NCERT Solutions class 10 maths chapter 15

Probability

NCERT Solutions class 10 maths chapter 15 Exercise 15.1

Question 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 __________.

Answer

(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

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.

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

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

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

Answer

(i) It is not have equally likely outcomes as the car may or may not start depending upon various factors like fuel or whether car state or not, etc.

(ii) this statement does not have equally likely outcomes as it depends the player's ability or player may shoot or miss the shot.

(iii) This statement has equally likely.

(iv) This statement also has 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?

Answer

Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 head or tail.So tossing is unpredictable and is considered to be completely fair.

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

the above option (B) -1.5 cannot be the probability of an event.As probability lie between 1 and 0

Question 5

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

Answer

Here We know that,

P(E)+P(not E) = 1

P(E) = 0.05

P(not E) = 1-P(E)

P(not E) = 1-0.05

∴ P(not E) = 0.95

Question 6

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) Number of orange flavored candies in the = 0

∴ The probability of orange flavored candies = 0/1 = 0

(ii) As only lemon flavored candies in the bag , P(lemon flavored candies) = 1 (or 100%)

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?

Answer

Given, P(E) = 0.992

We know that

P(E)+P(not E) = 1

Or, P(not E) = 1–0.992 = 0.008

∴ The probability that the 2 students have the same birthday is 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?

Answer

the total Number of balls = 5+3 = 8

P(E) =
(i) Probability of drawing red balls = P (red balls) = =
(ii) Probability of drawing black balls = P (black balls) = =

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?

Answer

The Total no. of balls = 5+8+4 = 17

P(E) =

(i) Total number of red balls = 5

P (red ball) = = 0.29

(ii) Total number of white balls = 8

P (white ball) = = 0.47
(iii) Total number of green balls = 4
P (green ball) = = 0.23
∴ P (not green) = 1-P(green ball) = 1-( ) = 0.77

Question 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?

Answer

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

(i) Total number of 50 p coin = 100

P (50 p coin) = 100/180 = 5/9 = 0.55

(ii) Total number of ₹5 coin = 10

P (₹5 coin) = 10/180 = 1/18 = 0.055

∴ P (not ₹5 coin) = 1-P (₹5 coin) = 1-0.055 = 0.945

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

Answer

The total number of fish in the tank = 5+8 = 13

Total number of male fish = 5

P (male fish) = 5/13 = 0.38

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. 15.5), 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?

Answer

Total number of possible outcomes = 8

(i) Total number of favourable events (i.e. 8) = 1

∴ P (of 8) = 1/8 = 0.125

(ii) Total number of odd numbers = 4 (1, 3, 5 and 7)

P (pointing at an odd number) = 4/8 = 1/2 = 0.5

(iii) Total numbers greater than 2 = 6 (3, 4, 5, 6, 7 and 8)

P (pointing at a number greater than 4) = 6/8 = 3/4 = 0.75

(iv) Total numbers less than 9 = 8 (1, 2, 3, 4, 5, 6, 7, and 8)

P (pointing at a number less than 9) = 8/8 = 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.

Answer

Total possible number when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)

i) Total number of prime numbers = 3 (2, 3 and 5)

P (getting a prime number) = 3/6 = 1/2 = 0.5

(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)

P (getting a number between 2 and 6) = 3/6 = 1/2 = 0.5

(iii) Total number of odd numbers = 3 (1, 3 and 5)

P (getting an odd number) = 3/6 = 1/2 = 0.5

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

Answer

Total number of possible outcomes = 52

(i) Total numbers of king of red colour = 2

P (king of red colour) = 2/52 = 1/26 = 0.038

(ii) Total numbers of face cards = 12

P (getting a face card) = 12/52 = 3/13 = 0.23

(iii) Total numbers of red face cards = 6

P ( red face cards ) = 6/52 = 3/26 = 0.11

(iv) Total numbers of jack of hearts = 1

P (getting a jack of hearts) = 1/52 = 0.019

(v) Total numbers of king of spade = 13

P (getting a king of spade ) = 13/52 = 1/4 = 0.25

(vi) Total numbers of queen of diamonds = 1

P (getting a queen of diamonds) = 1/52 = 0.019

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?

Answer

Total numbers of cards = 5

(i) Numbers of queen = 1

P (picking a queen) = 1/5 = 0.2

(ii) the total numbers of cards left after drawn queen is = 4

(a) Total numbers of ace = 1

P (picking an ace) = 1/4 = 0.25

(b) Total numbers of queen = 0

P (picking a queen) = 0/4 = 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.

Answer

Numbers of pens = defective pens + good pens

∴ Total number of pens = 132+12 = 144 pens

P(picking a good pen) = 132/144 = 11/12 = 0.916

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?

(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?

Answer

(i)Numbers of defective bulbs = 4

The total numbers of bulbs = 20

∴ Probability of getting a defective bulb = P (defective bulb) = 4/20 = 1/5= 0.2

(ii) the total numbers of bulbs left when 1 non-defective bulb is drawn is 9

So, the total numbers of events (or outcomes) = 19

Numbers of defective bulbs = 19-4 = 15

So, the probability that the bulb is not defective = 15/19 = 0.789

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.

Answer

The total numbers of discs = 90

(i) Total number of discs having two digit numbers = 81

P (bearing a two-digit number) = 81/90 = 9/10 = 0.9

(ii) perfect square numbers(1, 4, 9, 16, 25, 36, 49, 64 and 81) = 9

P (getting a perfect square number) = 9/90 = 1/10 = 0.1

(iii) Total numbers which are divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)

P (getting a number divisible by 5) = 18/90 = 1/5 = 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?

Answer

The total number of possible events = 6

(i) The total number of faces having A = 2

P (getting A) = 2/6 = 1/3 = 0.33

(ii) The total number of faces having D = 1

P (getting D) = 1/6 = 0.166

Question 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?

Answer

Lenght= 3m
Breadth =2m

So, the area of the rectangle = (3×2) m² = 6 m²

and,

The area of the circle = πr² = π(½)² m² = π/4 m² = 0.78

∴ The probability that die will land inside the circle = [(π/4)/6] = π/24 or, 0.78/6 = 0.13

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?

Answer

The total numbers of outcomes i.e. pens = 144

Given, numbers of defective pens = 20

∴ The numbers of non defective pens = 144-20 = 124

(i) P (buying) = 124/144 = 31/36 = 0.86

(ii) she will not buy them = 20

So, P (not buying) = 20/144 = 5/36 = 0.138

Question 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 Solution:.

Answer

If 2 dices are thrown, the possible events 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)

So, the total events: 6×6 = 36

(i)P(sum of 2) = 1/36

E (sum 3) = (1,2) and (2,1)

P(sum 3) = 2/36

Similarly,

E (sum 4) = (1,3), (3,1), and (2,2)

P (sum 4) = 3/36

E (sum 5) = (1,4), (4,1), (2,3), and (3,2)

P (sum 5) = 4/36

E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)

P (sum 6) = 5/36

E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)

P (sum 7) = 6/36

E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)

P (sum 8) = 5/36

E (sum 9) = (3,6), (6,3), (4,5), and (5,4)

P (sum 9) = 4/36

E (sum 10) = (4,6), (6,4), and (5,5)

P (sum 10) = 3/36

E (sum 11) = (5,6), and (6,5)

P (sum 11) = 2/36

E (sum 12) = (6,6)

So, P (sum 12) = 1/36

(ii) The argument is not correct as the possible outcomes is 36 and not 11.

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.

Answer

The total number of outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)

Total outcomes in which Hanif will lose the game = 6 (HHT, HTH, THH, TTH, HTT, THT)

P (losing the game) = 6/8 = 3/4= 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?

[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Answer

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)

So, the total number of outcome = 6×6 = 36

(i) Method 1:

P(5 comes in first throw) = 6/36,

P( comes in second throw) = 6/36 and

P(not B) = 5/6

So, P(not A) = 1-(6/36) = 5/6

∴ The required probability = (5/6)×(5/6) = 25/36

(ii) Number of events when 5 comes at least once = 11

∴ The r probability = 11/36

Question 25

Which of the following arguments are correct and which are not correct? Give reasons for your Solution:

(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

Answer

(i) All the possible events are (H,H); (H,T); (T,H) and (T,T)

So, P ( heads) = 1/4= 0.25

and, P (getting one of the each) = 2/4 = 1/2 = 0.5

∴ This statement is incorrect.

(ii)Since the two outcomes are equally likely,
this statement is correct.