NCERT Solutions Class 10 Maths Chapter 14 Probability Exercise 14.1
NCERT Solutions Class 10 Maths Chapter 14 Exercise 14.1 helps students hone their knowledge of an important topic of class 10, Probability. The exercise will begin with a simple introduction to the concept and go on to some important facts derived specifically from real-life situations. This exercise will help in building a strong base for solving complex problems in higher classes of mathematical studies. Here, you can find the complete NCERT Solutions Class 10 Maths Chapter 14 Exercise 14.1 PDF, along with concepts used in the exercise explained in brief.
1.0Download NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1 : Free PDF
2.0Introduction to Probability
Before starting our exploration of the formulas and calculations involved in probability, let’s briefly discuss the meaning and application of probability. Probability is a measurement of the likelihood of a certain event occurring. In other words, the probability is simply the chance of an event happening. Probability is applied to a vast range of disciplines, such as statistics, economics, and science. As per this chapter, we will address two kinds of probability: Experimental (empirical) probability and Theoretical (classical) probability.
3.0Class 10 Maths Chapter 14 Exercise 14.1 Overview: Key Concepts
Now that we know what probability actually means let’s take a quick look at the key concepts used in solving NCERT Solutions of Class 10 Maths Chapter 14 Exercise 14.1:
General Formula of Probability
As mentioned earlier, probability can be divided into two types, so the general formula for each type can be written as:
Experimental Probability:
The experimental probability is calculated from actual data and provides an approximation of the probability that something will happen. However, it can differ every time the experiment is conducted. An example of this probability can be a coin which is tossed 100 times and gets heads 45 times. The probability for an experimental event can be found using the formula:
Theoretical Probability:
Theoretical (or classical) probability is derived through mathematical argument rather than experimental results. It presupposes that every result of an experiment is as likely as every other. Theoretical probability is usually employed when it is impossible to repeat the experiment or when there are obvious assumptions concerning the character of the experiment. This type of probability is calculated as:
Total and Complementary Probability of an Event
In probability theory, knowledge of the total probability and complementary probability of an event assists in making complicated probability calculations easier and in comprehending the connection between various possible results.
Total Probability of an Event:
Total probability is the sum of all the probabilities of the possible outcomes of an experiment. In every experiment, the total probability of all the combined outcomes will always be 1, since one of the possible outcomes will surely occur. For example, in an event, the probability of the first outcome is P(N) and the second one is P(M), then the total probability of this event is:
Complementary Probability of an Event:
The complementary probability is the probability that an event will not happen. That is, if you have an event E, then the complementary event EC is the event in which E does not occur. The complementary probability can be found with the formula:
Here,
- P(E) is the probability of the event occurring,
- P(EC) is the probability of the event not occurring.
Remark: Note that for an event with no possibility at all to happen, the probability of such events will be 0. While the probability of an event with complete chances of happening, the probability will be 1.
To better grasp the above concepts, begin solving the NCERT Solutions Class 10 Maths Chapter 14 Probability Exercise 14.1 and try applying the probability concepts to many real-life scenarios.
4.0NCERT Solutions Class 10 Maths Chapter 14 Probability Exercise 14.1 : Detailed Solutions
1. Complete the following statements :
(i) Probability of an event E + probability of the event 'not E' = _____.
(ii) 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 _____.
Sol. (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 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.
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.
Sol. (i) The car will start if all its parts are working perfectly but if there is some defect in its parts, it will not start. So it is not equally likely.
(ii) The player may shoot or miss the shot. Therefore, the outcomes are not equally likely.
(iii) The outcome in this trial of a true-false question is either true or false, i.e., one out of the two and both have equal chances to happen. Hence, the two outcomes are equally likely.
(iv) In advance it is known that a newly born baby has to be either a boy or a girl. Therefore, the outcomes of either a boy or a girl are equally likely to occur.
3. 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?
Sol. Since on tossing a coin, the outcomes 'head' and 'tail' are equally likely, the result of tossing a coin is completely unpredictable and so it is a fair way.
4. Which of the following cannot be the probability of an event?
(A) 2/3
(B) -1.5
(C) 15%
(D) 0.7
Sol. The probability of an event cannot be less than 0 and greater than 1. Therefore, the correct option is (B) because -1.5 is less than 0.
5. If P(E) = 0.05, what is the probability of 'not E'?
Sol. We have P(E) = 0.05
Therefore, P(not E) = 1 - P(E) = 1 - 0.05 = 0.95
Therefore, P(not E) = 0.95.
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?
Sol. (i) Since there are lemon flavoured candies only in the bag. Taking out an orange flavoured candy is not possible. Probability of taking out an orange flavoured candy = 0. This is an impossible event.
(ii) Also probability of taking out a lemon flavoured candy = 1. This is a sure (certain) event.
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?
Sol. Event E is that two students out of 3 have the same birthday. We are given that P(not E) = 0.992.
Then P(E) = 1 - P(not E) = 1 - 0.992 = 0.008
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?
Sol. Total number of balls = 3 + 5 = 8
Therefore, number of possible outcomes = 8
(i) There are 3 red balls. Therefore, number of favourable outcomes = 3
Therefore, P(red) = (Number of favourable outcomes) / (Number of all possible outcomes) = 3/8
(ii) Probability of the ball drawn which is not red = 1 - P(red) = 1 - 3/8 = (8 - 3) / 8 = ⅝
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?
Sol. Total number of marbles in the box = 5 + 8 + 4 = 17.
Let A1 = Event of getting a red marble,
A2 = Event of getting a white marble and
A3 = Event of getting a green marble.
(i) There are 5 red marbles in the box.
Hence, required probability = P(A1) = P(getting a red marble) = 5/17.
(ii) There are 8 white marbles in the box.
Hence, required probability = P(A2) = P(getting a white marble) = 8/17.
(iii) There are 4 green marbles in the box.
Therefore, P(A3) = P(getting a green marble) = 4/17.
Hence, required probability = P(not getting a green marble) = 1 - P(getting a green marble) = 1 - P(A3) = 1 - 4/17 = 13/17.
10. A piggy bank contains hundred 50 p coins, fifty Rs. 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 50 p coin (ii) will not be a Rs 5 coin?
Sol. Number of: 50 p coins = 100, ₹ 1 coins = 50, ₹ 2 coins = 20, ₹ 5 coins = 10.
Total number of coins = 100 + 50 + 20 + 10 = 180.
Therefore, total possible outcomes = 180.
(i) For a 50 p coin:
Favourable events = 100
Therefore, P(50 p) = 100/180 = 5/9.
(ii) For not a ₹ 5 coin:
Since the number of ₹ 5 coins = 10, the number of 'not ₹ 5' coins = 180 - 10 = 170.
Favourable outcomes = 170
Therefore, P(not 5 rupee coin) = 170/180 = 17/18.
11. Gopi buys 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?
Solution:
Number of male fishes = 5
Number of female fishes = 8
Therefore, Total number of fishes = 8 + 5 = 13
Total number of outcomes = 13
For a male fish:
Number of favourable outcomes = 5
Therefore, P(male fish) = 5/13.
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, 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 ?
Sol. Total numbers marked = 8
Therefore, Total number of possible outcomes = 8
(i) When the pointer points at 8:
Number of favourable outcomes = 1
Therefore, P(8) = (Number of favourable outcomes) / (Total number of possible outcomes) = 1/8
(ii) When the pointer points at an odd number:
Odd numbers are 1, 3, 5, and 7.
Therefore, Number of favourable outcomes = 4
Therefore, P(odd) = (Number of favourable outcomes) / (Total number of possible outcomes) = 4/8 = 1/2
(iii) When the pointer points at a number greater than 2:
The numbers 3, 4, 5, 6, 7, and 8 are greater than 2.
Therefore, Number of numbers greater than 2 = 6
Number of favourable outcomes = 6
Therefore, P(greater than 2) = (Number of favourable outcomes) / (Total number of possible outcomes) = 6/8 = 3/4
(iv) When the pointer points at a number less than 9:
The numbers 1, 2, 3, 4, 5, 6, 7, and 8 are less than 9.
Therefore, Number of numbers less than 9 = 8
Number of favourable outcomes = 8
P(less than 9) = (Number of favourable outcomes) / (Total number of possible outcomes) = 8/8 = 1
13. 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.
Sol. There are 6 outcomes when a die is thrown. These are 1, 2, 3, 4, 5, 6, and all have equally likely chances.
(i) 2, 3, 5 are prime numbers.
Therefore, P(getting a prime) = 3/6 = 1/2
(ii) 3, 4, 5 are the numbers lying between 2 and 6.
Therefore, P(getting a number between 2 and 6) = 3/6 = 1/2
(iii) 1, 3, 5 are the odd numbers.
Therefore, P(getting an odd number) = 3/6 = ½
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
Sol. Number of cards in deck = 52
Therefore, Total number of possible outcomes = 52
(i) Number of red colour kings = 2
(Since King of diamond and heart is red)
Number of favourable outcomes = 2
P(red king) = (Number of favourable outcomes) / (Total number of possible outcomes) = 2/52 = 1/26
(ii) For a face card:
Since 4 kings, 4 queens and 4 jacks are face cards,
Therefore, Number of face cards = 12
Number of favourable outcomes = 12
P(face card) = (Number of favourable outcomes) / (Total number of possible outcomes) = 12/52 = 3/13
(iii) Since, cards of diamond and heart are red,
Therefore, there are 2 kings, 2 queens, 2 jacks i.e., 6 cards are red face cards.
Favourable outcomes = 6
P(red face card) = (Number of favourable outcomes) / (Total number of possible outcomes) = 6/52 = 3/26
(iv) Since, there is only 1 jack of hearts.
Therefore, Number of favourable outcomes = 1
P(jack of hearts) = (Number of favourable outcomes) / (Total number of possible outcomes) = 1/52
(v) There are 13 spades in a pack of 52 cards.
Therefore, Favourable outcomes = 13.
P(spade) = (Number of favourable outcomes) / (Total number of possible outcomes) = 13/52 = 1/4
(vi) Since, there is only one queen of diamond.
Therefore, Number of favourable outcomes = 1
P(queen of diamonds) = (Number of favourable outcomes) / (Total number of possible outcomes) = 1/52
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?
Sol.
(i) There is only one queen out of the five cards.
Therefore, P(getting queen) = 1/5
(ii) When the queen is drawn, four cards are left which are a ten, a jack, a king, and an ace.
(a) P(an ace) = 1/4
(b) P(a queen) = 0/4 = 0
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.
Sol. We have the number of good pens = 132 and the number of defective pens = 12.
Therefore, the total number of pens = 132 + 12 = 144.
Total possible outcomes = 144.
There are 132 good pens.
Therefore, the number of favorable outcomes = 132.
P(good pens) = (Number of favorable outcomes) / (Total number of possible outcomes) = 132/144 = 11/12
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?
Sol.
(i) P(Bulb drawn is defective) = 4/20 = 1/5
(ii) When a good bulb is drawn out from the 20 bulbs, then there are 4 defective and 15 good bulbs left in the remaining 19 bulbs. Now, P(Bulb drawn out of 19 bulbs is not defective) = 15/19
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.
Sol. We have the total number of discs = 90.
Therefore, the total number of possible outcomes = 90.
(i) Since, two-digit numbers are 10, 11, 12, ..., 90.
Therefore, the number of two-digit numbers = 90 - 9 = 81.
Number of favorable outcomes = 81.
Therefore, P(Two-digit number) = (Number of favorable outcomes) / (Total number of possible outcomes) = 81/90 = 9/10
(ii) Perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81.
Therefore, the number of perfect squares = 9.
Number of favorable outcomes = 9.
Therefore, P(Perfect square) = (Number of favorable outcomes) / (Total number of possible outcomes) = 9/90 = 1/10
(iii) Numbers divisible by 5 from 1 to 90 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90.
i.e., There are 18 numbers from (1 to 90) which are divisible by 5.
Therefore, the number of favorable outcomes = 18.
P(Numbers divisible by 5) = (Number of favorable outcomes) / (Total number of possible outcomes) = 18/90 = 1/5
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?
Sol. Since, there are six faces of the given die and these faces are marked with letters
Therefore, Total number of letters = 6
(i) Two faces have the letter A.
Therefore, Number of favourable outcomes = 2
Now, P(letter A) = (No. of favourable outcomes) / (Total number of possible outcomes) = 2/6 = 1/3
(ii) Number of D's = 1
Therefore, Number of possible outcomes = 1
=> P(letter D) = (No. of favourable outcomes) / (Total number of possible outcomes) = 1/6
20. Suppose you drop a die at random on the rectangular region shown in fig. What is the probability that it will land inside the circle with diameter 1 m ?
Solution:
Area of the rectangle = 3 m × 2 m = 6 m²
Area of the circle = π × (1/2)² m² = π/4 m²
Probability (die will land inside the circle) = (π/4) / 6 = π/24
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 it 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:
Total number of ball pens = 144
=> Total number of possible outcomes = 144
(i) Since there are 20 defective pens
=> Number of good pens = 144 - 20 = 124
=> Number of favourable outcomes = 124
=> Probability that she will buy it = 124/144 = 31/36
(ii)Probability that she will not buy it = 1 - [Probability that she will buy it]
= 1 - 31/36
= (36 - 31) / 36
= 5/36.
22. Two dice are thrown at the same time.
(i) Complete the following table
(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 of 1/11. Do you agree with this argument? Justify your answer.
Solution:
Since two dice are thrown together, the following are the possible outcomes:
(1, 1); (1, 2); (1, 3); (1, 4); (1, 5); (1, 6);
(2, 1); (2, 2); (2, 3); (2, 4); (2, 5); (2, 6);
(3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6);
(4, 1); (4, 2); (4, 3); (4, 4); (4, 5); (4, 6);
(5, 1); (5, 2); (5, 3); (5, 4); (5, 5); (5, 6);
(6, 1); (6, 2); (6, 3); (6, 4); (6, 5); (6, 6)
The total number of possible outcomes is 6 x 6 = 36.
(i)
(a) The sum of two dice is 3 for (1, 2) and (2, 1).
Therefore, the number of favorable outcomes = 2.
P(3) = 2/36 = 1/18
(b) The sum of two dice is 4 for (1, 3), (2, 2), and (3, 1).
Therefore, the number of favorable outcomes = 3.
P(4) = 3/36 = 1/12
(c) The sum of two dice is 5 for (1, 4), (2, 3), (3, 2), and (4, 1).
Therefore, the number of favorable outcomes = 4.
P(5) = 4/36 = 1/9
(d) The sum of two dice is 6 for (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1).
Therefore, the number of favorable outcomes = 5.
P(6) = 5/36
(e) The sum on two dice is 7 for (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1).
Therefore, the number of favorable outcomes = 6.
P(7) = 6/36 = 1/6
(f) The sum of two dice is 9 for (3, 6), (4, 5), (5, 4), and (6, 3).
Therefore, the number of favorable outcomes = 4.
P(9) = 4/36 = 1/9
(g) The sum of two dice is 10 for (4, 6), (5, 5), and (6, 4).
Therefore, the number of favorable outcomes = 3.
P(10) = 3/36 = 1/12
(h) The sum of two dice is 11 for (5, 6) and (6, 5).
Therefore, the number of favorable outcomes = 2.
P(11) = 2/36 = 1/18
Thus, the complete table is as under:
(ii) No, the eleven sums are not equally likely.
∴ The argument is not correct.
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.
Sol. Possible outcomes are:
HHH, TTT, HHT, HTH, THH, TTH, THT, HTT
i.e., in all 8 outcomes.
Out of these, 6 outcomes other than HHH and TTT favor the event E than Hanif will lose the game.
Therefore, P (Hanif will lose the game) = 6/8 = 3/4
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?
Sol. Since, throwing a die twice or throwing two dice simultaneously is the same.
∴ All possible outcomes are :
(1, 1); (1, 2); (1, 3); (1, 4); (1, 5); (1, 6);
(2, 1); (2, 2); (2, 3); (2, 4); (2, 5); (2, 6);
(3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6);
(4, 1); (4, 2); (4, 3); (4, 4); (4, 5); (4, 6);
(5, 1); (5, 2); (5, 3); (5, 4); (5, 5); (5, 6);
(6, 1); (6, 2); (6, 3); (6, 4); (6, 5); (6, 6)
∴ All possible outcomes = 36
(i) Let E be the event that 5 does not come up either time, then the favorable outcomes are = [36 - (5 + 6)] = 25
∴ P(E) = 25/36
(ii) Let N be the event that 5 will come up at least once, then number of favorable outcomes = 5 + 6 = 11
∴ P(N) = 11/36
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 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.
Sol.
(i) Incorrect. The given argument is not correct. Because, if two coins are tossed simultaneously then four outcomes are possible (HH, HT, TH, TT). So the total outcomes is 4.
∴ Probability = 1/4.
(ii) Correct. Because the two outcomes are possible. Total outcomes = 6 and odd numbers = 3 and even numbers = 3. So, favorable outcomes = 3 (in both the cases even or odd).
∴ Probability = 3/6 = 1/2
5.0Benefits of Class 10 Maths Chapter 14 Probability Exercise 14.1
- Builds a strong foundation in basic probability concepts.
- Enhances logical thinking and analytical skills.
- Prepares students for probability-based board exam questions.
- Encourages critical thinking in decision-making scenarios.
- Strengthens probability concepts for students attempting competitive exams like NTSE and Olympiads.
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