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Probability is a fundamental concept in Maths class 10 that helps in predicting the likelihood of an event occurring. It is a branch of mathematics that deals with studying random events and finding the chances of their occurrence. In this chapter, we explore different types of probability and how to calculate it using various methods. This article aims to simplify the chapter for students of class 10 by providing clear explanations and solutions. We will discuss the concepts and will resolve your doubts, such as 'What is the definition of probability?', 'how we calculate the probability of a single event or multiple events', 'What are the different types of probability?'.
NCERT Solutions for Class 10 Maths Chapter 14 Probability, created by experts at Careers360, offers detailed study material for students preparing for the CBSE Class 10 board exam. The solutions cover all NCERT Solutions for Class 10 and are readily available for download. With step-by-step explanations for various types of questions, these solutions will help students grasp the concepts of probability class 10 and gain confidence in solving related problems. The Class 10 Math Chapter 14 NCERT solutions are a valuable resource for students aiming to master this topic. Students can click on the above link for solutions to other classes and subjects. It is highly recommended for students to refer to the NCERT Class 10 Maths books and thoroughly study all the topics without missing any.
Also, Read,
Empirical Probability - Empirical probability gives us the likelihood of events based on real-world experiments.
Empirical Probability = (Number of Trials with Expected Outcome)/(Total Number of Trials)
Theoretical Probability -Theoretical probability deals with event likelihood we get from experiments. The formulation is as follows.
Theoretical Probability = (Number of Favourable Outcomes to Event E)/(Total Number of Possible Outcomes in the Experiment)
Free download NCERT Solutions for Class 10 Maths Chapter 14 Probability PDF for CBSE Exam.
NCERT Probability Class 10 Solutions:
Exercise: 14.1
Q1. Complete the following statements:
Q1 (i).Probability of an event E + Probability of the event ‘not E’ = ________
Answer:
Probability of an event E + Probability of the event ‘not E’ = 1
So,
Q1 (ii). The probability of an event that cannot happen is ______ . Such an event is called ______.
Answer:
The probability of an event that cannot happen is 0. Such an event is called an impossible event.
When there is no favorable outcome, i.e., the number of outcomes is zero.
Q1 (iii). The probability of an event that is certain to happen is_____. Such an event is called _______.
Answer:
The probability of an event that is certain to happen is 1. Such an event is called a sure/certain event
When the number of favorable outcomes is the same as the number of all possible outcomes, it is a sure event.
Q1 (iv). The sum of the probabilities of all the elementary events of an experiment is _______.
Answer:
The sum of the probabilities of all the elementary events of an experiment is 1.
Q1 (v). The probability of an event is greater than or equal to and less than or equal to ________ .
Answer:
The probability of an event is greater than or equal to 0 and less than or equal to 1.
Q2. Which of the following experiments have equally likely outcomes? Explain.
Q2 (i). A driver attempts to start a car. The car starts or does not start.
Answer:
It is not an equally likely event since it depends on various factors like there is no fuel, engine malfunctioning etc, that are not alike for both outcomes.
Q2 (ii). A player attempts to shoot a basketball. She/he shoots or misses the shot.
Answer:
It is not an equally likely event because it depends on the ability and amount of practice of the player. If he is a professional player, he will more likely have a successful shot. Whereas an amateur player will more likely miss the shot.
Q2 (iii). A trial is made to answer a true-false question. The answer is right or wrong.
Answer:
It is an equally likely event. The only options are true or false, and only one of them is correct.
Q2 (iv). A baby is born. It is a boy or a girl.
Answer:
It is an equally likely event. The only possibilities of gender are boy and girl. Hence, if not boy, then girl and vice versa.
Answer:
The tossing of the coin is considered to be a fair way of deciding because the only possible outcomes are heads and tails. Hence, they are equally likely events.
Q4. Which of the following cannot be the probability of an event?
(A)
(B) –1.5
(C) 15%
(D) 0.7
Answer:
We know that the probability of an event is either greater than or equal to 0 and always less than or equal to 1. Hence, the probability of an event can never be negative.
Therefore, (B)
Also, (A) :
(C):
(D): 0.7
Hence, (A), (C), and (D) all lie between 0 and 1.
Q5. If P(E) = 0.05, what is the probability of ‘not E’?
Answer:
Given,
We know,
Hence, the probability of 'not E' is 0.95
Q6 . A bag contains lemon-flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange-flavored candy?
(ii) a lemon-flavored candy?
Answer:
(i) According to the question, the bag contains only lemon-flavored candies. It does not contain any orange-flavored candy. Hence, every time, only lemon-flavored candy will come out. Therefore, P(an orange-flavored candy) = 0. i.e., the event of taking out an orange-flavored candy is impossible.
(ii) According to the question, the bag contains only lemon-flavored candies. So the event that Malini takes out a lemon-flavored candy is sure. Therefore, P(a lemon-flavored candy) = 1.
Answer:
Given,
The probability of two students not having the same birthday
Hence, the probability that the 2 students have the same birthday is 0.008.
Answer:
Total number of balls in the bag = 8
No. of red balls = 3
No. of black balls = 5
(i) Let E be the event of getting a red ball
n(E) = No. of red balls = 3
n(S) = No. of total balls = 8
(ii) We know,
where
Q9. 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?
Answer:
Given,
Total number of balls in the bag = 5 + 8 + 4 = 17
(i) Let R be the event that the ball taken out is red
The number of possible outcomes = 17
The number of outcomes favorable to the event R = 5
(ii) Let W be the event that the ball taken out is white
The number of possible outcomes = 17
The number of outcomes favorable to the event W = 8
(iii) Let G be the event that the ball taken out is green
The number of possible outcomes = 17
The number of outcomes favorable to the event G = 4
The required probability of not getting a green ball is
Q10. A piggy bank contains a hundred 50p 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,
(i) What is the probability that the coin will be a 50 p coin?
(ii) Will not be a Rs 5 coin?
Answer:
Total number of coins in the piggy bank
(i) Let E be the event of getting a 50p coin.
Number of possible outcomes = 180
Number of outcomes favorable to event E = 100
Therefore, the probability of getting a 50p coin is
(ii) Let F be the event of getting a Rs. 5 coin.
Number of possible outcomes = 180
Number of outcomes favorable to event F = 10
Therefore, the probability of not getting a Rs. 5 coin is
Answer:
Total number of fishes in the tank = 5 (male) + 8 (female) = 13
Let E be the event that the fish taken out is a male fish.
Number of possible outcomes = 13
Number of outcomes favorable to E = 5
Therefore, the probability that the fish taken out is a male fish is
Q12. A game of chance consists of spinning an arrow that comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig.), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(i) a number greater than 2?
(i) a number less than 9?
Answer:
Total possible outcomes = {1,2,3,4,5,6,7,8}
Number of possible outcomes = 8
(i) Let E be the event of getting 8.
Therefore, the probability that it will point at 8 is
(ii) Let E be the event of pointing at an odd number.
Total number of odd numbers = n({1,3,5,7}) = 4
Therefore, the probability of getting an odd number is
(iii) Let E be the event of pointing at a number greater than 2
Number of favorable outcomes= n({3,4,5,6,7,8}) = 6
Therefore, the probability of pointing at a number greater than 2 is
(iv) Let E be the event of pointing at a number less than 9
Since all the numbers on the wheel are less than 9, this is a sure event.
Number of favorable outcomes = 8
Therefore, the probability of pointing at a number less than 9 is $1.
Q13. A die is thrown once. Find the probability of
(ii) a number lying between 2 and 6.
(iii) getting an odd number.
Answer:
Possible outcomes when a die is thrown = {1,2,3,4,5,6}
Number of possible outcomes once = 6
(i) Let E be the event of getting a prime number.
Prime numbers on the die are = {2,3,5}
Number of favorable outcomes = 3
Therefore, the probability of getting a prime number is
(ii) Let F be the event of getting a number lying between 2 and 6
Numbers lying between 2 and 6 on the die are = {3,4,5}
Number of favorable outcomes = 3
Therefore, the probability of getting a number lying between 2 and 6 is
(iii) Let O be the event of getting an odd number.
Odd numbers on the die are = {1,3,5}
Number of favorable outcomes = 3
Therefore, the probability of getting an odd number is
Q14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(ii) a face card.
(iii) a red-face card
(iv) the Jack of hearts
(v) a spade
Answer:
The total number of cards in a well-shuffled deck = 52
Hence, the total possible outcomes = 52
(i) Let E be the event of getting a king of red color.
There are only red color kings: Hearts and diamonds
Hence, the number of favorable outcomes = 2
Therefore, the probability of getting a king of red color is
(ii) Let E be the event of getting a face card.
Face cards: (J, Q, K) of each four suits
Hence, the number of favorable outcomes = 12
Therefore, the probability of getting a face card is
(iii) Let E be the event of getting a red face card.
Face cards: (J, Q, K) of hearts and diamonds
Hence, the number of favorable outcomes =
Therefore, the probability of getting a red face card is
(iv) Let E be the event of getting the jack of hearts
Hence, the number of favorable outcomes = 1
Therefore, the probability of getting the jack of hearts is
(v) Let E be the event of getting a spade.
There are 13 cards in each suit.
Hence, the number of favourable outcomes = 13
Therefore, the probability of getting a spade is
(vi) Let E be the event of getting the queen of diamonds
Hence, the number of favorable outcomes = 1
Therefore, the probability of getting the queen of diamonds is
(ii) the second card picked up is (a) an ace? (b) a queen?
Answer:
Total number of cards = 5
Hence, the total possible outcomes = 5
(i) There is only one queen.
Hence, favorable outcome = 1
(ii-a) There is only one ace.
Hence, favorable outcome = 1
Therefore, the probability of getting an ace is 0.25.
(ii-b) Since there is no queen left.
Hence, favorable outcome = 0
Therefore, the probability of getting a queen is 0. Thus, it is an impossible event.
Answer:
Total number of pens = 132 (good) + 12 (defective)
Hence, the total possible outcomes = 144
Number of good pens = number of favorable outcomes = 132
17. (i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
Answer:
(i) Total number of bulbs = 20
Hence, the total possible outcomes = 20
Number of defective bulbs = 4
Hence, the number of favorable outcomes = 4
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced.
Number of remaining bulbs
Number of remaining non-defective bulbs
The probability that this bulb is not defective
Q18. 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
(iii) a number divisible by 5.
Answer:
(i) Total number of discs = 90
Number of discs having a two-digit number between 1 and 90 = 81
(ii) Total number of discs = 90
Perfect square numbers from 1 to 90 are {1, 4, 9, 16, 25, 36, 49, 64, 81}
Therefore, the total number of discs having perfect squares = 9.
(iii) Total number of discs = 90
Numbers between 1 and 90 that are divisible by 5 are
Therefore, the total number of discs having numbers that are divisible by 5 = 18.
Q19. (i) 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
Answer:
The six faces of the die contain: {A, B, C, D, E, A}
Total number of letters = 6
(i) Since there are two A's,
Number of favorable outcomes = 2
Therefore, the probability of getting A is
(ii) Since there is only one D,
Number of favorable outcomes = 1
Therefore, the probability of getting D is
Answer:
Here, the Total outcome is the area of the rectangle, and the favorable outcome is the area of the circle.
Area of the rectangle =
Area of the circle =
Answer:
Total number of pens = 144
Total number of defective pens = 20
She will buy it if the pen is good.
(i) Therefore, the probability that she buys = probability that the pen is good =
(ii) The probability that she will not buy = probability that the pen is defective
Q22. Refer to Example 13. (i) Complete the following table:
Event: 'Sum on 2 dice' | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
1/36 | 5/36 | 1/36 |
Answer:
Number of possible outcomes to get the sum as 2 = (1,1) = 2
Number of possible outcomes to get the sum as 3 = (2,1), (1, 2) = 2
Number of possible outcomes to get the sum as 4 = (2, 2), (1, 3), (3,1) = 3
Number of possible outcomes to get the sum as 5 = (3, 2), (2, 3), (4,1), (1, 4) = 4
Number of possible outcomes to get the sum as 6 = (5,1), (1, 5), (3, 3), (4, 2), (2, 4) = 5
Number of possible outcomes to get the sum as 7 = (4, 3), (3, 4), (6,1), (1, 6), (5, 2), (2, 5) = 6
Number of possible outcomes to get the sum as 8 = (4, 4), (6, 2), (2, 6), (5, 3), (3, 5) = 5
Number of possible outcomes to get the sum as 9 = (5, 4), (4, 5), (6, 3), (3, 6) = 4
Number of possible outcomes to get the sum as 10 = (5, 5), (6, 4), (4, 6) = 3
Number of possible outcomes to get the sum as 11 = (6, 5), (5, 6) = 11
Number of possible outcomes to get the sum as 12 = (6, 6) = 1
The table becomes:
The sum of two dice | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Probability | 1/36 | 1/18 | 1/12 | 1/9 | 5/36 | 1/6 | 5/36 | 1/9 | 1/12 | 1/18 | 1/36 |
Answer:
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. We do not agree with this argument because there are a different number of possible outcomes for each sum. We can see that each sum has a different probability.
Answer:
The possible outcomes when a coin is tossed 3 times (Same as 3 coins tossed at once)
{HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}
Number of total possible outcomes = 8
For Hanif to win, there are only two favorable outcomes: {HHH, TTT}
Number of favorable outcomes = 2
Therefore, the probability that Hanif will lose is
Q24 (i) A die is thrown twice. What is the probability that 5 will not come up either time?
Answer:
When a die is thrown twice, the possible outcomes =
Total number of possible outcomes =
The outcomes when 5 comes up at least once =
{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}
Number of such favorable outcomes = 11
Therefore, the probability that 5 will not come either time is
24 (ii) A die is thrown twice. What is the probability that 5 will come up at least once?
Answer:
When a die is thrown twice, the possible outcomes =
Total number of possible outcomes =
The outcomes when 5 comes up at least once =
{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}
Number of such favorable outcomes = 11
Therefore, the probability that 5 comes at least once is
Answer:
The possible outcomes when two coins are tossed = {HH, HT, TH, TT}
Total number of possible outcomes = 4
Hence, the given statement is not correct. This is because one of each can occur in two different ways. Hence the mentioned events are not equally likely.
Answer:
The possible outcomes when a die is thrown= {1,2,3,4,5,6}
Total number of possible outcomes = 6
Number of odd numbers, {1,3,5} = 3
And the number of even numbers {2,4,6} = 3
Hence, both these events are equally likely
Similarly,
So, this argument is correct.
We have provided the solutions for the type of questions given below:
The basic formula for finding out the probability
Questions based on tossing a coin multiple times
Cards based questions
Question related to probability when a dice is rolled
Word problems related to real-life situations
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
NCERT solutions are also available for classes from 6 to 12.
Also Read,
Comprehensive Coverage: These class 10 math chapter 14 solutions offer comprehensive coverage of Chapter 14, ensuring that all important topics and concepts are thoroughly covered.
Clarity and Accuracy: The probability solution class 10 is presented with clarity and accuracy, making it easier for students to understand and apply mathematical principles.
Step-by-Step Approach: Each solution for class 10th chapter probability follows a step-by-step approach, guiding students through the problem-solving process.
In-Depth Explanation: Complex concepts are explained in-depth, helping students grasp the underlying principles and logic in class 10th chapter probability solutions.
CBSE Guidelines: The solutions adhere to the CBSE curriculum and guidelines, making them a reliable resource for exam preparation.
First of all, go through the NCERT Class 10 Maths syllabus of chapter 14 Probability.
Go through the theory given in the chapter from the NCERT Class 10 textbook.
Then, come to the practice exercises and list down those questions with which you are having trouble in class 10th probability.
Now, go through Chapter 14 Probability Class 10 ncert Solutions to know the solution to those questions.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 | Probability |
Chapter 14 of Class 10 Maths is about Probability, and the topics we covered were the introduction to probability, theoretical probability, experiments and outcomes, complementary events, and some basic theorems of probability. Also, we have covered questions based on:
1. Cards
2. Dice
3. Coins
4. Numbers
When solving probability questions, we have to remember some steps, such as identifying the experiment, defining the total outcomes, and using the probability formula.
Use this formula:
Where n(E) = number of favourable events.
and, n(S) = number of total outcomes.
You can find the NCERT Solutions for Class 10 Maths Chapter 14 PDF by clicking on the given download link where Career 360 provides all solutions to the given question of ex 14.1 Download.
In class 10, we only use the given basic formulae for given problems, which are:
1.
2. P( not A) = 1- P(A)
To find the probability of an event, first, we determine the total number of outcomes known as sample space [n(S)]. We calculate the number of outcomes of the event [n(E)]. At last, by applying the given formula, we get our answer.
Hello
Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.
1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.
2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.
3. Counseling and Seat Allocation:
After the KCET exam, you will need to participate in online counseling.
You need to select your preferred colleges and courses.
Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.
4. Required Documents :
Domicile Certificate (proof that you are a resident of Karnataka).
Income Certificate (for minority category benefits).
Marksheets (11th and 12th from the Karnataka State Board).
KCET Admit Card and Scorecard.
This process will allow you to secure a seat based on your KCET performance and your category .
check link for more details
https://medicine.careers360.com/neet-college-predictor
Hope this helps you .
Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
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