land inside the circle with diameter 1m?

View Solution

Probability

Medium

Q.

A 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.

View Solution

Probability

Medium

Q.

A die is thrown twice. What is the probability that
(i)  5 will not come up either time?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

View Solution

Probability

Medium

Q.

A 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 in two throws 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?

View Solution

The solutions are explained in a clear, step-by-step manner so students can confidently identify outcomes, apply formulas correctly, and present answers properly in exams.

Q1. Complete the following statements:

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

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

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 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.

$$0 \le P(E) \le 1$$

0≤P(E)≤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 (battery, engine, 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 (if answered by guessing), because the two outcomes are possible with equal chance.

(iv) A baby is born. It is a boy or a girl.
Solution: Equally likely (in basic probability), 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: Tossing a coin is fair because it has two equally likely outcomes: Head or Tail.
So, each team gets an equal chance of winning the toss.

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

(A)

$\frac{2}{3}$

32​ (B) -1.5 (C) 15% (D) 0.7

Solution: Probability always lies between 0 and 1.
So -1.5 cannot be a probability because probability cannot be negative.
✅ Correct option: (B)

Q5. If

$P(E)=0.05$

$P(E)=0.05$, what is the probability of ‘not E’?

Solution:

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

P(E)+P(not E)=1

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

0.05+P(not E)=1

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

P(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.
(ii) Only lemon candies are present, so probability = 1.

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

Solution:
Let

$E$

E = “2 students have the same birthday”
Then

$\text{not }E$

not E = “2 students do not have the same birthday”

Given:

$$P(\text{not }E)=0.992$$

P(not E)=0.992

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

P(E)=1−P(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 the ball drawn is:

(i) red
(ii) not red

Solution:
Total balls

$=3+5=8$

=3+5=8

(i)

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

P(red)=83​

(ii) Not red = black balls = 5

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

P(not red)=85​

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

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

Solution:
Total marbles

$=5+8+4=17$

=5+8+4=17

(i)

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

P(red)=175​

(ii)

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

P(white)=178​

(iii) Not green = red or white =

$17-4=13$

17−4=13

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

P(not green)=1713​

Q10. A piggy bank has 100 (50p), 50 (Re 1), 20 (Rs 2), 10 (Rs 5) coins. 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)

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

P(50p)=180100​=95​

(ii)

$$P(\text{Rs 5})=\frac{10}{180}=\frac{1}{18}$$

P(Rs 5)=18010​=181​

$$P(\text{not Rs 5})=1-\frac{1}{18}=\frac{17}{18}$$

P(not Rs 5)=1−181​=1817​

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

Solution:
Total fish

$=5+8=13$

=5+8=13

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

P(male)=135​

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)

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

P(8)=81​

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

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

P(odd)=84​=21​

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

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

P(>2)=86​=43​

(iv) Less than 9: all 1 to 8 (8 numbers)

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

P(<9)=88​=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: 2,3,5 (3 numbers)

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

P(prime)=63​=21​

(ii) Between 2 and 6: 3,4,5 (3 numbers)

$$P=\frac{3}{6}=\frac{1}{2}$$

P=63​=21​

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

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

P(odd)=63​=21​

Q14. One card drawn from a well-shuffled deck of 52 cards. Find 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

Solution: Total cards = 52
Face cards = J, Q, K in each suit ⇒

$3\times4=12$

3×4=12

(i) Red kings: King of hearts, King of diamonds = 2

$$P=\frac{2}{52}=\frac{1}{26}$$

P=522​=261​

(ii) Face cards = 12

$$P=\frac{12}{52}=\frac{3}{13}$$

P=5212​=133​

(iii) Red face cards: (hearts 3) + (diamonds 3) = 6

$$P=\frac{6}{52}=\frac{3}{26}$$

P=526​=263​

(iv) Jack of hearts = 1

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

P=521​

(v) Spades = 13

$$P=\frac{13}{52}=\frac{1}{4}$$

P=5213​=41​

(vi) Queen of diamonds = 1

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

P=521​

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

(i) Probability the 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)

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

P(Q)=51​

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

(a)

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

P(A)=41​

(b) Queen is not present now

$$P(Q)=0$$

P(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

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

P(good)=144132​=1211​

Q17.

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

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

(i)

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

P(defective)=204​=51​

(ii) One good bulb removed ⇒ remaining bulbs = 19, remaining good = 15

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

P(not defective)=1915​

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 = 90

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

$$P=\frac{81}{90}=\frac{9}{10}$$

P=9081​=109​

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

$$P=\frac{9}{90}=\frac{1}{10}$$

P=909​=101​

(iii) Multiples of 5 from 1 to 90 = 18

$$P=\frac{18}{90}=\frac{1}{5}$$

P=9018​=51​

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 twice, D appears once.

(i)

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

P(A)=62​=31​

(ii)

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

P(D)=61​

Q20. A die is dropped randomly on a rectangle 3 m × 2 m. Find probability it lands inside a circle of diameter 1 m.

Solution:
Area of rectangle

$=3\times2=6$

=3×2=6
Radius of circle

$= \frac{1}{2}$

=21​
Area of circle

$=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}$

=π(21​)2=4π​

$$P=\frac{\text{area of circle}}{\text{area of rectangle}} =\frac{\pi/4}{6}=\frac{\pi}{24}$$

P=area of rectanglearea of circle​=6π/4​=24π​

✅ Answer:

$\frac{\pi}{24}$

24π​

Q21. Lot has 144 ball pens; 20 defective, others good. Find probability:

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

Solution:
Good pens

$=144-20=124$

=144−20=124

(i)

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

P(buy)=144124​=3631​

(ii)

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

P(not buy)=14420​=365​

Q22.

(i) Complete the table for sum of 2 dice:

Sum: 2 3 4 5 6 7 8 9 10 11 12
Probability:

$$\frac{1}{36},\frac{2}{36},\frac{3}{36},\frac{4}{36},\frac{5}{36},\frac{6}{36}, \frac{5}{36},\frac{4}{36},\frac{3}{36},\frac{2}{36},\frac{1}{36}$$

361​,362​,363​,364​,365​,366​,365​,364​,363​,362​,361​

(ii) Student says each outcome has probability

$1/11$

1/11.
Solution: Not correct, because sums are not equally likely (7 has more combinations than 2, etc.).

Q23. Toss a coin 3 times. Hanif wins if all same (HHH or TTT). Find probability Hanif loses.

Solution:
Total outcomes

$=2^3=8$

=23=8
Winning outcomes = 2

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

P(win)=82​=41​

$$P(\text{lose})=1-\frac{1}{4}=\frac{3}{4}$$

P(lose)=1−41​=43​

Q24. A die is thrown twice. Find probability:

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

Solution:

$$P(\text{no 5 in one throw})=\frac{5}{6}$$

P(no 5 in one throw)=65​

(i)

$$P(\text{no 5 in both})=\left(\frac{5}{6}\right)^2=\frac{25}{36}$$

P(no 5 in both)=(65​)2=3625​

(ii)

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

P(at least one 5)=1−3625​=3611​

Q25. Check which arguments are correct:

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

$$P(HH)=\frac{1}{4},\quad P(TT)=\frac{1}{4},\quad P(\text{one of each})=\frac{2}{4}=\frac{1}{2}$$

P(HH)=41​,P(TT)=41​,P(one of each)=42​=21​

So not

$1/3$

1/3.

(ii) One die thrown → odd or even → probability odd = 1/2.
Solution: Correct.
Odd: 1,3,5 (3 outcomes)

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

P(odd)=63​=21​

FAQs – Class 10 Maths Chapter 15 Exercise 15.2

Q1. What is the focus of Exercise 15.2?
Exercise 15.2 focuses on calculating probability of events and understanding the concept of complementary events.

Q2. What is a complementary event?
If E is an event, then the probability of its complement (not E) is:
P(not E) = 1 – P(E)

Q3. Why is Exercise 15.2 important for board exams?
Questions based on complementary probability are commonly asked in CBSE board exams and are scoring when solved carefully.

Q4. How can students prepare effectively for Exercise 15.2?
Students should practice identifying favourable outcomes correctly, revise the complement formula, and solve multiple probability-based questions for better confidence.