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Page No 708:

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 708:

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 708:

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 709:

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 709:

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) = ​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 ​P(E1) + ​P(E2) = (0.71 + 0.29) = 1

Page No 709:

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 709:

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 709:

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 709:

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 710:

Question 10:

12 Packets of salt, each marked 2 kg, actually contained the following weights (in kg) of salt:
1.950, 2.020, 2.060, 1.980, 2.030, 1.970,
2.040, 1.990, 1.985, 2.025, 2.000, 1.980.
Out of these packets, one packet is chosen at random.
What is the probability that the chosen packet contains more than 2 kg of salt?

Answer:


Total number of salt packets = 12

Number of packets which contains more than 2 kg of salt = 5

∴ P(Chosen packet contains more than 2 kg of salt) = Number of packets which contains more than 2 kg of saltTotal number of salt packets=512

Thus, the probability that the chosen packet contains more than 2 kg of salt is 512.

Page No 710:

Question 11:

In a circket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that he did not hit a boundary.

Answer:


Number of balls played by the batsman = 30

Number of balls in which he hits boundaries = 6

∴ Number of balls in which he did not hit a boundary = 30 − 6 = 24

P(Batsman did not hit a boundary) = Number of balls in which he did not hit a boundaryNumber of balls played by the batsman=2430=45

Thus, the probability that he did not hit a boundary is 45.

Page No 710:

Question 12:

An organisation selected 2400 families at random and surveyed them to determine a relationship between the income level and the number of vehicles in a family. The information gathered is listed in the table below:

Monthly income 
(in ₹)
Number of vehicles per family
0 1 2 3 or more
 Less than ₹ 25000 10 160 25 0
 ₹ 25000 – ₹ 30000 0 305 27 2
 ₹ 30000 – ₹ 35000 1 535 29 1
 ₹ 35000 – ₹ 40000 2 469 59 25
 ₹ 40000 or more 1 579 82 88

Suppose a family is chosen at random. Find the probability that the family chosen is
(i) earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles.
(ii) earning ₹ 40000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 25000 per month and not owning any vehicle.
(iv) earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles.
(v) owning not more than 1 vehicle.

Answer:


Number of families surveyed = 2400

(i) Number of families earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles = 27

∴ P(Family chosen is earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles)

= Number of families earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles / Number of families surveyed

=272400=9800
(ii) Number of families earning ₹ 40000 or more per month and owning exactly 1 vehicle = 579

∴ P(Family chosen is earning ₹ 40000 or more per month and owning exactly 1 vehicle)

= Number of families earning ₹ 40000 or more per month and owning exactly 1 vehicle / Number of families surveyed

=5792400=193800
(iii) Number of families earning less than ₹ 25000 per month and not owning any vehicle = 10

∴ P(Family chosen is earning less than ₹ 25000 per month and not owning any vehicle)

= Number of families earning less than ₹ 25000 per month and not owning any vehicle / Number of families surveyed

=102400=1240
(iv) Number of families earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles = 59 + 25 = 84

∴ P(Family chosen is earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles)

= Number of families earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles / Number of families surveyed

=842400=7200
(v) Number of families owning not more than 1 vehicle 

= Number of families owning 0 vehicle + Number of families owning 1 vehicle

= 10 + 0 + 1 + 2 + 1 + 160 + 305 + 535 + 469 + 579 = 2062

∴ P(Family chosen is owning not more than 1 vehicle)

= Number of families owning not more than 1 vehicle / Number of families surveyed

=20622400=10311200

Page No 710:

Question 13:

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

Marks
(Class interval)
1 – 10 11 – 20 21 – 30 31 – 40 41 – 50
Number of students
(Frequency)
7 10 6 4 3

Out of these students, one is chosen at random. What is the probability that the marks of the chosen student
(i) are 30 or less?
(ii) are 31 or more?
(iii) lie in the interval 21 – 30?

Answer:


Total number of students = 30

(i) Number of students whose marks are 30 or less = 7 + 10 + 6 = 23

∴ P(Marks of the chosen student are 30 or less) = Number of students whose marks are 30 or lessTotal number of students=2330

(ii) Number of students whose marks are 31 or more = 4 + 3 = 7

∴ P(Marks of the chosen student are 31 or more) = Number of students whose marks are 31 or moreTotal number of students=730

(iii) Number of students whose marks lie in the interval 21–30 = 6

∴ P(Marks of the chosen student lie in the interval 21–30) = Number of students whose marks lie in the interval 21-30Total number of students=630=15



Page No 711:

Question 14:

The table given below shows the ages of 75 teachers in a school.
Age (in years) 18 – 29 30 – 39 40 – 49 50 – 59
Number of teachers 3 27 37 8
A teacher from this school is chosen at random. What is the probability that the selected teacher is
(i) 40 or more than 40 years old?
(ii) of an age lying between 30 – 39 years (including both)?
(iii) 18 years or more and 49 years or less?
(iv) 18 years or more old?
(v) above 60 years of age?
Note Here 18 – 29 means 18 or more but less than or equal to 29.

Answer:


Total number of teachers = 75

(i) Number of teachers who are 40 or more than 40 years old = 37 + 8 = 45

∴ P(Selected teacher is 40 or more than 40 years old) = Number of teachers who are 40 or more than 40 years oldTotal number of teachers=4575=35

(ii) Number of teachers of an age lying between 30 – 39 years (including both) = 27

∴ P(Selected teacher is of an age lying between 30 – 39 years (including both))

Number of teachers of an age lying between 30-39 years including bothTotal number of teachers=2775=925

(iii) Number of teachers 18 years or more and 49 years or less = 3 + 27 + 37 = 67 

∴ P(Selected teacher is 18 years or more and 49 years or less) = Number of teachers 18 years or more and 49 years or lessTotal number of teachers=6775

(iv) Number of teachers 18 years or more old = 3 + 27 + 37 + 8 = 75 

∴ P(Selected teacher is 18 years or more old) = Number of teachers 18 years or more oldTotal number of teachers=7575=1

(v) Number of teachers above 60 years of age = 0 

∴ P(Selected teacher is above 60 years of age) = Number of teachers above 60 years of ageTotal number of teachers=075=0

Page No 711:

Question 15:

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.
What is 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) 10 years or more but less than 40 years?
(iv) 10 years or more?
(v) less than 10 years?

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) Let Ebe the event that the selected patient is 10 years or more but less than 40 years.

Number of patients whose age is 10 years or more but less than 40 years = 90 + 50 + 60 = 200
∴ P(Patient's age is 10 years or more but less than 40 years) = P(E3) = 200360=59

(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
(v)
Number of patients whose age is less than 10 years = 0
     Let E5 be the event that the selected patient's age is less than 0.
 
   ∴ P(patient's age is less than 10 years)= P(E5) =  0360 =0

Page No 711:

Question 16:

The marks obtained by 90 students of a school in mathematics out of 100 are given as under:
Marks 0 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 and above
Number 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:


Total number of students = 90

(i) Number of students who gets 20% or less marks = Number of students who gets 20 or less marks = 7

∴ P(Student gets 20% or less marks) = Number of students who gets 20% or less marksTotal number of students=790

(ii) Number of students who gets 60% or more marks = Number of students who gets 60 or more marks = 10 + 9 = 19

∴ P(Student gets 60% or more marks) = Number of students who gets 60% or more marksTotal number of students=1990

Page No 711:

Question 17:

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 electric bulbs in the box = 800

Number of defective electric bulbs in the box = 36

∴ Number of non-defective electric bulbs in the box = 800 − 36 = 764

P(Bulb chosen is non-defective) = Number of non defective electric bulbs in the boxTotal number of electric bulbs in the box=764800=191200

Thus, the probability that the bulb chosen is non-defective is 191200.
 



Page No 712:

Question 18:

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) ≤ ....... .

Answer:


(i) Probability of an impossible event =     0   
(ii) Probability of a sure event =     1    
(iii) Let E be an event. Then, P(not E) =     1 − P(E)​     .
(iv) P(E) + P(not E) =     1    
(v)     0    ≤ P(E) ≤     1    

Page No 712:

Question 1:

In a sample survey of 645 people, it was found that 516 people have a high school certificate. If a person is chosen at random, what is the probability that he / she has a high school certificate?
(a) 12

(b) 35

(c) 710

(d) 45

Answer:


Total number of people surveyed = 645

Number of people who have a high school certificate = 516

∴ P(Person has a high school certificate) = Number of people who have a high school certificateTotal number of people surveyed=516645=45

Hence, the correct answer is option (d).

Page No 712:

Question 2:

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

Blood group A B AB O
Number of students 11 15 8 6
From this class, a student is chosen at random. What is the probability that the chosen student has blood group AB?

(a) 1320

(b) 38

(c) 15

(d) 1140

Answer:

(c) 15

Explanation:
Total number of students = 40
 Number of students with blood group AB = 8

 Let E be the event that the selected student's blood group is AB.
 
   ∴ Required probability = P(E) =  840 = 15
1440 = 0.35



Page No 713:

Question 3:

80 bulbs are selected at random from a lot and their lifetime is hours is recorded as under.

Lifetime (in hours) 300 500 700 900 1100
Frequency 10 12 23 25 10
One bulb is selected at random from the lot. What is the probability that its life is 1150?
(a) 180

(b) 716

(c) 1

(d) 0

Answer:

(d) 0
Maximum lifetime a bulb has is 1100 hours. There is no bulb with lifetime 1150 hours.

Page No 713:

Question 4:

In a survey of 364 children aged 19 – 36 months, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he / she does not like to eat potato chips is
(a) 14

(b) 12

(c) 34

(d) 45

Answer:


Total number of children surveyed = 364

Number of children who liked to eat potato chips = 91

Number of children who do not liked to eat potato chips = 364 − 91 = 273

∴ P(Child does not like to eat potato chips) = Number of children who do not liked to eat potato chipsTotal number of children surveyed=273364=34

Hence, the correct answer is option (c).

Page No 713:

Question 5:

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

Number of heads 2 1 0
Frequency 200 550 250
Now, if two coins are tossed at random, what is the probability of getting at most one head?

(a) 34

(b) 45

(c) 14

(d) 15

Answer:


Total number of times two coins are tossed = 1000

Number of times of getting at most one head = Number of times of getting 0 heads + Number of times of getting 1 head = 250 + 550 = 800

∴ P(Getting at most one head) = Number of times of getting at most one headTotal number of times two coins are tossed=8001000=45

Hence, the correct answer is option (b).

Page No 713:

Question 6:

80 bulbs are selected at random from a lot and their lifetime in hours is recorded as under.

Lifetime (in hours) 300 500 700 900 1100
Frequency 10 12 23 25 10
One bulb is selected at random from the lot. What is the probability that the selected bulb has a life more than 500 hours?

(a) 2740

(b) 2940

(c) 516

(d) 1140

Answer:


(b) 2940

Explanation:
 Total number of bulbs in the lot  = 80
 Number of bulbs with life time of more than 500 hours = (23 + 25 + 10) = 58

 Let E be the event that the chosen bulb's life time is more than 500 hours.
 
∴ Required probability = P(E) =  5880 = 2940

Page No 713:

Question 7:

To know the opinion of the students about the subject Sanskrit, a survey of 200 students was conducted. The data is recorded as under.

Opinion like dislike
Number of students 135 65
What is the probability that a student chosen at random does not like it?

(a) 1327

(b) 2740

(c) 1340

(d) 2713

Answer:


Total number of students surveyed = 200

Number of students who does not like the subject Sanskrit = 65

∴ P(Student chosen at random does not like the subject Sanskrit) = Number of students who does not like the subject SanskritTotal number of students surveyed=65200=1340

Hence, the correct answer is option (c).



Page No 714:

Question 8:

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 714:

Question 9:

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) 0.35
(d) 0.15

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 ​P(E)  ⇒ 1 0.7 = 0.3

Page No 714:

Question 10:

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 714:

Question 11:

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 

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Question 12:

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

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Question 13:

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 

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Question 14:

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



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Question 15:

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

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Question 16:

Two coins are tossed simultaneously 600 times to get 2 heads : 234 times, 1 head : 206 times, 0 head : 160 times.
If two coins are tossed at random, what is the probability of getting at least one head?

(a) 103300     

(b) 39100

(c) 1115

(d) 415

Answer:


Number of times two coins are tossed simultaneously = 600

Number of times of getting at least one head = Number of times of getting 1 head + Number of times of getting 2 heads = 206 + 234 = 440

∴ P(Getting at least one head) = Number of times of getting at least one headNumber of times two coins are tossed simultaneously=440600=1115

Hence, the correct answer is option (c).



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