**NCERT Solutions Class 10 Maths Chapter 15 Probability** – Here are all the NCERT solutions for Class 10 Maths Chapter 15. This solution contains questions, answers, images, explanations of the complete Chapter 15 titled Probability of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 15 Probability. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 15 Probability in one place.

## NCERT Solutions Class 9 Maths Chapter 15 Probability

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 15 Probability , Maths, Class 9.

Class | 9 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 15 |

Chapter Name |
Probability |

### NCERT Solutions Class 9 Maths chapter 15 Probability

Class 9, Maths chapter 15, Probability solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Probability

Q.1:In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Ans :Number of times the batswoman hits a boundari = 6 Total number of balls played = 30 Number of times that the batwoman does not hit a boundary = 30 -6 = 24 p(she does not hit the boundary)=\( \begin{array}{c} \frac {{\text { Number of times when she does not hit boundary }} }{ {\text { Total number of balls played }}}\end{array} \) \( =\frac{24}{30}=\frac{4}{5}\)

Q.2:1500 families with 2 children were selected randomly, and the following data were recorded: Compute the probability of a family,chosen at random,having (i) 2 girls (ii) 1 girl (iii) no girl Also check weather the sum of these probabilities is 1.

Ans :Total number of family=475+814+211=1500 (i) number of families of 2 girls =475 p1(she does not hit the boundary)=\( \begin{array}{c} \frac {{\text { Number of families having 2 girls}} }{ {\text { Total number of balls played }}}\end{array} \) \( =\frac{475}{1500}=\frac{19}{60}\) (ii) Number of families having 1 girl =814 p2(she does not hit the boundary)=\( \begin{array}{c} \frac {{\text { Number of families having 1 girl }} }{ {\text { Total number of balls played }}}\end{array} \) \( =\frac{814}{1500}=\frac{407}{750}\) (iii) Number of families having no girl = 211 p3(she does not hit the boundary)=\( \begin{array}{c} \frac {{\text { Number of families having no girl }} }{ {\text { Total number of balls played }}}\end{array} \) \( =\frac{211}{1500}\) Sum of all these probabilities = \( =\frac{19}{60}+\frac{407}{750}+\frac{211}{1500}\) \( =\frac{475+814+211}{1500}\) \( =\frac{1500}{1500}=1\)

Q.3:Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.

Ans :Number of students born in the month of August = 6 Total number of students = 40 \( \mathrm{P} \text { (Students born in the month of August) }=\frac{\text { Number of students born in August }}{\text { Total number of students }}\) \( =\frac{6}{40}=\frac{3}{20}\)

Q.4:Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes: If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up,

Ans :Number Of times 2 heads come up = 72 Total number Of times the coins were tossed = 200 \( \mathrm{P}(2 \text { heads will come up })=\frac{\text { Number of times } 2 \text { heads come up }}{\text { Total number of times the coins were tossed }}\) \( =\frac{72}{200}=\frac{9}{25}\)

Q.5:An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below: Suppose a family is chosen, find the probability that the family chosen IS (i) earning Rs 10000 — 13000 per month and owning exactly 2 vehicles. (ii) earning Rs 16000 or more per month and owning exactly I vehicle. (iii) earning less than Rs 7000 per month and does not own any vehicle. (iv) earning Rs 13000 — 16000 per month and owning more than 2 vehicles. (v) owning not more than 1 vehicle.

Ans :Number of total families surveyed = 10 + 160 + 25 + 0 + 0 + 305 + 27 +2+1+535 +29+1+2 + 469 +59+ 25 +1 + 579 + 82 +88 = 2400 (i) Number of families earning Rs 10000 - 13000 per month and owning exactly 2 vehicles = 29 Hence, required probability, p=\( \mathrm{P}=\frac{579}{2400}\) (ii) Number Of families earning Rs 16000 or more per month and owning exactly 1 vehicle = 579 Hence, required probability, p=\( \frac{10}{2400}=\frac{1}{240}\) (iii) Number of families earning less than Rs 7000 per month and dcmas not own any vehicle = 10 Hence, required probability, p= \(\frac{10}{2400}=\frac{1}{240}\) (iv) Number of families earning Rs 13000 - 16000 per month and owning more than 2 vehicles = 25 Hence, required probability, p= \( \frac{25}{2400}=\frac{1}{96}\) (v) Number of families owning not more than 1 vehicle = 10+ 160+0+ 305 + 535 +2+ 469 +1 + 579 =2062 Hence, required probability, p= \( \frac{2062}{2400}=\frac{1031}{1200}\)

Q.6:Refer to Table 14.7, Chapter 14. (i) Find the probability that a student obtained less than 20% in the mathematics test. (ii) Find the probability that a student obtained marks 60 or above.

Ans :Total number Of students born in the year = 3 + 4 + 2 + 2 + 5 + 1 + 2 + 6 + 3 + 4 + 4 + 4 = 40 Number of students born in August :. Probability that a student of the class was born in August = \( \frac{6}{40}=\frac{3}{20} \).

Q.7:To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table. Find the probability that a student chosen at random (i) likes statistics, (ii) does not like it.

Ans :Total number of students = 135 + 65 = 200 (i) Number of students liking statistics =135 P(student liking statistics)=\( \frac{135}{200}=\frac{27}{40}\)

Q.8:Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives: (i) less than 7 km from her place of work? (ii) more than or equal to 7 km from her place of work? (iii) within 1 2 km from her place of work?

Ans :(i) Total number of engineers = 40 Number of engineers living less than 7 km from their place of work 9 Hence, required probability that an engineer lives less than 7 km from her place Of P=\( \frac{9}{40}\) (ii) Number of engineers living more than or equal to 7 km from their place of work = 40 -9 = 31 Hence, required probability that an engineer lives more than or equal to 7 km from Her place of work, \( \mathrm{P}=\frac{31}{40}\) Hence, required probability that an engineer lives within \( \frac{1}{2}\) km from her place of work, P = 0

Q.9:Activity : Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.

Ans :DIY

Q.10:Activity : Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.

Ans :DIY

Q.11:Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg): 4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00 Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Ans :Number of total bags 11 Number of bags containing more than 5 kg of flour = 7 Hence ,required probability \( \mathrm{P}=\frac{7}{11}\)

Q.12:In Q.5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 - 0.16 on any of these days.

Ans :Number days for which the concentration Of sulphur dioxide was in the interval Of 0.12 - 0.16 = 2 Total number Of days = 30 Hence,required probability \( P=\frac{2}{30}=\frac{1}{15}\)

Q.13:In Q.1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.

Ans :Number of students having blood group AB=3 Total number of students = 30 Hence,required probability \( P=\frac{3}{30}=\frac{1}{10}\)

Q.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 ___

Ans :(i) 1 (ii) 0, impossible event (iii) 1,sure event or certain event (iv) 1 (v) 0,1

Q.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

Ans :(i) It is not an equally likely event, as it depends on various factors such as whether the car will start or not. And factors for both the conditions are not the same. (ii) It is not an equally likely event, as it depends on the player's ability and there is no information given about that. (iii) It is an equally likely event. (iv) It is an equally likely event.

Q.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?

Ans :When we toss a coin, the possible outcomes are only two, head Or tail, which are equally likely outcomes. Therefore, the result of an individual toss is completely unpredictable.

Q.4:Which of the following cannot be the probability of an event? (A) ⅔ (B) -1.5 (C) 15 % (D) 0.7

Ans :Probability of an event (E) is always greater than or equal to 0. Also, it is always less than or equal to one. This implies that the probability of an event cannot be negative or greater than 1. Therefore, out of these alternatives, -1.5 cannot be a probability of an event. Hence, (B)

Q.5:If P(E) = 0.05, what is the probability of ‘not E’?

Ans :We Know that, \( \mathbf{P}(\overline{\mathbf{E}})=1-\mathrm{P}(\mathrm{E}) \) \( \mathbf{P}(\overline{\mathbf{E}})=1-0.05 \) \( =0.95 \) Therefore, the probability of ‘not E’ is 0.95

Q.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?

Ans :(i) The bag contains lemon flavoured candies only, It does not contain any orange flavoured candies. This implies that every time, she will take out only lemon flavoured candies. Therefore, event that Malini will take out an orange flavoured candy is an impossible event. Hence, P (an orange flavoured candy) = 0 (ii) As the bag has lemon flavoured candies, Malini will take out only lemon flavoured candies. Therefore, event that Malini will take out a lemon flavoured candy is a sure event. P (a lernon flavoured candy) =1

Q.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?

Ans :Probability that two students are not having same birthday \( P(\overline{E})=0.992 \) Probability that two students are having same birthday \( P(E)=1-P(\overline{E}) \) \( =1-0.992 \) \( =0.008 \)

Q.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?

Ans :(i) Total number of balls in the bag= 8 Probability of getting a red ball \(=\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) =⅜ (ii) Probability of not getting red ball = 1 - Probability of getting a red ball = \( 1-\frac{3}{8} \) = \( \frac{5}{8} \)

Q.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?

Ans :Total number of marbles =5+8+4 =17 (i) Number of red marble\( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) = 5/17 (ii) Number of white marbles=8 Number of white marble\( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) =8/17 (iii) Number of green marble \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }}\) =4/17 Probability of not getting a green marble \( =1-\frac{4}{17}=\frac{13}{17} \)

Q.10: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?

Ans :Total number of coins in a piggy bank =100+50+20+10 =180 (i) Probability of getting a 50p coin \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) \( =\frac{100}{180}=\frac{5}{9} \) (ii) Number of Rs 5 Coins =10 Probability of getting a 5 Rs coin \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) \(=\frac{10}{180}=\frac{1}{18} \) Probability of not getting a 5Rs coin \( =1-\frac{1}{18} \) \( =\frac{17}{18} \)

Q.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 Fig.) . What is the probability that the fish taken out is a male fish?

Ans :Total number of fishes in a tank = Number of male fishes + number of female fishes = 5+8=1 Probability of getting a male fish \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) \( =\frac{5}{13} \)

Q.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 Fig. ), 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 number of possible outcomes =8 (i) Probability of getting 8 \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }}=\frac{1}{8} \) (ii) Total number of odd numbers on spinner= 4 Probability of getting an odd number \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) \( =\frac{4}{8}=\frac{1}{2} \) (iii) The numbers greater than 2 are 3,4,5,6,7 and 8 Therefore , total numbers greater than 2 =6 Probability of getting a number greater than 2 \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }}=\frac{6}{8}=\frac{3}{4} \) (iv) The numbers less than 9 are 1,2,3,4,6,7 and 8 Therefore total numbers less than 9 =8 Probability of getting a number less than 9= 8/8=1

Q.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

Ans :The possible outcomes when a dice is thrown =\( \{1,2,3,4,5,6\} \) Number Of possible outcomes of a dice = 6 (i) Prime numbers on a dice are 2, 3, and 5. Total prime numbers on a dice= 3 Probability of getting a prime number =\(\frac{3}{6}=\frac{1}{2} \) (ii) Numbers lying between 2 and 6 =3,4,5 Total numbers lying between 2 and 6=3 Probability of getting a number lying between 2 and 6 \( =\frac{3}{6}=\frac{1}{2} \) (iii) Odd numbers on a dice= 1,3 and 5 Total odd numbers on a dice =3 Probability of getting an odd number \( =\frac{3}{6}=\frac{1}{2} \)

Q.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

Ans :(i) Total number of cards in a well-shuffled deck =52 Total number of kings of red colour =2 P (getting a king of red colour ) \(=\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) \( =\frac{2}{52}=\frac{1}{26} \) (ii) Total number of face cards =12 P(getting a face cards) \(=\frac{\text { Number of favourable outcomes. }}{\text { Number of total possible outcomes }} \) \(=\frac{12}{52}=\frac{3}{13} \) (iii) Total number of red face cards =6 P(getting a red face card) \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) \(=\frac{6}{52}=\frac{3}{26} \) (iv) Total number of Jack of hearts =1 p(getting a jack of hearts) \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) =1/52 (v) Total number of spade cards =13 P (getting a spade card) \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) =13/52 =1/4 (vi) Total numbers of queen of diamonds =1 P(getting a queen of diamond) \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) =1/52

Q.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?

Ans :Total numbers of cards =5 Total numbers of queens= 1 P(getting a queen) \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) \( =\frac{1}{5} \) (ii) When the queen is drawn and put aside, the total number of remaining cards will be 4 (a) Total number of aces=1 P(getting an ace) =¼ (b) As queen is already drawn, therefore, the number of queens will be 0. P(getting a queen) =0/4=0

Q.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

Ans :Total number of pens =12+132=144 Total number of good pens =132 P(getting a good pen) \( =\frac{\text { Number of favourable outcomes }}{\text { Number of total possible outcomes }} \) \( =\frac{132}{144}=\frac{11}{12} \)

Q.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 (iii) 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 ?

Ans :Total numbers of bulbs=20 Total number of defective bulbs=4 P(getting a defective bulb) \( =\frac{\text { Number of favourable outcomes. }}{\text { Number of total possible outcomes }} \) \(=\frac{4}{20}=\frac{1}{5} \) (ii) Remaining total number of bulbs=19 Remaining total number of non-defective bulbs=16-1=15 P(getting a not defective bulb) =15/19

Q.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.

Ans :Total number of discs=90 (i) Total number of two digit numbers between 1 and 90 =81 P(getting a two digit number) \( =\frac{81}{90}=\frac{9}{10} \) (ii) Perfect squares between 1 and 90 are 1,4,9,16,25,36,49,64, and 81. Therefore, total number of perfect squares between 1 and 90 is 9 P(getting a perfect square) \( =\frac{9}{90}=\frac{1}{10}\) (iii) Numbers that are between 1 and 90 and divisible by 5 are 5,10,15,20,25,30,35,40,45,50,55,60,65,70,80,85, and 90.Therefore a total numbers divisible by 5 \( =\frac{18}{90}=\frac{1}{5} \)

Q.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?

Ans :Total number of possible outcomes on the dice=6 (i) Total number of faces having A on it =2 P(getting A) \( =\frac{2}{6}=\frac{1}{3} \) (ii) Total number of faces having D on it = 1 P(getting D) =1/6

Q.20:Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?

Ans :Area of rectangle \( =l \times b=3 \times 2=6 \mathrm{m}^{2} \) Area of circle (of diameter 1 m) \( =\pi r^{2}=\pi\left(\frac{1}{2}\right)^{2}=-\frac{1}{4} \mathrm{m}^{2} \) P (die will land inside the circle) \( =\frac{\frac{\pi}{4}}{6}=\frac{\pi}{24} \)

Q.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 ?

Ans :Total number of pens 144 Total number of defective pens = 20 Total number of good pens \( 144-20=124 \) (i) Probability of getting a good pen = \(=\frac{124}{144}=\frac{31}{36} \) P (Nuri buys a pen) \( =\frac{31}{36} \) (ii) P (Nuri will not buy a pen) \( =1-\frac{31}{36}=\frac{5}{36} \)

Q.22:Refer to Example 13. (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 1/11. Do you agree with this argument? Justify your answer.

Ans :(i) It can be observed that, To get the sum as 2, possible outcomes \( =(1,1)\) To get the sum as 3, possible outcomes \( =(2,1) \text { and }(1,2) \) To get the sum as 4, possible outcomes\( =(3,1),(1,3),(2,2) \) To get the sum as 5, possible outcomes\( =(4,1),(1,4),(2,3),(3,2) \) To get the sum as 6, possible outcomes\( =(5,1),(1,5),(2,4),(4,2) (3,3) \) To get the sum as 7, possible outcomes\(=(6,1),(1,6),(2,5),(5,2)(3,4),(4,3) \) To get the sum as 8, possible outcomes\(=(6,2),(2,6),(3,5),(5,3) (4,4) \) To get the sum as 9, possible outcomes\( =(3,6),(6,3),(4,5),(5,4) \) To get the sum as 10, possible outcomes \(=(4,6),(6,4),(5,5) \) To get the sum as 11, possible outcomes \( =(5,6),(6,5) \) To get the sum as 12, possible outcomes = (6, 6) (ii) Probability of each of these sums will not be 1/11 as these sums are not equally likely

Q.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

Ans :The possible outcomes are {HHH, TTT, HHT , HTH , THH, TTH ,THT, HTT} Number of total possible outcomes=8 Numbers of favourable outcomes= 2 { i.e TTT and HHH} P ( Hanif will win the game) \( =\frac{2}{8}=\frac{1}{4} \) P(Hanif will lose the game ) \( =1-\frac{1}{4}=\frac{3}{4} \)

Q.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]

Ans :Total number of outcomes = 6 x 6 =>36 (i) Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2), (5,3) ,\( (5,4),(5,5),(5,6),(1,5),(2,5),(3,5),(4,5),(6,5) \) Hence, total number of favourable cases 11 P (5 will come up either time) \( =\frac{11}{36} \) P (5 will not come up either time) \( =\frac{11}{36} \) (ii)Total number of cases, when 5 can come at least once =11 P (5 will come at least once) \( =\frac{11}{36} \)

Q.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 .

Ans :(i) Incorrect When two coins are tossed, the possible outcomes are (H, H), (H, T), (T, H), and (T, T). It can be observed that there can be one Of each in two possible ways — (H, T), (T,H) Therefore, the probability of getting two heads is 4 , the probability of getting two tails is 1/4 , and the probability of getting one of each is 1/2 It can be observed that for each outcome, the probability is not 1/3 (ii) Correct When a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5, and 6. Out of these, 1, 3, 5 are odd and 2, 4, 6 are even numbers. Therefore the probability of getting an odd number is ½

Q.1:Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?

Ans :There are a total of 5 days. Shyam can go to the shop in 5 ways and Ekta can go to the shop in 5 ways. Therefore, total number Of outcomes = 5 x 5 = 25 (i) They can reach on the same day in 5 ways. i.e., (t, t), (w, w), (th, th), (f, f), (s, s) P (both will reach on same day) \( =\frac{5}{25}=\frac{1}{5} \) (ii) They can reach on consecutive days n these 8 ways - (t, w), (w, th), (th, f), (f, s), (w, t), (th, w), (f, th), (s, f) Therefore, P (both will reach on consecutive days) \( =\frac{8}{25} \) (iii) P (both will reach on same day) \( =\frac{1}{5} \) (From (i)] P (both will reach on different days) \( =1-\frac{1}{5}=\frac{4}{5} \)

Q.2:A die is numbered in such a way that ts faces show the number 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?

Ans :Total number of possible outcomes when two dice are thrown =\( =6 \times 6=36 \) (i) Total times when the sum is even 18 P (getting an even number) \( =\frac{18}{36}=\frac{1}{2} \) (ii) Total times when the sum is 6 = 4 P (getting sum as 6) \(=\frac{4}{36}=\frac{1}{9} \) (iii) Total times when the sum is at least 6 (i.e., greater than 5) =15 P (getting sum at least 6) \( =\frac{15}{36}=\frac{5}{12} \)

Q.3:A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag

Ans :Let the number of blue balls be x Number of red balls 5 Total number of balls = x + 5 P (getting a red ball) \( =\frac{5}{x+5} \) P (getting a blue ball) \( =\frac{x}{x+5} \) Given that, \( 2\left(\frac{5}{x+5}\right)=\frac{x}{x+5} \) \( 10(x+5)=x^{2}+5 x \) \( x^{2}-5 x-50=0 \) \( x^{2}-10 x+5 x-50=0 \) \( x(x-10)+5(x-10)=0 \) Either \(x-10=0 \text { or } x+5=0 \) \( x=10 \text { or } x=-5 \) However, the number of balls cannot be negative. Hence, number of blue balls =10

Q.4:A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x

Ans :Total number of balls =12 Total number Of black balls = x P (getting a black ball) \( =\frac{x}{12} \) If 6 more black balls are put in the box, then Total number of balls 12 + 6 18 Total number of black balls \( =x+6 \) P (getting a black ball now) \( =\frac{x+6}{18} \) According to the condition given in the question, \( 2\left(\frac{x}{12}\right)=\frac{x+6}{18} \) \( 3 x=x+6 \) \( 2 x=6 \) \( x=3 \)

Q.5:A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2 /3 ⋅ Find the number of blue balls in the jar.

Ans :Total number of marbles = 24 Let the total number of green marbles be x. Then, total number of blue marbles =24-x P (getting a given marble) \( =\frac{x}{24} \) According to the condition given in the question, \( \frac{x}{24}=\frac{2}{3} \) \( x=16 \) Therefore, total number of green marbles in the jar = 16 Hence, total number of blue marbles \( =24-x=24-16=8 \)

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