Master Class 10 Maths Ch Probability

Probability is an important topic in Class 10 Mathematics that deals with the chance or likelihood of an event occurring. In the NCERT textbook, Chapter 15 introduces students to the basic concepts of theoretical probability through real-life examples and simple experiments like tossing a coin or rolling a die. This chapter helps students develop logical thinking and decision-making skills based on mathematical reasoning. Our NCERT Solutions Class 10 Maths Ch Probability provide step-by-step answers to all the questions in the textbook, helping students understand the concepts clearly and prepare effectively for exams.

Class 10 Maths Ch Probability- Textbook Ex 14.1 (CBSE) 15.1 (SEBA)

Question 1.
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 ………..

Solution:
(i) Probability of an event E + Probability of the event ‘not E’ = 1
(ii) The probability of an event that cannot happen is 0. Such an event is called impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called sure event 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.

Question 2.
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.

Solution:
(i) A driver attempts to start a car. The car starts or does not start.
Answer: Not equally likely outcomes
Explanation: The probability of the car starting or not starting depends on many factors like the condition of the car, weather, battery, etc. These outcomes do not have equal chances of occurring.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Answer: Not equally likely outcomes
Explanation: The chances of success or failure depend on the player’s skill, distance from the basket, and other factors. The outcomes are not equally likely.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.
Answer: Equally likely outcomes
Explanation: If the person answering has no prior knowledge and guesses randomly, both outcomes (true or false) are equally likely, each with a probability of 0.5.

(iv) A baby is born. It is a boy or a girl.
Answer: Equally likely outcomes
Explanation: Generally, the probability of a baby being a boy or a girl is considered equal (about 50%), so the outcomes are equally likely.

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the bail at the beginning of a football game?

Solution:
Tossing a coin is considered a fair way of deciding which team should get the ball at the beginning of a football game because:

Two Equally Likely Outcomes:
A fair coin has two sides — heads and tails — and each side has an equal probability of ½ (or 50%) of showing up.
No Bias:
The coin toss is not influenced by any team’s skill, strength, or strategy. It is a random process, free from human control or judgment.
Transparency:
The process is simple, visible to both teams and referees, and easy to understand. Everyone can see the result and accept it.
Tradition and Universality:
Coin tossing is a widely accepted and time-tested method in sports to make neutral decisions.

Conclusion:

Because the outcomes are equally likely and unbiased, tossing a coin ensures fairness in deciding which team starts the game.

Question 4.
Which of the following cannot be the probability of an event?
(A) 23
(B) -1.5
(C) 15%
(D) 0.7
Solution: Answer: (A) 23 and (B) -1.5

Explanation:
The probability of any event lies between 0 and 1.
So, only values between 0 and 1 (inclusive) are valid probabilities.

(A) 23 is greater than 1, so it is not a valid probability.
(B) -1.5 is less than 0, so it is not a valid probability.
(C) 15% = 15/100 = 0.15, which lies between 0 and 1.
(D) 0.7 is already between 0 and 1.

Therefore, 23 and -1.5 cannot be probabilities of an event.

Ex 15.1 Class 10 Maths Question 5.
If P (E) = 0.05, what is the probability of ‘not E’?
Solution:
If P(E) = 0.05, then the probability of ‘not E’ is:

P(not E) = 1 − P(E)
= 1 − 0.05
= 0.95

Question 6.
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?

Solution:
(i) Probability of taking out an orange flavoured candy:
Since the bag has no orange flavoured candies,
P(orange flavoured candy) = 0

(ii) Probability of taking out a lemon flavoured candy:
Since all candies are lemon flavoured,
P(lemon flavoured candy) = 1

Question 7.
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?

Solution:
Given:
The probability that 2 students do not have the same birthday = 0.992

We need to find the probability that 2 students have the same birthday.

We know:
Probability of event E + Probability of event ‘not E’ = 1

Let event E = “2 students have the same birthday”
Then,
P(E) = 1 − P(not E)
= 1 − 0.992
= 0.008

Question 8.
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?

Solution: A bag contains 3 red balls and 5 black balls.
Total number of balls = 3 + 5 = 8

(i) Probability that the ball drawn is red:
Number of red balls = 3
Probability = Number of red balls / Total number of balls
= 3 / 8

(ii) Probability that the ball drawn is not red:
Number of non-red (black) balls = 5
Probability = Number of black balls / Total number of balls
= 5 / 8

Question 9.
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?

Solution:
Given:

Red marbles = 5
White marbles = 8
Green marbles = 4
Total marbles = 5 + 8 + 4 = 17

(i) Probability that the marble is red:
P(red) = 5/17

(ii) Probability that the marble is white:
P(white) = 8/17

(iii) Probability that the marble is not green:
P(not green) = 13/17 (since there are 13 marbles that are not green, i.e., 5 red + 8 white)

Question 10.
A piggy bank contains hundred 50 p 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?
Solution:
Given:

Number of 50 p coins = 100
Number of ₹ 1 coins = 50
Number of ₹ 2 coins = 20
Number of ₹ 5 coins = 10
Total number of coins = 180

(i) Probability that the coin will be a 50 p coin:
P(50 p coin) = 100/180 = 5/9

(ii) Probability that the coin will not be a ₹ 5 coin:
Not ₹ 5 coin = 100 + 50 + 20 = 170
P(not ₹ 5 coin) = 170/180 = 17/18

Question 11.
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 figure). What is the probability that the fish taken out is a male fish?

Solution:
Given:

Number of male fish = 5
Number of female fish = 8
Total number of fish = 13

Probability that the fish taken out is a male fish:
P(male fish) = 5/13

Question 12.
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 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?

Solution:
Given:
The game consists of spinning an arrow that points to one of the numbers 1, 2, 3, 4, 5, 6, 7, or 8. The total number of outcomes is 8.

Total number of possible outcomes = 8

(i) Probability that it will point at 8:
P(8) = 1/8

(ii) Probability that it will point at an odd number:
P(odd number) = 4/8 = 1/2

(iii) Probability that it will point at a number greater than 2:
P(number > 2) = 6/8 = 3/4

(iv) Probability that it will point at a number less than 9:
P(number < 9) = 8/8 = 1

Question 13.
A die is thrown once. Find the probability of getting
(i) a prime number
(ii) a number lying between 2 and 6
(ill) an odd number

Solution:
Given:
A standard die has six faces, numbered 1, 2, 3, 4, 5, and 6.
Total number of possible outcomes = 6

(i) Probability of getting a prime number:
Prime numbers between 1 and 6 are 2, 3, and 5.
P(prime number) = 3/6 = 1/2

(ii) Probability of getting a number lying between 2 and 6:
The numbers lying between 2 and 6 are 3, 4, and 5.
P(number between 2 and 6) = 3/6 = 1/2

(iii) Probability of getting an odd number:
The odd numbers between 1 and 6 are 1, 3, and 5.
P(odd number) = 3/6 = 1/2

Question 14.
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 diamonds
Solution:
Given:
A standard deck of 52 cards consists of 4 suits: Spades, Hearts, Diamonds, and Clubs.
Each suit contains 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.

(i) Probability of getting a king of red colour:
There are 2 red kings (King of Hearts and King of Diamonds).
P(king of red colour) = 2/52 = 1/26

(ii) Probability of getting a face card:
Face cards are the Jack, Queen, and King of each suit. There are 12 face cards in total.
P(face card) = 12/52 = 3/13

(iii) Probability of getting a red face card:
There are 6 red face cards (Jack, Queen, King of Hearts and Diamonds).
P(red face card) = 6/52 = 3/26

(iv) Probability of getting the jack of hearts:
There is only 1 Jack of Hearts.
P(jack of hearts) = 1/52

(v) Probability of getting a spade:
There are 13 spades in a deck.
P(spade) = 13/52 = 1/4

(vi) Probability of getting the queen of diamonds:
There is only 1 Queen of Diamonds.
P(queen of diamonds) = 1/52

Question 15.
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?
Solution:
Given:

Five cards: the ten, jack, queen, king, and ace of diamonds.
These cards are well shuffled and placed face downwards.

Total number of possible outcomes = 5 (since there are 5 cards).

(i) Probability that the card is the queen:

There is 1 queen in the deck of 5 cards.

The probability is:
P(queen) = 1/5

(ii) If the queen is drawn and put aside:

After the queen is put aside, 4 cards remain in the deck.

(a) Probability that the second card picked up is an ace:

There is 1 ace remaining out of the 4 cards.

The probability is:
P(ace) = 1/4

(b) Probability that the second card picked up is a queen:

Since the queen has already been put aside, there are no queens left in the remaining 4 cards.

The probability is:
P(queen) = 0/4 = 0

Question 16.
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.
Solution:
Given:

Number of defective pens = 12
Number of good pens = 132
Total number of pens = 12 + 132 = 144

Probability that the pen taken out is a good one:
P(good pen) = 132/144 = 11/12

Question 17.
(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?
Solution:
(i) Probability that the bulb drawn is defective:

Given:

Total number of bulbs = 20
Number of defective bulbs = 4

P(defective bulb) = 4/20 = 1/5

(ii) Probability that the second bulb drawn is not defective:

After the first bulb is drawn and found not defective, there are 19 bulbs remaining. The number of non-defective bulbs is 16.

P(not defective) = 16/19

Question 18.
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.
Solution:
Given:
The box contains 90 discs numbered from 1 to 90.
Total number of discs = 90

(i) Probability that the disc bears a two-digit number:

The two-digit numbers range from 10 to 90.
The number of two-digit numbers = 81.
P(two-digit number) = 81/90 = 9/10

(ii) Probability that the disc bears a perfect square number:

The perfect square numbers between 1 and 90 are: 1, 4, 9, 16, 25, 36, 49, 64, 81.
The number of perfect square numbers = 9.
P(perfect square number) = 9/90 = 1/10

(iii) Probability that the disc bears a number divisible by 5:

The numbers divisible by 5 between 1 and 90 are: 5, 10, 15, 20, …, 90.
The number of terms = 18.
P(divisible by 5) = 18/90 = 1/5

Ex 15.1 Class 10 Maths Question 19.
A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting
(i) A?
(ii) D?
Solution:
Given:
A die has six faces, each showing a letter (assuming the letters are A, B, C, D, E, and F).

Total number of possible outcomes = 6

(i) Probability of getting A:
There is 1 face with the letter A on it.
P(A) = 1/6

(ii) Probability of getting D:
There is 1 face with the letter D on it.
P(D) = 1/6

Question 20.
Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?

Solution:
Given:

The dimensions of the rectangular region are:
Length = 3 meters,
Breadth = 2 meters.
A circle with a diameter of 1 meter, so the radius is 0.5 meters.

The radius of the circle is 0.5 meters.
The area of the circle is:
Area(circle) = π × (0.5)^2 = π/4 square meters.

The area of the rectangle is:
Area(rectangle) = Length × Breadth = 3 × 2 = 6 square meters.

The probability of the die landing inside the circle is:
P(inside the circle) = Area(circle) / Area(rectangle) = (π/4) / 6 = π/24.

Question 21.
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?
Solution:
Given:

Total number of ball pens = 144
Number of defective pens = 20
Number of good pens = 144 – 20 = 124

(i) Probability that Nuri will buy the pen:

Nuri will buy a pen if it is good.
P(buy) = 124/144 = 31/36

(ii) Probability that Nuri will not buy the pen:

Nuri will not buy a pen if it is defective.
P(not buy) = 20/144 = 5/36

Question 22.
Two dice, one blue and one grey, are thrown at the same time. Now:

(i) Complete the following table:

Event (Sum on 2 dice)	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) A student argues that there are 11 possible outcomes (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12) and therefore each has a probability of 1/11. Do you agree with this argument? Justify your answer.

Solution :Given:
Two dice, one blue and one grey, are thrown at the same time. The task is to complete the probability table for the sum of the numbers on the two dice.

(i) To complete the table:

The probability of a particular sum depends on how many possible combinations of numbers on the two dice can result in that sum.

Sum of 2: Only 1 combination, (1, 1).
Probability = 1/36
Sum of 3: 2 combinations, (1, 2) and (2, 1).
Probability = 2/36 = 1/18
Sum of 4: 3 combinations, (1, 3), (2, 2), (3, 1).
Probability = 3/36 = 1/12
Sum of 5: 4 combinations, (1, 4), (2, 3), (3, 2), (4, 1).
Probability = 4/36 = 1/9
Sum of 6: 5 combinations, (1, 5), (2, 4), (3, 3), (4, 2), (5, 1).
Probability = 5/36
Sum of 7: 6 combinations, (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).
Probability = 6/36 = 1/6
Sum of 8: 5 combinations, (2, 6), (3, 5), (4, 4), (5, 3), (6, 2).
Probability = 5/36
Sum of 9: 4 combinations, (3, 6), (4, 5), (5, 4), (6, 3).
Probability = 4/36 = 1/9
Sum of 10: 3 combinations, (4, 6), (5, 5), (6, 4).
Probability = 3/36 = 1/12
Sum of 11: 2 combinations, (5, 6) and (6, 5).
Probability = 2/36 = 1/18
Sum of 12: 1 combination, (6, 6).
Probability = 1/36

Completed Table:

Event (Sum on 2 dice)	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’s Argument:

The student argues that there are 11 possible outcomes (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12) and therefore each has a probability of 1/11.

Answer:

This argument is incorrect. The total number of possible outcomes for rolling two dice is 36, not 11.

The probability for each sum varies depending on how many combinations of dice rolls result in that sum. For example, there are 6 possible combinations that give a sum of 7, so the probability of getting a sum of 7 is 6/36 = 1/6, not 1/11.

The probabilities are not equal for all sums, as shown in the completed table.

Question 23.
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.
Solution:
Given:

A one rupee coin is tossed 3 times.
Hanif wins if all tosses give the same result, i.e., either three heads (HHH) or three tails (TTT).
Hanif loses the game otherwise.

Total possible outcomes:
Since the coin is tossed 3 times, each toss has 2 possible outcomes: Head (H) or Tail (T).
The total number of possible outcomes for 3 tosses is:
2 × 2 × 2 = 8

So, the total number of possible outcomes is 8.

Winning outcomes:
Hanif wins if the result is either three heads (HHH) or three tails (TTT).
So, there are 2 winning outcomes:

Losing outcomes:
Hanif loses if the result is anything other than three heads or three tails.
Thus, the remaining outcomes are:

These are 6 losing outcomes.

Probability of losing:
The probability of Hanif losing the game is the ratio of the number of losing outcomes to the total number of possible outcomes:
P(lose) = 6/8 = 3/4

Question 24.
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.]
Solution:
Given:

A die is thrown twice.
The total number of possible outcomes for each throw of the die is 6 (since the die has 6 faces).

So, for two throws, the total number of possible outcomes is:
6 × 6 = 36

(i) Probability that 5 will not come up either time:

The probability that 5 will not come up on a single throw is:
P(not 5) = 5/6
This is because there are 5 outcomes (1, 2, 3, 4, 6) where a 5 does not appear.

The probability that 5 will not come up either time (on both throws) is:
P(not 5 on both throws) = 5/6 × 5/6 = 25/36

(ii) Probability that 5 will come up at least once:

The event “5 comes up at least once” is the complement of the event “5 does not come up either time.”
Thus, the probability of 5 coming up at least once is:
P(at least one 5) = 1 – P(not 5 on both throws) = 1 – 25/36 = 11/36

Question 25.
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 13.
(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 12.
Solution:
Answer:
This argument is incorrect. When two coins are tossed, there are four possible outcomes:

Head on the first coin and head on the second coin (HH)
Tail on the first coin and tail on the second coin (TT)
Head on the first coin and tail on the second coin (HT)
Tail on the first coin and head on the second coin (TH)

So, there are 4 possible outcomes, not 3. The correct probability for each of these outcomes is 1/4, not 1/3. Therefore, the argument is not correct.

(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.

Answer:
This argument is correct. When a die is thrown, the outcomes are the numbers 1, 2, 3, 4, 5, and 6.

Odd numbers: 1, 3, 5 (3 outcomes)
Even numbers: 2, 4, 6 (3 outcomes)

There are 3 odd numbers and 3 even numbers, so the probability of getting an odd number is:
P(odd number) = 3/6 = 1/2
Therefore, the argument is correct.

Class 10 Maths Ch Probability – Ex 14.1 (CBSE) 15.1 (SEBA)

Class 10 Maths Ch Probability- Textbook Ex 14.1 (CBSE) 15.1 (SEBA)

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