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Page No 619:

Question 1:

A coin is tossed 500 times and we get
head: 285 times, tail: 215 times.
When a coin is tossed at random, what is the probability of getting
(i) a head?
(ii) a tail?

Answer:

Total number of tosses =  500
Number of heads = 285
Number of tails = 215
(i) Let E be the event of getting a head.
   P(getting a head) = P (E) = Number of heads coming upTotal number of trials = 285500 = 0.57

(ii) Let F be the event of getting a tail.
   P(getting a tail) = P (F) = Number of tails coming upTotal number of trials = 215500 = 0.43

Page No 619:

Question 2:

Two coins are tossed 400 times and we get
two heads: 112 times; one head: 160 times; 0 head: 128 times.
When two coins are tossed at random, what is the probability of getting
(i) 2 heads?
(ii) 1 head?
(iii) 0 head?

Answer:

 Total number of tosses =  400
Number of times 2 heads appear = 112
Number of times 1 head appears = 160
Number of times 0 head appears = 128

In a random toss of two coins, let E1, E2, E3 be the events of getting 2 heads, 1 head and 0 head, respectively. Then,

(i) P(getting 2 heads) =  P(E1) = Number of times 2 heads appearTotal number of trials = 112400 = 0.28

(ii)  P( getting 1 head) =  P(E2) = Number of times 1 head appearsTotal number of trials = 160400 = 0.4

(iii) P( getting 0 head) =  P(E3) = Number of times 0 head appearsTotal number of trials = 128400 = 0.32

Remark: Clearly, when two coins are tossed, the only possible outcomes are E1E2 and E3 and P(E1) + P(E2) +  P(E3) = (0.28 + 0.4 + 0.32) = 1

Page No 619:

Question 3:

Three coins are tossed 200 times and we get
three heads: 39 times; two heads: 58 times;
one head: 67 times; 0 head: 36 times.
When three coins are tossed at random, what is the probability of getting
(i) 3 heads?
(ii) 1 head?
(iii) 0 head?
(iv) 2 heads?

Answer:

Total number of tosses =  200
Number of times 3 heads appear = 39
Number of times 2 heads appear = 58
Number of times 1 head appears = 67
Number of times 0 head appears = 36

In a random toss of three coins, let E1E2E3 and E4  be the events of getting 3 heads, 2 heads, 1 head and 0 head, respectively. Then;
(i) P(getting 3 heads) =  P(E1) = Number of times 3 heads appearTotal number of trials = 39200 = 0.195


(ii)  P(getting 1 head) =  P(E2) = Number of times 1 head appearsTotal number of trials = 67200 = 0.335

(iii) P(getting 0 head) =  P(E3) = Number of times 0 head appearsTotal number of trials = 36200 = 0.18

(iv) P(getting 2 heads) =  P(E4) = Number of times 2 heads appearTotal number of trials = 58200 = 0.29
 
Remark: Clearly, when three coins are tossed, the only possible outcomes are E1E2E3 and E4 and P(E1) + P(E2) +  P(E3) + P(E4) = (0.195 + 0.335 + 0.18 + 0.29) = 1



Page No 620:

Question 4:

A dice is thrown 300 times and the outcomes are noted as given below.

Outcome 1 2 3 4 5 6
Frequency 60 72 54 42 39 33
When a dice is thrown at random, what is the probability of getting a
(i) 3?
(ii) 6?
(iii) 5?
(iv) 1?

Answer:

Total number of throws =  300
In a random throw of a dice, let E1, E2, E3, E4, be the events of getting 3, 6, 5 and 1, respectively. Then,
(i) P(getting 3) = P(E1)  = Number of times 3 appearsTotal number of trials = 54300 = 0.18

(ii) P(getting 6) = P(E2)  = Number of times 6 appearsTotal number of trials = 33300 =0.11

(iii) P(getting 5) = P(E3)  = Number of times 5 appearsTotal number of trials = 39300 = 0.13

(iv) P(getting 1) = P(E4)  = Number of times 1 appearsTotal number of trials = 60300 = 0.20

Page No 620:

Question 5:

In a survey of 200 ladies, it was found that 142 like coffee, while 58 dislike it.
Find the probability that a lady chosen at random
(i) likes coffee
(ii) dislikes coffee

Answer:

Total number of ladies = 200
Number of ladies who like coffee = 142

Number of ladies who dislike coffee = 58

Let E1 and E2 be the events that the selected lady likes and dislikes coffee, respectively.Then,
(i) P(selected lady likes coffee) = P(E1) =  Number of ladies who like coffeeTotal number of ladies = 142200 = 0.71

(ii)​ (selected lady dislikes coffee) = P(E2) = â€‹Number of ladies who dislike coffeeTotal number of ladies = 58200 = 0.29


REMARK: In the given survey, the only possible outcomes are E1 and E2 and â€‹P(E1) + â€‹P(E2) = (0.71 + 0.29) = 1

Page No 620:

Question 6:

The percentages of marks obtained by a student in six unit tests are given below.

Unit test I II III IV V VI
Percentage of marks obtained 53 72 28 46 67 59
A unit test is selected at random. What is the probability that the student gets more than 60% marks in the test?

Answer:

Total number of unit tests = 6
Number of tests in which the student scored more than 60% marks = 2

Let E be the event that he got more than 60% marks in the unit tests.Then,
 
required probability = P(E) =  Number of unit tests in which he got more than 60% marksTotal number of unit tests = 26 = 13

Page No 620:

Question 7:

On a particular day, in a city, 240 vehicles of various types going past a crossing during a time interval were observed, as shown:

Types of vehicle Two-wheelers Three-wheelers Four-wheelers
Frequency 84 68 88
Out of these vehicles, one is chosen at random. What is the probability that the given vehicle is a two-wheeler?

Answer:

Total number of vehicles going past the crossing = 240
Number of two-wheelers = 84

Let E be the event that the selected vehicle is a two-wheeler. Then,
 
required probability = P(E) =  84240= 0.35Number of unit tests in which he got more than 60% marksTotal number of unit tests = 26 = 13

Page No 620:

Question 8:

On one page of a telephone directory, there are 200 phone numbers. The frequency distribution of their units digits is given below:

Units digit 0 1 2 3 4 5 6 7 8 9
Frequency 19 22 23 19 21 24 23 18 16 15
One of the numbers is chosen at random from the page. What is the probability that the units digit of the chosen number is
(i) 5?
(ii) 8?

Answer:

Total phone numbers on the directory page  = 200

(i) Number of numbers with units digit 5 = 24

     Let E1 be the event that the units digit of selected number is 5.
 
∴ Required probability = P(E1) =  24200 = 0.12

(ii) 
Number of numbers with units digit 8 = 16
     Let E2 be the event that the units digit of selected number is 8.
 
∴  Required probability = P(E2) =  16200 = 0.08

Page No 620:

Question 9:

The following table shows the blood groups of 40 students of a class.

Blood group A B O AB
Number of students 11 9 14 6
One student of the class is chosen at random. What is the probability that the chosen student's blood group is
(i) O?
(ii) AB?

Answer:

Total number of students  = 40
(i) Number of students with blood group O = 14

     Let E1 be the event that the selected student's blood group is O.
 
   ∴ Required probability = P(E1) =  1440 = 0.35

(ii)
 Number of students with blood group AB = 6
     Let E2 be the event that the selected student's blood group is AB.
   ∴  Required probability = P(E2) =  640 = 0.15



Page No 621:

Question 10:

The table given below shows the marks obtained by 30 students in a test.

Marks
(Class interval)
Number of students (Frequency)
1−10 7
11−20 10
21−30 6
31−40 4
41−50 3
One of these students is chosen at random. What is the probability that his marks lie in the interval 2130?

Answer:

   Total number of students  = 30
   Number of students whose marks lie in the interval 21-30 = 6

   Let E be the event that the selected student's marks lie in the interval 21- 30 .

 ∴ Required probability = P(E) =  630 = 15 = 0.2

Page No 621:

Question 11:

Following are the ages (in years) of 360 patients, getting medical treatment in a hospital:

Age (in years) 10−20 20−30 30−40 40−50 50−60 60−70
Number of patients 90 50 60 80 50 30
One of the patients is selected at random.
Find the probability that his age is
(i) 30 years or more but less than 40 years
(ii) 50 years or more but less than 70 years
(iii) less than 10 years
(iv) 10 years or more

Answer:

Total number of patients  = 360
(i) Number of patients whose age is 30 years or more but less than 40 years = 60

     Let E1 be the event that the selected patient's age is in between 30 - 40.
 
   ∴ P(patient's age 30 is years or more but less than 40 years) = P(E1) =  60360 = 16

(ii)
  Number of patients whose age is 50 years or more but less than 70 years = (50 +30) = 80
     Let E2 be the event that the selected patient's age is in between 50 - 70.
   ∴ P(patient's age is 50 years or more but less than 70 years) = P(E2) =  80360 = 29

(iii) Number of patients whose age is less than 10 years = 0
     Let E3 be the event that the selected patient's age is less than 0.
 
   ∴ P(patient's age is less than 10 years)= P(E3) =  0360 =0

(iv) Number of patients whose age is 10 years or more = 90 + 50 + 60 + 80 + 50 + 30 =  360
     Let E4 be the event that the selected patient's age is 10 years or more. Then
   ∴ P(patient's age is 10 years or more) = P(E4) =  360360 = 1



Page No 623:

Question 1:

A coin is tossed 100 times with the following outcomes:
head 43 times and tail 57 times.
In a single throw of a coin, what is the probability of getting a head?
(a) 4357

(b) 5743

(c) 43100

(d) 750

Answer:

(c) 43100

Explanation:
Total number of trials =  100
Number of heads = 43
Number of tails = 57
Let E be the event of getting a head.
 P(getting a head) = P (E) = Number of heads coming upTotal number of trials = 43100 

Page No 623:

Question 2:

A coin is tossed 200 times with the following outcomes:
head 112 times and tail 88 times.
What is the probability of getting a tail in a single throw of a coin ?
(a) 1125

(b) 1425

(c) 1114

(d) 1411

Answer:

(a) 1125

Explanation:
Total number of trials = 200
Number of heads = 112
Number of tails = 88
Let E be the event of getting a tail.
   P(getting a tail) = P (E) = Number of tails coming upTotal number of trials = 88200 = 1125

Page No 623:

Question 3:

A survey of 200 persons of a locality shows the like and dislike for tea.

No. of persons who like tea 148
No. of persons who dislike tea 52
Out of these persons one was chosen at random. What is the probability that the chosen person likes tea?
(a) 1337

(b) 3713

(c) 1350

(d) 3750

Answer:

(d) 3750

Explanation: 
Total number of persons = 200
Number of persons who like tea = 148

Let E be the event that the selected person likes tea. Then,

 P(E) =  Number of persons who like teaTotal number of persons = 148200 = 3750Number of ladies who like coffeeTotal number of ladies = 142200 = 0.71

Page No 623:

Question 4:

In a locality, 100 families were chosen at random and the following data was collected.

Number of children in each family 0 1 2 3 4 or more
Number of families 6 184 672 127 11
Out of these families, a family was chosen at random. What is the probability that the chosen family has 2 children?
(a) 1336

(b) 84125

(c) 41125

(d) 164375

Answer:

(b) 84125

Explanation: 
Total number of families = 1000
Number of families with 2 children = 672

Let E be the event that the chosen family has 2 children.Then,

 P(E) =  Number of families with 2 childrenTotal number of persons = 6721000 = 84125Number of ladies who like coffeeTotal number of ladies = 142200 = 0.71



Page No 624:

Question 5:

The table given below shows the month of birth of 36 students of a class.

Month of birth Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec
No. of students 4 3 5 0 1 6 1 3 4 3 4 2
A student is chosen at random from the class. What is the probability that the chosen student was born in October?
(a) 13

(b) 23

(c) 14

(d) 112

Answer:

(d) 112

Explanation: 
Total number of students = 36
Number of students born in October = 3

Let E be the event that the chosen student was born in October. Then,

 P(E) =  Number of students born in OctoberTotal number of students = 336 = 112

Page No 624:

Question 6:

In 50 tosses of a coin, tail appears 32 times. If a coin is tossed at random, what is the probability of getting a head?
(a) 132

(b) 118

(c) 1625

(d) 925

Answer:

(d) 925

Explanation:
Total number of tosses = 50
Number of times tail appears = 32
So, the number of times head appears = 50 − 32 = 18
 Let E be the event of getting a head.
 P(getting a head) = P (E) = Number of heads coming upTotal number of trials = 1850 =925

Page No 624:

Question 7:

In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. What is the probability that in a given delivery, the ball does not hit the boundary?
(a) 14

(b) 15

(c) 45

(d) 34

Answer:

(c) 45

Explanation: 
Total number of balls faced = 30
Number of times the ball hits the boundary = 6

Number of times the ball does not hit the boundary = (30 − 6 )= 24

Let E be the event that the ball does not hit the boundary. Then,

 P(E) =  Number of times ball does not hit the boundaryTotal number of balls = 2430 = 45

Page No 624:

Question 8:

A dice is thrown 40 times and each time the number on the uppermost face is noted as shown:

Outcome 1 2 3 4 5 6
Number of times 5 6 8 10 7 6
A dice is thrown at random. What is the probability of getting a 5?
(a) 57

(b) 75

(c) 18

(d) 740

Answer:

(d) 740

Explanation:
Total number of trials =  40
Number of times 5 appears = 7
P(getting 5) =  Number of times 5 appearsTotal number of trials = 740 

Page No 624:

Question 9:

In 50 throws of a dice, the outcomes were noted as shown below:

Outcome 1 2 3 4 5 6
Number of times 8 9 6 7 12 8
A dice is thrown at random. What is the probability of getting an even number?
(a) 1225

(b) 350

(c) 18

(d) 12

Answer:

(a) 1225

Explanation:
Total number of trials =  50
Let E be the event of getting an even number.
Then, E contains 2, 4 and 6, i.e. 3 even numbers.

∴ P(getting an even number) = P(E)  = Number of times even numbers appearTotal number of throws = (9+7+8)50 = 2450 =1225



Page No 625:

Question 10:

The outcomes of 65 throws of a dice were noted as shown below:

Outcome 1 2 3 4 5 6
Number of times 8 10 12 16 9 10
A dice is thrown at random. What is the probability of getting a prime number?
(a) 335

(b) 35

(c) 3165

(d) 3665

Answer:

(c) 3165

Explanation:
Total number of throws = 65
Let E be the event of getting a prime number. 
Then, E contains 2, 3 and 5, i.e. three numbers.

∴ P(getting a prime number) = P(E)  = Number of times prime numbers occurTotal number of throws = (10+12+9)65 = 3165 

Page No 625:

Question 11:

On one page of a directory, there are 160 telephone numbers. The frequency distribution of the unit place digit is shown below:

Unit place digit 0 1 2 3 4 5 6 7 8 9
Frequency 19 16 18 21 14 11 15 16 13 17
From this page, one number is chosen at random. What is the probability that the units place digit of the chosen number is 6?
(a) 25

(b) 332

(c) 380

(d) 2932

Answer:

(b) 332

Explanation:
 Total phone numbers on a page of the directory  = 160
 Numbers with units digit 6 = 15

 Let E be the event that the units digit of selected number is 6.
∴ Required probability = P(E) =  15160 = 332
24200 = 0.12

Page No 625:

Question 12:

Two coins are tossed 1000 times and the outcomes are recorded as shown below:

Number of heads 2 1 0
Frequency 266 540 194
A coin is thrown at random. What is the probability of getting at most one head?
(a) 403500

(b) 2750

(c) 367500

(d) 97500

Answer:

(c)367500 

Explanation:
Total number of tosses = 1000
Number of times 2 heads appear = 266
Number of times 1 head appears = 540
Number of times 0 head appears = 194

If at most 1 head appears, then either 1 head appears or no heads appear = 540 + 194 = 734

 P(getting at most 1 head) = 7341000 =367500 

Page No 625:

Question 13:

80 bulbs are selected at random from a lot and their lifetime is recorded in the form of a frequency table given below:

Lifetime (in hours) 300 500 700 900 1100
Frequency 10 15 23 25 7
A bulb is chosen at random from the lot. What is the probability that the bulb chosen has a lifetime of less than 900 hours?
(a) 7380

(b) 35

(c) 516

(d) 2380

Answer:


(b) 35

Explanation:
 Total number of bulbs in the lot  = 80
 Number of bulbs with life time of less than 900 hours = (10 + 15 + 23) = 48

 Let E be the event that the chosen bulb's life time is less than 900 hours.
 
∴ Required probability = P(E) =  4880 = 35



Page No 626:

Question 14:

In a medical examination of 40 students of a class, the following blood groups were recorded:

Blood group A B AB O
No. of students 11 15 9 5
From this group, a student is chosen at random. What is the probability that the chosen student's blood group is B?
(a) 35

(b) 58

(c) 35

(d) 83

Answer:

(a) 38

Explanation:
Total number of students = 40
 Number of students with blood group B = 15

 Let E be the event that the selected student's blood group is B.
 
   ∴ Required probability = P(E) =  1540 = 38
1440 = 0.35

Page No 626:

Question 15:

In a group of 60 persons, 35 like coffee. Out of this group, if one person is chosen at random, what is the probability that he or she does not like coffee?
(a) 712

(b) 512

(c) 57

(d) 312

Answer:

(b) 512

Explanation:
Total number of persons = 60
Number of persons who like coffee = 35

Number of persons who does not like coffee = 60 − 35 = 25

Let E be the event that the selected person does not like coffee.Then,
 
P(selected person does not like coffee) = P(E) =  2560 = 512Number of ladies who like coffeeTotal number of ladies = 142200 = 0.71

Page No 626:

Question 16:

A dice is thrown 50 times and the outcomes are recorded as shown below.

Outcome 1 2 3 4 5 6
No. of times 11 9 8 5 7 10
If a dice is thrown at random, what is the probability of getting 8?
(a) 15

(b) 350

(c) 13

(d) 0

Answer:

(d) 0

Explanation:
Total number of trials =  50
Number of times 8 appears = 0
P(getting 8) =  Number of times 8 appearsTotal number of trials = 050  = 0

(Remark: A dice is marked from 1 to 6. So, it is impossible to get 8 and probability of an impossible event is 0.)
  

Page No 626:

Question 17:

It is given that the probability of winning a game is 0.7. What is the probability of losing the game?
(a) 0.8
(b) 0.3
(c) 0.7
(d) 0.07

Answer:

(b) 0.3

Explanation:
Let E be the event of winning the game. Then,
P(E) =  0.7
P(not E) = P(losing the game) = 1 â€‹P(E)  ⇒ 1 0.7 = 0.3

Page No 626:

Question 18:

A coin is tossed 60 times and the tail appears 35 times. What is the probability of getting a head?
(a) 712

(b) 127

(c) 512

(d) 125

Answer:

(c) 512
​
Explanation:
Total number of trials = 60
Number of times tail appears = 35

Number of times head appears = 60 35 = 25
Let E be the event of getting a head.
  ∴ P(getting a head) = P (E) = Number of times head appearsTotal number of trials = 2560 = 512



Page No 627:

Question 19:

Assertion: In a cricket match, a batsman hits 9 boundaries out of 45 balls he plays. The probability that in a given throw he does not hit the boundary is 45.
Reason: P(E) + P(not E) = 1.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Explanation: 
Total number of balls = 45
Number of times the ball hits the boundary = 9

Number of times the ball does not hit the boundary = (45
9 ) = 36

Let E be the event that the ball hits the boundary.Then,

 P(E) = 945 =15

 P(not E) = P( he does not hit the boundary) = 1-15 = 45                                  [ P(not E)​ = 3645 = 45
Also, 
 
P(E)  +  P(not E) â€‹ = 1 is true.

Thus, Assertion ( A) and Reason (R) are true and Reason ( R) is a correct explanation of Assertion (A).
So, the correct answer is (a).

Page No 627:

Question 20:

Assertion: The probability of a sure event is 1.
Reason: Let E be an event. Then, 0  P(E)  1.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(b) Both Assertion and Reason are true and Reason is not a correct explanation of Assertion.

Explanation:

Clearly, Assertion (A) and Reason ( R) are both true. But, Reason (R) is not a correct explanation of Assertion (A).
 Hence, the correct answer is (b).

Page No 627:

Question 21:

Fill in the blanks
(i) Probability of an impossible event = ...... .
(ii) Probability of a sure event = ...... .
(iii) Let E be an event. Then, P(not E) = ...... .
(iv) P(E) + P(not E) = ...... .
(v) P(E) lies between ..... and ...... .

Answer:

(i)  0
(ii)  1
(iii) 1 P(E)
(iv) 1
(v) 0, 1

Page No 627:

Question 22:

The marks obtained by 90 students in mathematics out of 100 are given below.

Marks 0−20 20−30 30−40 40−50 50−60 60−70 70 and above
No. of students 7 8 12 25 19 10 9
From these students, a student is chosen at random.
What is the probability that the chosen student
(i) gets 20% or less marks?
(ii) gets 60% or more marks?

Answer:

   â€‹ Total number of students = 90
(i) Number of students who scored 20% marks or less = 7.

    Let E1 be the event that selected student got 20% marks or less.Then,
 
    P(E1) = 790
(ii)  Number of students who scored 60% marks or more = (10 + 9 ) = 19.
    Let E2 be the event that selected student got 60% marks or more.Then,
 
    P(E2) = 1990

Page No 627:

Question 23:

It is known that a box of 800 electric bulbs contains 36 defective bulbs. One bulb is taken at random out of the box. What is the probability that the bulb chosen is non-defective?

Answer:

 Total number of bulbs in the box  = 800
 Number of defective bulbs  = 36

 Number of non-defective bulbs  = 800 36 = 764

 Let E be the event that the chosen bulb is non-defective.
 
∴ Required probability = P(E) = 764800 =191200  4880 = 35

Page No 627:

Question 24:

The table given below shows the ages of 75 teachers in a school.

Age (in years) 18−29 30−39 40−49 50−59
No. of teachers 5 25 35 10
Here 18−29 means from 18 to 29, including both.
A teacher from the school is chosen at random. What is the probability that the teacher chosen is
(i) 40 or more than 40 years old?
(ii) 49 or less than 49 years old?
(iii) 60 or more than 60 years old?

Answer:

    Total number of teachers  = 75
(i) Number of teachers who are 40 or more than 40 years old = 35 + 10 = 45

     Let E1 be the event that the selected teacher's age is 40  years or more.
 
   ∴ Required probability = P(E1) =  4575 = 3560360 = 16

(ii)
  Number of teachers who are 49 or less than 49 years old = (35 + 25 + 5) = 65
     Let E2 be the event that the selected teacher's age is 49 years or less.
   ∴ Required probability P(E2) =  6575 = 131580360 = 29

(iii) Number of teachers who are 60 or more than 60 years old = 0
     Let E3 be the event that the selected teacher's age is more than 60 years.
 
   ∴ Required probability P(E3) =  075 = 00360 =0360360 = 1



Page No 631:

Question 1:

There are 600 electric bulbs in a box, out of which 20 bulbs are defective. If one bulb is chosen at random from the box, what is the probability that it is defective?
(a) 119

(b) 120

(c) 130

(d) 2930

Answer:

(c) 130

Explanation:
Total number of bulbs in the box  = 600
 Number of defective bulbs  = 20

 Let E be the event that the chosen bulb is defective.
 
∴ Required probability = P(E) = 20600 = 130

Page No 631:

Question 2:

A bag contains 5 red, 8 black and 7 white balls. One ball is chosen at random. What is the probability that the chosen ball is black?
(a) 23

(b) 25

(c) 35

(d) 13

Answer:

(b) 25

Explanation:
Total number of balls in the bag =  5 + 8 + 7 = 20
 Number of black balls  = 8

 Let E be the event that the chosen ball is black.
 
∴ Required probability = P(E) = 820 = 25

Page No 631:

Question 3:

A bag contains 16 cards bearing numbers 1, 2, 3, ..., 16 respectively. One card is chosen at random. What is the probability that the chosen card bears a number divisible by 3?
(a) 316

(b) 516

(c) 1116

(d) 1316

Answer:

(b) 516 

Explanation:
Total number of cards in the bag =  16
Numbers on the cards that are divisible by 3 are 3, 6, 9, 12 and 15.

Number of cards with numbers divisible by 3 = 5
​Let E be the event that the chosen card bears a number divisible by 3.
 
∴ Required probability = P(E) = 516 

Page No 631:

Question 4:

In a cricket match, a batsman hits a boundary 4 times out of the 32 balls he faces. In a given ball, what is the probability that he does not hit a boundary?
(a) 18

(b) 78

(c) 17

(d) 67

Answer:

(b) 78

Total number of balls that the batsman faces = 32
Number of times the batsman hits a boundary = 4

Number of times the batsman does not hit a boundary = (32 − 4 ) = 28

Let E be the event that the batsman does not hit a boundary.
∴ Required probability =  P(E) =  2832 = 78

Page No 631:

Question 5:

Define the probability of an event E.

Answer:

The probability of an event E is the ratio of number of trials in which event E happened to total number of trials.
P(E) = Number of trials in which event happenedTotal number of trials
Thus, 0P(E) 1.

Page No 631:

Question 6:

A coin is tossed 60 times with the following outcomes:
Head 28 times and Tail 32 times.
Find the probability of getting a head in a single throw of the coin.

Answer:

Total number of tosses = 60
Number of times tail appears = 32

Number of times head appears = 28
Let E be the event of getting a head. Then,
   P(getting a head) = P (E) = Number of times head appearsTotal number of trials = 2860 = 715

Page No 631:

Question 7:

Write down all the possible outcomes when a dice is thrown.

Answer:

When a dice is thrown, the possible outcomes are 1, 2, 3 ,4, 5 and 6.

Page No 631:

Question 8:

Fill in the blanks:
(i) Probability of a sure event = ..... .
(ii) Probability of an impossible event = ..... .
(iii) If E be an event, then P(E) + P(not E) = ..... .
(iv) If E is an event, then ....... P (E)  ...... .

Answer:

 (i) 1
(ii) 0
(iii) 1 
(iv) 0, 1



Page No 632:

Question 9:

A die was thrown 80 times and the outcomes were noted as shown:

Outcome 1 2 3 4 5 6
Frequency 14 11 9 13 18 15
If the die is thrown at random, what is the probability of getting a 2?

Answer:

Total number of trials =  80
Number of times 2 appears = 11
Let E be the event of getting 2 on the die.
 P( getting 2) = P(E) =  Number of times 2 appearsTotal number of trials=1180
Number of times 8 appearsTotal number of trails = 050  

Page No 632:

Question 10:

The probability of losing a game is 0.6. What is the probability of winning the game?

Answer:

Let E be the event of losing the game. Then,
P
(E) =  0.6
P( not E) = P(winning the game) = 1− â€‹P(E)
 ⇒ 1 −  0.6 = 0.4
Hence, probability of winning the game is 0.4.

Page No 632:

Question 11:

A coin is tossed 50 times and the tail appears 28 times. In a single throw of a coin, what is the probability of getting a head?

Answer:

Total number of trials = 50
Number of times the tail appears = 28

So, number of times the head appears = 50 28 = 22
Let E be the event of getting a head. Then,
   P(getting a head) = P (E) = Number of times head appearsTotal number of trials = 2250 = 1125

Page No 632:

Question 12:

In a one-day cricket match, a batsman hits the boundary 8 times out of the 48 balls that he faced. Find the probability that he does not hit the boundary.

Answer:

Total number of balls faced by the batsman = 48
Number of times the batsman hits the boundary = 8

Number of times the he does not hit the boundary = (48
8 ) = 40

Let E be the event that the batsman does not hit the boundary.
∴ Required probability =  P(E) =  4048 = 562832 = 78

Page No 632:

Question 13:

Two coins were tossed simultaneously 80 times and the outcomes were recorded as under:

Outcome Two heads One head No head
Frequency 24 36 20
If two coins are tossed at random, find the probability of getting
(i) two heads
(ii) one head
(iii) no head

Answer:

Total number of trials =  80
Number of times 2 heads appear = 24
Number of times 1 head appears = 36
Number of times 0 head appears = 20

In a random toss of two coins, let E1E2E3 be the events of getting 2 heads, 1 head and 0 head, respectively. Then,

(i) P(getting 2 heads) =  P(E1) = Number of times 2 heads appearTotal number of trials = 2480 =310 =  0.3

(ii)  P(getting 1 head) =  P(E2) = Number of times 1 head appearsTotal number of trials = 3680 =920 =  0.45

(iii) P(getting 0 head) =  P(E3) = Number of times 0 head appearsTotal number of trials = 2080 =14 =  0.25

Clearly, when two coins are tossed, the only possible outcomes are E1E2 and E3 and P(E1) + P(E2) +  P(E3) = (0.3 + 0.45 + 0.25) = 1

Page No 632:

Question 14:

Marks obtained by 90 students of Class IX in a test are given below:

Marks 0−20 20−40 40−60 60−80 80−100
No. of students 8 15 32 26 9
Out of these students one is chosen at random. Find the probability that the chosen student obtains
(i) less than 20% marks
(ii) 80% or more marks
(iii) 60% or more marks

Answer:

   Total number of students = 90
(i) Number of students who scored less than 20% marks = 8

     Let E1 be the event that selected student got less than 20% mark .
     ∴ P(E1) = 890 = 445
 
(ii)  Number of students who scored 80% marks or more = 9
      Let E2 be the event that selected student got 60% marks or more .
 
       P(E2) = 990 = 110
(iii)  Number of students who scored 60% marks or more = (26 + 9 ) = 35
      Let E3 be the event that selected student got 60% marks or more.
 
        ∴​ P(E3) =  3590 = 7181990

Page No 632:

Question 15:

The blood groups of 30 students of a class were recorded as under:

Blood Group A B AB O
No. of students 9 11 4 6
If a student of this class is chosen at random, what is the probability that the chosen student has blood group
(i) O?
(ii) A?
(iii) AB?

Answer:

   Total number of students  = 30
(i) Number of students with blood group O = 6

    Let E1 be the event that the selected student has blood group O. 
 
   ∴ Required probability = P(E1) =  630 = 15

(ii) Number of students with blood group A = 9
     Let E2 be the event that the selected student has blood group A.
 
   ∴ Required probability = P(E2) =  930 = 310

(iii) Number of students with blood group AB = 4
      Let E3 be the event that the selected student has blood group AB. 
 
   ∴ Required probability = P(E3) =  430 = 215
1540 = 38



Page No 633:

Question 16:

In a survey of 100 families with 2 or less boys, the following data were obtained:

Number of boys in a family 0 1 2
Number of families 18 46 36
If one of the given families is chosen at random, what is the probability that the chosen family has
(i) 1 boy?
(ii) 2 boys?
(iii) no boys?

Answer:

 â€‹  Total number of families  = 100
(i) Number of families with 1 boy  = 46

    Let E1 be the event that the selected family has 1 boy.
 
   ∴ Required Probability = P(E1) =  46100 = 2350

(ii) Number of families with 2 boys = 36
     Let E2 be the event that the selected families have 2 boys
 
   ∴ Required Probability = P(E2) =  36100 = 925

(iii) Number of families without boys = 18
      Let E3 be the event that the selected families have no boys
 
   ∴ Required Probability = P(E3) =  18100 = 950
1540 = 38

Page No 633:

Question 17:

A die was thrown 100 times and the following observations were recorded:

Outcome 1 2 3 4 5 6
Frequency 12 18 14 26 14 16
If a die is thrown at random, find the probability of getting
(i) a number less than 3
(ii) a number greater than 4
(iii) an even number

Answer:

Total number of trials =  100

(i) Let E1 be the event of getting a number less than 3. Then,
    P(getting a number less than 3)   = P(E1)  = 12+18100 = 30100 = 310
Number of times 3 appearsTotal number of trails = 54300 = 0.18

(ii)  Let E2 be the event of getting a number greater  than 4. Then,  
     P(getting a number greater than 4) = P(E2)  = 14+16100 =30100 = 310
Number of times 6 appearsTotal number of trails = 33300 =0.11

(iii)  Let E3 be the event of getting an even number. Then,  
      P(getting an even number) = P(E3)  = 18 + 26 +16100 = 60100 = 35
mber of times 5 appearsTotal number of trails = 39300 = 0.1
Number of times 1 appearsTotal number of trails = 60300 = 0.2

Page No 633:

Question 18:

A survey of 600 students on their preference for coffee was conducted and recorded as shown:

Opinion Like Dislike
No. of students 360 140
Out of these students one is chosen at random. What is the probability that the chosen student
(i) likes coffee?
(ii) dislikes coffee?

Answer:

   Total number of students = 600
(i) Number of
students who like coffee = 360
    Let E1 be the event that the selected person likes coffee.Then,
    P ( selected person likes coffee) = P(E1) =  360600 = 35

(ii) 
Number of students who dislike coffee =  140
     Let E2 be the event that the selected person dislikes coffee.Then,
    P ( selected person dislike coffee) = P(E2) = 140600 = 730  2560 = 512

Page No 633:

Question 19:

Two coins are tossed 1000 times and the outcomes are recorded as given below:

Number of heads 0 1 2
Frequency 240 450 310
In a simple throw of two coins, what is the probability of getting
(i) at most one head?
(ii) at least one head?

Answer:

Total number of tosses =  1000
Number of times 2 heads appear = 310
Number of times 1 head appears = 450
Number of times 0 head appears = 240


(i) Let E1 be the event of getting at most 1 head.
   Then, number of favourable outcomes for E1​ (either 1 head or  0 head) = (450 + 240) = 690   
       P(getting at most 1 head) = P(E1) =  (450 + 240)1000 =6901000 = 69100 = 0.69

(ii)  Let E2 be the event of getting at least 1 head.
     Then, number of favourable outcomes for E2 â€‹â€‹(either 1 head or 2 heads appear) = 450 + 310 = 760   
       P( getting at least 1 head) = P(E2) =    7601000 =1925 =  0.76

Page No 633:

Question 20:

A coin is tossed 80 times with the following outcomes:
Head: 35 times and Tail: 45 times.
In a single throw of a coin, if E be the event of getting a head, verify that P(E) + P(not E) = 1.

Answer:

Total number of trials = 80
Number of times the tail appears = 45

So, number of times the head appears = 35
Let E be the event of getting a head. Then, 
   P(getting a head) = P (E) = Number of times head appearsTotal number of trials = 3580 = 716

P(not E) = P (getting a tail) = â€‹Number of times tail appearsTotal number of trials = 4580 = 916

P (E) + â€‹P(not E) = 716 + 916 = 1616 =1
Hence, for an event E of getting a head, P(E) + P(not E) = 1



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