NCERT solutions for class 10 maths exercise 15.1 of chapter 15 Probability
NCERT solutions for class 10 maths exercise 15.1 of chapter 15 Probability will help to clear doubts of students when they would solve unsolved questions of chapter 15 Probability in their NCERT maths textbook. The expert of maths has contributed his full efforts in the explanation of each question therefore every student will understand the solutions and the concept of Probability that is required in solving the questions of maths question paper in the CBSE board exams.
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NCERT solutions for class 10 maths of chapter 15 Probability
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NCERT solutions for class 10 maths exercise 15.1 of chapter 15 Probability
Q1. Complete the following statement
(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 ……………
(iv) The sum of the probabilities of all elementary events of an experiment is ……………
(v) The probability of an event is greater than or equal to ……………
Ans.
(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 sure event
(iv) The sum of the probabilities of all 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
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.
(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.
Ans. (i) If a driver attempts to start a car, there are more chances that it will start, therefore this event has not equally likely outcomes
(ii) If a player attempts to shoot a basketball since opponents are also involved in the game, therefore this event has not equally likely outcomes
(iii) A trial is made to answer a true-false question, the chances that answer is true or false is 50-50, therefore this event has equally likely outcomes
(iv) A baby is born, there are 50-50 chances of that born baby is boy or girl, therefore this event has equally likely outcomes
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?
Ans. The event of tossing a coin has two outcomes head and tail,both outcomes have 50-50 chances since it is unpredictable that it will appear head or tail that’s why tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game.
Q4. Which of the following cannot be the probability of an event?
(a) 23
(b) – 1.5
(c) 15%
(d) 0.7
Ans. (b) – 1.5 since the probability is greater or equal to 0.
Q5. If P(E) = 0.05, what is the probability of ‘not E’?
Ans. Probability of an event E + Probability of the event ‘not E’ =1
0.05 + Probability of the event ‘not E’ = 1
Probability of the event ‘not E’ = 1 – 0.05 = o.95
Q6. A bag contains lemon flavoured candies only. Malini takes out without looking into the bag. What is the probability that she takes out
(a) An orange flavoured candy?
(b) A lemon flavoured candy?
Ans.(a) A bag contains lemon flavoured candies only, the event of taking out an orange flavoured candy is an impossible event, therefore its probability is 0.
(b) A bag contains lemon flavoured candies only, the event of taking out a lemon flavoured candy is a sure event, therefore its probability is 1.
Q7. 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?
Ans. Let the event of the birthday of two friends that lies on the same day is ‘E’ and the event of the birthday of two friends that don’t lie on the same day is ‘not E’
We know
P(E) + P(not E) = 1
0.992 + P(not E) = 1
P(not E) = 1 – 0.992 = 0.008
Q8. 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
(a) red?
(b) not red?
Ans. Number of red balls in the bag = 3
Number of black balls in the bag = 5
Total ball in the bag = 5+3 = 8
We know the probability P(E) of an event, E is
P(E) = (Favourable outcomes)/(Total possible outcomes)
(a) Favourable outcomes = Number of red balls =3
Total possible outcomes = Total ball in the bag = 8
P(red ball) = The number of red ball/Total number of balls = 3/8
(b) Favourable outcomes = Number of balls which are not red = 5
Total possible outcomes = Total ball in the bag = 8
P(red ball) = Number of balls which are not red /Total number of balls = 5/8
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Q9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. 1 marble is taken out of the box at random. What is the probability that the marble taken out will be
(a) red?
(b) white?
(c) not green?
Ans. A box contains red marbles = 5,white marbles = 8,green marbles =4
Total marbles = 5+8 +4 = 17
We know the probability P(E) of an event, E is
P(E) = (Favourable outcomes)/(Total possible outcomes)
(a) Favourable outcomes =Number of red marbles =5
Total possible outcomes = Total marbles in the box = 17
P(red marbles) = The number of red marbles /Total number of marbles= 5/17
(b) Favourable outcomes = Number of white marbles = 8
Total possible outcomes = Total marbles = 17
P(white marbles) = Number of white marbles /Total number of marbles = 8/17
(c) The marbles which are not green= red marbles +white marbles =5+8=13
Total possible outcomes = Total marbles = 17
P(not green) = The marbles which are not green/Total number of balls=13/17
Q10. A piggy bank contains hundred 50 p coins, fifty Rs 1 coin, twenty 2 rupee 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 the coin
(a) will be a 50 p coins?
(b) will not be a rupee 5 coin?
Ans. 50 p coins = 100, Rs 1 coin =50,Rs 2 coin = 20,Rs 5 = 10
Total number of coins = 100+50+ 20+ 10 = 180
(a) P(50 p coins ) = Number of 50 p coins/Total number of coins =100/180 =5/9
(b) P(not 5 coin) = Number of coins which are not Rs 5/Total number of coins =(100+50+20)/180 =170/180 =17/18
Q11. 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?
Ans. Number of male fish in the tank = 5
Number of female fish in the tank = 8
Total number of fish in the tank = 8+5=13
P(male fish) = Number of male fish in the tank/Total number of fish in the tank =5/13
Q12.A game of chance consists of spinning an arrow which comes to rest pointing 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?
Ans. Total possible outcomes=Total number of digits = 8 (1,2,3,4,5,6,7,8)
(i) Fvourable outcomes =The number of 8 digits =1
P(getting 8) = 1/8
(ii) Fvourable outcomes =The number odd digits =4(1,3,5,7)
P(getting odd number) = 4/8 =1/2
(iii) Fvourable outcomes = Numbers greater than 2 = 6(3,4,5,6,7,8)
P(getting numbers greater than 2) =6/8 =3/4
(iv) Fvourable outcomes =Number less than 9= all the numbers(1,2,3….8)
P(getting number less than 9) = 8/8 =1
Q13. 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
Ans. Total possible outcomes=Total number of digits in the dice = 6 (1,2,3,4,5,6)
(i) Fvourable outcomes =Number of prime numbers =3(2,3,5)
P(getting prime number ) = 3/6 =1/2
(ii Fvourable outcomes =Numbers lying between 2 and 6 =3(3,4,5)
P(getting numbers lying between 2 and 6 ) = 3/6 =1/2
(iii) Fvourable outcomes =Number of odd numbers =3(1,3,5)
P(getting odd numbers) = 3/6 =1/2
Q14. One card is drawn from a well-shuffled 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
Ans. Total possible outcomes= Total number of cards =52
(i) Fvourable outcomes=Number of red kings = 2
P(getting a red king) = 2/52 =1/26
(ii) Total possible outcomes= Total number of cards =52
(i) Fvourable outcomes=Number of face cards = 12
P(getting a face cards) = 12/52 =3/13
(iii) Favourable outcomes=Number of red face cards = 6(3 diamond and 3 heart)
P(getting a red face cards) =6/52 =2/26
(iv) Favourable outcomes=The jack of hearts =1
P(getting jack of heart) =1/52
(v) Favourable outcomes=The number of spades =13
P(getting a spade) =13/52 =1/4
(vi)Favourable outcomes= The number of queen of diamonds=1
P(getting a queen of diamond) =1/52
Q15. 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, the probability that the second card picked up is
(a) an ace?
(b) a queen?
Ans. The number of cards = 5( The ten, jack, queen, king and ace of diamonds)
(i) Favourable outcomes= The number of queen =1
P(getting a queen) = 1/5
(ii) If the queen is drawn and put aside, the total possible outcomes =4
(a) Favourable outcomes= Number of ace =1
P(getteng an ace) =1/4
(b) Favourable outcomes= Number of queen =0 (since the queen is kept aside)
P(getting a queen) = 0/4 =0
The concept of probability actually came into existence from gambling or betting. Probability is the measurement of how much extent an event is likely to happen. Probability depends on two factor
(i) Favourable outcome (ii) Total possible outcome
Q16. 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.
Ans. Defective pens are = 12
The pens which are good ones = 132
Total pens are = 132 +12 = 144
P(pen which are good one) =Favourable outcomes/Total possible= The pens which are good ones /Total pens = 132/144 =11/12
Probability formula
The probability of an event is represented by P(E)
P(E) = Favourable outcomes/Total possible outcomes
(i) Favourable outcomes
Favourable outcomes is a set of sample space of an event that shows a number of outcomes that are likely to happen, such a list of outcomes is known as favourable outcomes.
(i)Total possible outcomes
All the possible outcomes of an event are known as total possible outcomes.
Example: Let there is an event of throwing a coin twice.
After throwing the coin two times, we get the total possible outcomes (HH),(TT),(H, T),(T, H).
The occurrence of each outcome is as follows
Two head =1, two tale = 1, one head and one tale =2
Here number of favourable outcomes of the event if there comes two head are=1 and similarly,we can find others.
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| Chapter 1-Sets | Chapter 9-Sequences and Series |
| Chapter 2- Relations and functions | Chapter 10- Straight Lines |
| Chapter 3- Trigonometry | Chapter 11-Conic Sections |
| Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
| Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
| Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
| Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
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| Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
| Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
| Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
| Chapter 4-Determinants | Chapter 12-Linear Programming |
| Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
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| Chapter 7- Integrals | |
| Chapter 8-Application of Integrals |
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