Class 10 - Chapter 14 Archives – Mathemafia http://sh016.global.temp.domains/~mathemaf/category/class-10-chapter-14/ Mathematics is like music - Enjoy it!!! Mon, 28 Apr 2025 07:17:44 +0000 en-US hourly 1 https://wordpress.org/?v=7.0 https://mathemafia.com/wp-content/uploads/2020/08/cropped-M__Logo__SM-1-32x32.png Class 10 - Chapter 14 Archives – Mathemafia http://sh016.global.temp.domains/~mathemaf/category/class-10-chapter-14/ 32 32 Exercise 14.1 (NCERT) – Probability, Class 10 – All Solutions https://mathemafia.com/exercise-14-1-ncert-probability-class-10-all-solutions/ Fri, 21 Feb 2025 10:54:08 +0000 https://mathemafia.com/?p=12528 Complete solution of Exercise 14.1, Chapter 14, Probability of NCERT book of Mathematics for Class 10. The questions are solved in the way you should write in your final exams. Most of the questions have similar solution that makes this chapter really easy. You will also find a one-shot quick revision video along with an interesting probability quiz on this webpage.

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Exercise 14.1 (NCERT) - Probability, Class 10

Chapter 14 Exercise 14.1 Class 10

Below are the Quick links for all questions of Exercise 14.1, Probability, Class 10.

Click a link to view the solution of corresponding question.

Probabiity Quiz

If you are confidant about your class 10 Probability skills, then test yourself by attempting the quizzes shown below. The questions are taken from the NCERT class 10 Mathematics book. Good luck!!!

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Quick Exam Revision @2X 

Just before your final exam, you can watch this tutorial, already created @2x speed, to quickly revise complete exercise 14.1, Probability. Best of luck!!!

Below are the solutions of every question of exercise 14.1 of chapter 14, Probability, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.  

Question 1

Complete the following statements:

(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.

(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.

Answer: Not equally likely

Reason: Starting of a car depends upon many factors. For example:

  • Petrol in the tank
  • Condition of the battery
  • Wiring
  • Weather

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

Answer: Not equally likely

Reason: This event also depends upon many factors. For example:

  • How far the basket is?
  • Skills of the player
  • Current form of the player

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

Answer: Equally likely

Reason: This is a bit tricky question for students to understand but it is really easy actually. A student has to answer either True or False just by tossing the coin. For example, Heads for True and Tails for False. The probability for answering True or False is 1/2 each. Hence it is equally likely. 

(iv) A baby is born. It is a boy or a girl.

Answer: Equally likely

Question 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?

A coin has only 2 sides – Head and Tail.

The probability of both is 1/2 hence both are equally likely. This gives equal chance to win the toss.

Question 4

Which of the following cannot be the probability of an event?

(A) 2/3 (B) –1.5 (C) 15% (D) 0.7

Answer: (B)

Reason: The probability of an event can never be less than 0. 

Imp: 15% is actually 15/100 = 0.15

Question 5

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

𝑃(𝐸)+𝑃(𝑛𝑜𝑡  𝐸)⇒𝑃(𝑛𝑜𝑡  𝐸)=1=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?

(i) P(an orange flavoured candy) = 0

Reason: There is no orange flavoured candy in the bag.

(i) P(a lemon flavoured candy) = 1

Reason: Every candy is a lemon flavoured candy. So whichever candy is taken out will be a lemon flavoured candy.

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?

Total no: of students = 3

Let Event E = 2 students not having same birthday

=> Event not E = 2 students have same birthday

Given P(E) = 0.992

𝑃(𝐸)+𝑃(𝑛𝑜𝑡  𝐸)⇒𝑃(𝑛𝑜𝑡  𝐸)=1=1−𝑃(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?

No: of Red balls = 3

No: of Black balls = 5

Total no: of balls = 3 + 5 = 8

One ball is drawn at random from the bag.

𝑃(𝐑𝐞𝐝 Ball)=Total no. of 𝐑𝐞𝐝 ballsTotal no. of balls=38
𝑃(𝐁𝐥𝐚𝐜𝐤 Ball)=Total no. of 𝐁𝐥𝐚𝐜𝐤 ballsTotal no. of balls=58

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?

No: of Red marbles = 5

No: of  White  balls = 8

No: of Green balls = 4

Total no: of balls = 5 + 8 + 4 = 17

One marble is taken out at random from the box.

(i) red 

𝑃(𝐑𝐞𝐝)=Total no. of 𝐑𝐞𝐝 ballsTotal no. of balls=517

(ii) white 

𝑃(𝐖𝐡𝐢𝐭𝐞)=Total no. of 𝐖𝐡𝐢𝐭𝐞 ballsTotal no. of balls=817

(iii) not green

Method 1

𝑃(𝐆𝐫𝐞𝐞𝐧)=Total no. of 𝐆𝐫𝐞𝐞𝐧 ballsTotal no. of balls=417
𝑃(Green)+𝑃(Not Green)⇒𝑃(Not Green)=1=1–𝑃(Green Ball)=1−417=1317

Method 2

𝑃(𝐍𝐨𝐭 𝐆𝐫𝐞𝐞𝐧)=Total no. of 𝐑𝐞𝐝 & 𝐖𝐡𝐢𝐭𝐞 ballsTotal no. of balls=5+817=1317

Question 10

A piggy bank contains 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, what is the probability that the coin

(i) will be a 50 p coin ?

(ii) will not be a Rs. 5 coin?

No: of 50p coins = 100

No: of Rs. 1 coins = 50

No: of Rs. 2 coins = 20

No: of Rs. 5 coins = 10

Total no: of coins = 100 + 50 + 20 + 10 =180

One coin fallen out of piggy bank.

(i) will be a 50 p coin ?

𝑃(50𝐩 𝐜𝐨𝐢𝐧)=Total no. of 50𝐩 coinsTotal no. of coins=100180=1018=59

(ii) will not be a Rs. 5 coin?

𝑃(𝐑𝐬. 5 𝐜𝐨𝐢𝐧)𝑃(𝐍𝐨𝐭 𝐑𝐬. 5 𝐜𝐨𝐢𝐧)=Total no. of 𝐑𝐬. 5 coinsTotal no. of coins=10180=118=1–𝑃(𝐑𝐬. 5 𝐜𝐨𝐢𝐧)=1−118=1718

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 Fig. 14.4). What is the probability that the fish taken out is a male fish?

No: of male fish = 5

No: of female fish = 8

Total no: of fish = 5 + 8 = 13

One fish is taken out at random from the tank.

𝑃(𝐌𝐚𝐥𝐞 𝑓𝑖𝑠ℎ)=Total no. of 𝐌𝐚𝐥𝐞 fishTotal no. of fish=513

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 Fig. 14.5 ), 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?

Total Numbers = 8

Total no. of 8s = 1

Total no. of Odd number (1,3,5,7) = 4

Total no. Greater than 2 (3,4,5,6,7,8) = 6

Total no. Less than 9 (All) = 8

Spinning the arrow will stop it randomly at one of the 8 numbers.

𝑃(𝐍𝐮𝐦𝐛𝐞𝐫 𝐢𝐬 8)𝑃(𝐀𝐧 𝐨𝐝𝐝 𝐧𝐮𝐦𝐛𝐞𝐫)𝑃(𝐍𝐮𝐦𝐛𝐞𝐫 > 2)𝑃(𝐍𝐮𝐦𝐛𝐞𝐫 <9)=Total no. of 8𝐬Total Numbers=18=Total no. of 𝐎𝐝𝐝 𝐍𝐮𝐦𝐛𝐞𝐫𝐬Total Numbers=48=12=Count of numbers >2Total Numbers=68=34=Count of numbers <9Total Numbers=88=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;

(iii) an odd number.

Throwing a dice has following outcomes:

1, 2, 3, 4, 5, 6

Total no. of outcomes = 6

Total no. of prime numbers (2, 3, 5) = 3

Count of numbers lying between 2 & 6 (3, 4, 5) = 3

Total no. of odd numbers (1, 3, 5) = 3

𝑃(𝐏𝐫𝐢𝐦𝐞 𝐧𝐮𝐦𝐛𝐞𝐫)𝑃(𝐍𝐮𝐦𝐛𝐞𝐫 𝐛/𝐰 2 & 6)𝑃(𝐎𝐝𝐝 𝐧𝐮𝐦𝐛𝐞𝐫)=Total no. of 𝐏𝐫𝐢𝐦𝐞 𝐍𝐮𝐦𝐛𝐞𝐫𝐬Total number of outcomes=36=12=Count of numbers 𝐛/𝐰 2 & 6Total number of outcomes=36=12=Total no. of 𝐎𝐝𝐝 𝐧𝐮𝐦𝐛𝐞𝐫𝐬Total number of outcomes=36=12

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

Total no: of cards = 52

One card is taken out at random from the deck.

(i) a king of red colour

Total no: of Red Kings = 2

(1 Heart, 1 Diamond)

𝑃(𝐊𝐢𝐧𝐠 𝐨𝐟 𝐫𝐞𝐝 𝐜𝐨𝐥𝐨𝐮𝐫)=Total no. of 𝐑𝐞𝐝 𝐊𝐢𝐧𝐠𝐬Total no. of cards=252=126

(ii) a face card

Total no: of face cards = 3 x 4 = 12

(4 Kings, 4 Queens, 4 Jacks)

𝑃(𝐀 𝐟𝐚𝐜𝐞 𝐜𝐚𝐫𝐝)=Total no. of 𝐅𝐚𝐜𝐞 𝐜𝐚𝐫𝐝𝐬Total no. of cards=1252=313

(iii) a red face card

Total no: of red face cards = 2 x 3 = 6

Heart and Diamond cards are red cards.

(2 Red Kings, 2 Red Queens, 2 Red Jacks )

𝑃(𝐀 𝐫𝐞𝐝 𝐟𝐚𝐜𝐞 𝐜𝐚𝐫𝐝)=Total no. of 𝐑𝐞𝐝 𝐟𝐚𝐜𝐞 𝐜𝐚𝐫𝐝𝐬Total no. of cards=652=326

(iv) the jack of hearts

Total no: of jack of hearts = 1

𝑃(𝐉𝐚𝐜𝐤 𝐨𝐟 𝐡𝐞𝐚𝐫𝐭𝐬)=Total no. of 𝐉𝐚𝐜𝐤𝐬 𝐨𝐟 𝐡𝐞𝐚𝐫𝐭𝐬Total no. of cards=152

(v) a spade

Total no: of spade cards = 13

𝑃(𝐀 𝐬𝐩𝐚𝐝𝐞)=Total no. of 𝐒𝐩𝐚𝐝𝐞 𝐜𝐚𝐫𝐝𝐬Total no. of cards=1352=14=0.25

(vi) the queen of diamonds

Total no: of queens of diamonds = 1

𝑃(𝐐𝐮𝐞𝐞𝐧 𝐨𝐟 𝐝𝐢𝐚𝐦𝐨𝐧𝐝𝐬)=Total no. of 𝐐𝐮𝐞𝐞𝐧𝐬 𝐨𝐟 𝐝𝐢𝐚𝐦𝐨𝐧𝐝𝐬Total no. of cards=152

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?

(i) What is the probability that the card is the queen?

Total no. of cards = 5

Total no. of Tens = 1

Total no. of Jacks = 1

Total no. of Queens = 1

Total no. of Kings = 1

Total no. of Aces = 1

𝑃(𝐂𝐚𝐫𝐝 𝐢𝐬 𝐭𝐡𝐞 𝐐𝐮𝐞𝐞𝐧)=Total no. of 𝐐𝐮𝐞𝐞𝐧𝐬Total no. of cards=15

(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?

Queen has been removed from the cards. Now we have:

Total no. of cards = 4

Total no. of Tens = 1

Total no. of Jacks = 1

Total no. of Queens = 0

Total no. of Kings = 1

Total no. of Aces = 1

𝑃(𝐂𝐚𝐫𝐝 𝐢𝐬 𝐚𝐧 𝐀𝐜𝐞)=Total no. of 𝐀𝐜𝐞𝐬Total no. of cards=14
𝑃(𝐂𝐚𝐫𝐝 𝐢𝐬 𝐭𝐡𝐞 𝐐𝐮𝐞𝐞𝐧)=Total no. of 𝐐𝐮𝐞𝐞𝐧𝐬Total no. of cards=04=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.

No: of Defective pens = 12

No: of Good pens = 132

Total no: of pens = 12 + 132 = 144

One pen is taken out at random from the lot.

𝑃(𝐆𝐨𝐨𝐝 𝐩𝐞𝐧)=Total no. of 𝐆𝐨𝐨𝐝 𝐩𝐞𝐧𝐬Total no. of cards=132144=3336=1112

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?

Total No: of bulbs = 20

No: of Defective bulbs = 4

No: of Non-Defective bulbs = 20 – 4 = 16

One bulb is taken out at random from the lot.

𝑃(𝐃𝐞𝐟𝐞𝐜𝐭𝐢𝐯𝐞 𝐛𝐮𝐥𝐛)=Total no. of 𝐃𝐞𝐟𝐞𝐜𝐭𝐢𝐯𝐞 𝐛𝐮𝐥𝐛𝐬Total no. of bulbs=420=15=0.2

Question 17 (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 ?

After removing 1 Non-Defective bulb from the lot. 

Total No: of bulbs = 19

No: of Defective bulbs = 4

No: of Non-Defective bulbs = 15

One bulb is taken out at random from the lot of 19 bulbs.

𝑃(𝐍𝐨𝐧-𝐃𝐞𝐟𝐞𝐜𝐭𝐢𝐯𝐞 𝐛𝐮𝐥𝐛)=Total no. of 𝐍𝐨𝐧-𝐃𝐞𝐟𝐞𝐜𝐭𝐢𝐯𝐞 𝐛𝐮𝐥𝐛𝐬Total no. of bulbs=1519

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.

Total No: of discs = 90

Single digit discs (1-9)  = 9

Double digit discs (10-90)  = 81

One disc is drawn at random from the box.

(i) a two-digit number

𝑃(𝐓𝐰𝐨-𝐝𝐢𝐠𝐢𝐭 𝐧𝐮𝐦𝐛𝐞𝐫)=Total no. of 𝐓𝐰𝐨-𝐝𝐢𝐠𝐢𝐭 𝐝𝐢𝐬𝐜𝐬Total no. of discs=8190=910=0.9

(ii) a perfect square number

Perfect squares from 1 to 90 are:

12, 22, 32, 42, 52, 62, 72, 82, 92 i.e.

1, 4, 9, 16, 25, 36, 49, 64, 81

Total no: discs with perfect squares = 9

𝑃(𝐏𝐞𝐫𝐟𝐞𝐜𝐭 𝐬𝐪𝐮𝐚𝐫𝐞 𝐧𝐮𝐦𝐛𝐞𝐫)=Total no. of 𝐩𝐞𝐫𝐟𝐞𝐜𝐭 𝐬𝐪𝐮𝐚𝐫𝐞 𝐧𝐮𝐦𝐛𝐞𝐫𝐞𝐝 𝐝𝐢𝐬𝐜𝐬Total no. of discs=990=110=0.1

(iii) a number divisible by 5.

Numbers divisible by 5 ranging from 1 to 90 are:

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90

Total no: of numbers divisible by 5 = 18

𝑃(𝐍𝐮𝐦𝐛𝐞𝐫 𝐝𝐢𝐯𝐢𝐬𝐢𝐛𝐥𝐞 𝐛𝐲 5)=Total no. of 5𝐱 𝐧𝐮𝐦𝐛𝐞𝐫𝐞𝐝 𝐝𝐢𝐬𝐜𝐬Total no. of discs=1890=15=0.2

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?

Total no. of faces = 6

Total no. of faces with A = 2

Total no. of faces with D = 1

𝑃(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐚𝐧 𝐀)𝑃(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐚 𝐃)=Total no. of 𝐟𝐚𝐜𝐞𝐬 𝐰𝐢𝐭𝐡 𝐀Total no. of faces=26=13=Total no. of 𝐟𝐚𝐜𝐞𝐬 𝐰𝐢𝐭𝐡 𝐃Total no. of faces=16

Question 20

Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter 1m?

𝐀𝐫𝐞𝐚 𝐨𝐟 𝐫𝐞𝐜𝐭𝐚𝐧𝐠𝐮𝐥𝐚𝐫 𝐫𝐞𝐠𝐢𝐨𝐧=𝑙×𝑏=3×2=6𝑚2

Diameter of circle = 1 m

=> Radius of the circle (r) = 1/2 m 

𝐀𝐫𝐞𝐚 𝐨𝐟 𝐜𝐢𝐫𝐜𝐮𝐥𝐚𝐫 𝐫𝐞𝐠𝐢𝐨𝐧𝑃(𝐃𝐢𝐞 𝐥𝐚𝐧𝐝𝐬 𝐢𝐧 𝐜𝐢𝐫𝐜𝐥𝐞)=𝜋𝑟2=𝜋×(12)2=𝜋4𝑚2=𝐀𝐫𝐞𝐚 𝐨𝐟 𝐜𝐢𝐫𝐜𝐮𝐥𝐚𝐫 𝐫𝐞𝐠𝐢𝐨𝐧𝐀𝐫𝐞𝐚 𝐨𝐟 𝐫𝐞𝐜𝐭𝐚𝐧𝐠𝐮𝐥𝐚𝐫 𝐫𝐞𝐠𝐢𝐨𝐧=𝜋4×16=𝜋24

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 ?

Total No: of ball pens = 144

No: of Defective pens = 20

No: of Good pens = 144 – 20 = 124 

Nuri will buy only a good pen.

Shopkeeper draws a pen at random from the lot.

(i) She will buy it ?

𝑃(𝐒𝐡𝐞 𝐰𝐢𝐥𝐥 𝐛𝐮𝐲 𝐢𝐭)=Total no. of 𝐆𝐨𝐨𝐝 𝐩𝐞𝐧𝐬Total no. of pens=124144=6272=3136

(ii) She will not buy it ?

Method 1

𝑃(𝐒𝐡𝐞 𝐰𝐢𝐥𝐥 𝐧𝐨𝐭 𝐛𝐮𝐲 𝐢𝐭)=Total no. of 𝐃𝐞𝐟𝐞𝐜𝐭𝐢𝐯𝐞 𝐩𝐞𝐧𝐬Total no. of pens=20144=1072=536

Method 2

𝑃(𝐒𝐡𝐞 𝐰𝐢𝐥𝐥 𝐛𝐮𝐲 𝐢𝐭)=1−𝑃(𝐒𝐡𝐞 𝐰𝐢𝐥𝐥 𝐧𝐨𝐭 𝐛𝐮𝐲 𝐢𝐭)=1−3136=536

Question 22 (i)

Refer to Example 13. (i) Complete the following table:

IMP: Create the table as created in the video embedded below.

Event, Sum of 2 dice 2 3 4 5 6 7 8 9 10 11 12
Probability
1/36
5/36
1/36

Below is the complete table:

Event, Sum of 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

Question 22 (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.

(ii) As we can see in the table displayed above, the probabilities of every number is not 1/11. Hence the student’s argument is not correct.

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.

Following are the possibilities of tossing a coin 3 times 

HHH

HHT

HTH

THH

TTH

THT

HTT

TTT

𝑃(𝐇𝐚𝐧𝐢𝐟 𝐥𝐨𝐬𝐞𝐬 𝐭𝐡𝐞 𝐠𝐚𝐦𝐞)=No. of outcomes other than 𝐇𝐇𝐇, 𝐓𝐓𝐓Total no. of outcomes=68=34=0.75

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?

IMP: Create the table as created in the video embedded below.

Total no. of pairs = 36

Pairs with either one 5 = 11

Pairs with no 5 = 25

(i) 5 will not come up either time

𝑃(5 𝐰𝐢𝐥𝐥 𝐧𝐨𝐭 𝐜𝐨𝐦𝐞 𝐮𝐩 𝐞𝐢𝐭𝐡𝐞𝐫 𝐭𝐢𝐦𝐞)=Total no. of pairs with 𝐧𝐨 5Total no. of pairs=2536

(ii) 5 will come up at least once

Method 1

𝑃(5 𝐜𝐨𝐦𝐞𝐬 𝐮𝐩 𝐚𝐭 𝐥𝐞𝐚𝐬𝐭 𝐨𝐧𝐜𝐞)=Total no. of pairs with 𝐚𝐭 𝐥𝐞𝐚𝐬𝐭 𝐨𝐧𝐞 5Total no. of pairs=1136

Method 2

𝑃(5 𝐜𝐨𝐦𝐞𝐬 𝐮𝐩 𝐚𝐭 𝐥𝐞𝐚𝐬𝐭 𝐨𝐧𝐜𝐞)=1–𝑃(5 𝐰𝐢𝐥𝐥 𝐧𝐨𝐭 𝐜𝐨𝐦𝐞 𝐮𝐩 𝐞𝐢𝐭𝐡𝐞𝐫 𝐭𝐢𝐦𝐞)=1−2536=1136

Question 25 (i)

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.

This statement is False.

Reason: Tossing 2 coins simultaneously has 4 possible outcomes.

HH, HT, TH, TT

𝑃(2 𝐇𝐞𝐚𝐝𝐬)𝑃(2 𝐓𝐚𝐢𝐥𝐬)𝑃(1 𝐇𝐞𝐚𝐝, 1 𝐓𝐚𝐢𝐥)=14=14=24=12

Question 25 (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.

This statement is True.

Reason: Throwing a die always gives either an odd number (1, 3 or 5) or Even number (2, 4 or 6).

𝑃(𝐎𝐝𝐝 𝐍𝐮𝐦𝐛𝐞𝐫)𝑃(𝐄𝐯𝐞𝐧 𝐍𝐮𝐦𝐛𝐞𝐫)=36=12=36=12

Probability Quiz – Test your skills

If you have read the chapter thoroughly, you should be able to attempt the following quiz easily. Click the video embedded below to test your knowledge about concepts of Probability. Best of luck!!!

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The post Exercise 14.1 (NCERT) – Probability, Class 10 – All Solutions appeared first on Mathemafia.

]]> Class 10 | Chapter 14 | All Solutions | Probability https://mathemafia.com/class-10-chapter-14-all-solutions-probability/ Fri, 17 Dec 2021 07:44:02 +0000 http://sh016.global.temp.domains/~mathemaf/?p=7351 This playlist covers all the questions, concepts and examples of chapter 15 (Probability) of class 10, NCERT book. Click a thumbnail to watch the tutorial.  Watch complete chapter on YouTube

The post Class 10 | Chapter 14 | All Solutions | Probability appeared first on Mathemafia.

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This playlist covers all the questions, concepts and examples of chapter 15 (Probability) of class 10, NCERT book. Click a thumbnail to watch the tutorial. 

Watch complete chapter on YouTube

The post Class 10 | Chapter 14 | All Solutions | Probability appeared first on Mathemafia.

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