Probability

About Probability – Class 9 Maths

The class-9 maths notes for the Chapter-Probability chapter helps students understand the basics of chance and prediction. Our Class 9 maths notes break down probability concepts through examples from real life, like dice and coins. The NCERT solutions for class 9 Maths guide learners in solving exercises step by step for complete comprehension.

Our Class 9 Maths tuition sessions include fun probability experiments that make learning interactive. Tutors help students develop analytical thinking and logical reasoning, ensuring a solid base for higher classes and competitive exams.

Theory of probability deals with measurement of uncertainty of the occurrence of same event or incident in terms of percentage or ratio.

Different Approaches

There are following three types of approaches to theory of probability:

1. Experimental approach, empirical approach or observed required approach.
2. Classical approach.
3. Axiomatic approach.

Note: In this chapter, we will study only the Empirical approach.

Important Terms and Definitions

Experiment: An activity which gives some well-defined outcomes is called experiment.
Example: "Tossing a coin" gives either head (H) or tail (T), is an experiment.

Random experiment: It is an experiment, in which we know about all the possible outcomes but not sure about a specific outcome.

For example: in throwing a dice, possible outcomes are 1, 2, 3, 4, 5, 6 but we are not sure that the no on dice is '4'.

Event: The possible outcomes of a trial are called events.
Example: when a die is rolled, showing the number 1 or 2 or 3 or 4 or 5 or 6 is an event.

Sure event: When all the outcomes of a random experiment favour an event, the event is called a sure event and its empirical probability is 1.

Impossible event: When no outcome of a random experiment favours an event, the event is called an impossible event and its empirical probability is 0.

Sample space: The collection of all possible outcomes in an experiment is called sample space.

Favourable events: The cases, which ensure the occurrence of an event, are called favourable cases to that event.

Some Special Sample Spaces

1. A die is thrown once

S = {1, 2, 3, 4, 5, 6}; n(S) = 6

2. A coin is tossed once

S = {H, T}; n(S) = 2

3. A coin is tossed twice OR Two coins are tossed simultaneously

S = {HH, HT, TH, TT}; n(S) = 4 = 2²

4. A coin is tossed three times OR Three coins are tossed simultaneously

S = {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT}; n(S) = 8 = 2³

5. Two dice are thrown together OR A die is thrown twice

S = {(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)}
n(S) = 6² = 36

Probability Formulas

1. Probability of an Event:

P(E) = Number of favourable cases / Total number of cases

2. Complementary Event:

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

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

3. Range of Probability:

0 ≤ P(E) ≤ 1

4. Sum of Probabilities:

Sum of the probabilities of all the outcomes of random experiment is 1.

Example 1: A card is drawn from a well shuffled deck of 52 cards. Find the probability of:

(i) A king

Number of kings = 4 (favorable cases)

P(A) = 4/52 = 1/13

(ii) A heart

P(A) = 13/52 = 1/4

(iii) A seven of heart

P(A) = 1/52

(iv) A jack, queen or a king

P(A) = 12/52 = 3/13

(v) A two of heart or a two of diamond

P(A) = 2/52 = 1/26

(vi) A face card

P(A) = 12/52 = 3/13

(vii) A black card

P(A) = 26/52 = 1/2

(viii) Neither a heart nor a king (13 heart + 4 king, but 1 common)

P(A) = 1 - 16/52 = (52-16)/52 = 36/52 = 9/13

(ix) Neither an ace nor a king

P(A) = 44/52 = 11/13

Example 2: Two coins are tossed simultaneously. Find the probability of getting:

On tossing two coins simultaneously, all the possible outcomes are: HH, HT, TH, TT.

(i) two heads

The probability of getting two heads = P(HH) = Event of occurrence of two heads / Total number of possible outcomes = 1/4

(ii) at least one head

The probability of getting at least one head = Favourable outcomes / Total no. of outcomes = 3/4

(iii) no head

The probability of getting no head P(TT) = 1/4

Example 3: A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is:

Number of red balls in the bag = 5

Number of white balls in the bag = 8

Number of green balls in the bag = 4

Number of black balls in the bag = 7

Total number of balls in the bag = 5 + 8 + 4 + 7 = 24.

Drawing balls randomly are equally likely outcomes.

Total number of possible outcomes = 24

(i) Black

There are 7 black balls, hence the number of such favourable outcomes = 7

Probability of drawing a black ball = Number of favourable outcomes / Total number of possible outcomes = 7/24

(ii) Not red

There are 5 red balls, hence the number of such favourable outcomes = 5.

Probability of drawing a red ball = Number of favourable outcomes / Total number of possible outcomes = 5/24

Probability of drawing not a red ball = P(Not Red ball) = 1 - 5/24 = 19/24

(iii) Green

There are 4 green balls.

Number of such favourable outcomes = 4

Probability of drawing a green ball = Number of favourable outcomes / Total number of possible outcomes = 4/24 = 1/6

Example 4: A card is drawn from a well-shuffled deck of playing cards. Find the probability of drawing:

Random drawing of cards ensures equally likely outcomes

(i) a face card

Number of face cards (King, Queen and jack of each suits) = 4 × 3 = 12

Total number of cards in deck = 52

Total number of possible outcomes = 52

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

(ii) a red face card

Number of red face cards = 2 × 3 = 6

Number of favourable outcomes of drawing red face card = 6

P(drawing of red face card) = 6/52 = 3/26

LET'S TRY - PRACTICE QUESTIONS

Questions

Q1. A coin is tossed 500 times with the following frequencies of two outcomes: Head: 240 times, tail: 260 times. Find the probability of occurrence of each of these events.

Solution 1: A coin is tossed 500 times

It is given that the coin is tossed 500 times.

Total number of cases = 500

The events of getting a head and of getting a tail are denoted by A and B respectively. Then,

Number of cases in which the event A happens = 240

and, Number of cases in which the event B happens = 260

P(A) = Number of cases in which the event A happens / Total number of cases = 240/500 = 0.48

and, P(B) = Number of cases in which the event B happens / Total number of cases = 260/500 = 0.52

Q2. Two coins are tossed simultaneously 1000 times with the following frequencies of different outcomes: Two heads: 210 times, One head: 550 times, No head: 240 times. Find the probability of occurrence of each of these events.

Solution 2: Two coins are tossed simultaneously 1000 times

Let A = Getting two heads
B = Getting one head
C = Getting no head.

Total number of cases = 1000

Number of cases in which event A happens = 210

Number of cases in which event B happens = 550

Number of cases in which event C happens = 240

P(A) = Number of cases in which the event A happens / Total number of cases = 210/1000 = 0.21

P(B) = Number of cases in which the event B happens / Total number of cases = 550/1000 = 0.55

P(C) = Number of cases in which the event C happens / Total number of cases = 240/1000 = 0.24

Q3. In a cricket match, a batsman hits a boundary 8 times out of 40 balls he plays. Find the probability that he didn't hit a boundary.

Solution 3: Cricket boundary problem

Let A denote the event that the batsman did not hit a boundary.

We have, Total number of cases = 40

Number of cases in which the event A happened = 40 − 8 = 32

P(A) = 32/40 = 4/5 = 0.8

Q4. The record of a weather station shows that out of the past 250 consecutive days, its weather forecast were correct 175 times. What is the probability that on a given day:

(i) it was correct?

(ii) it was not correct?

Solution 4: Weather forecast problem

We have,
Total number of days for which the weather forecast was made = 250
Number of days for which the forecast was correct = 175
Number of days for which the forecast was not correct = 250 − 175 = 75

Therefore,

(i) Probability that the forecast was correct on a given day

= Number of days for which the forecast was correct / Number of days for which the forecast was made

= 175/250 = 0.7

(ii) Probability that the forecast was not correct on a given day

= Number of days for which the forecast was not correct / Number of days for which the forecast was made

= 75/250 = 0.3

Q5. There are 40 students in a class and their results is presented as below:

Result (Pass/Fail)	Pass	Fail
Number of Students	30	10

If a student is chosen at random out of the class, find the probability that the student has passed the examination.

Solution 5: Student pass/fail problem

Total number of students = 40

Cases which favour a student to pass = 30

The probability of the required event, i.e., the student has passed the examination = 30/40 = 0.75

Q6. A coin is tossed 150 times and the outcomes are recorded. The frequency distribution of the outcomes H (i.e. head) and T (i.e., tail) is given below:

Outcome	H	T
Frequency	85	65

Find the value of P(H), i.e., probability of getting a head in a single case.

Solution 6: Coin tossed 150 times

Total number of cases = 150

Cases which favour the outcome H = 85.

P(H) = 85/150 = 0.567 (approx.)

Q7. 400 students of class X of a school appeared in a test of 100 marks in the subject of social studies and the data about the marks secured is as below:

Marks secured	0 – 25	26 – 50	51 –75	Above 75	Total no. of student
Number of Students	50	220	100	30	400

If the result card of a student be picked up at random, what is the probability that the student has secured more than 50 marks?

Solution 7: Student marks problem

Total number of students, i.e., the total frequency = 400

The total number of students who secured more than 50 marks = 100 + 30 = 130

Probability that the marks secured are more than 50 = 130/400 = 0.325

Q8. A die is thrown 1000 times with the following frequencies for the outcomes 1, 2, 3, 4, 5, and 6 as given below:

Outcomes:	1	2	3	4	5	6
Frequency:	179	150	157	149	175	190

Solution 8: Die thrown 1000 times

P(E₁) = Probability of getting outcome 1
= Frequency of 1 / Total number of times the die is thrown
= 179/1000 = 0.179

P(E₂) = Probability of getting outcome 2
= Frequency of 2 / Total number of times the die is thrown
= 150/1000 = 0.15

Similarly, we have,

P(E₃) = 157/1000 = 0.157
P(E₄) = 149/1000 = 0.149
P(E₅) = 175/1000 = 0.175, and
P(E₆) = 190/1000 = 0.19

Q9. 1000 families with 2 children were selected randomly and the following data were recorded:

Number of boys in a family :	0	1	2
Number of families :	140	560	300

If a family is chosen at random, find the probability that it has:

(i) No boy

(ii) one boy

(iii) 2 boys

(iv) at least one boy

(v) at most one boy.

Solution 9: Families with 2 children

Total number of families with two children = 1000

(i) Number of families having no boy child = 140

Probability that a family chosen at random does not have a boy child = 140/1000 = 0.14

(ii) Number of families having one boy child = 560

Probability that a family chosen at random has one boy child = 560/1000 = 0.56

(iii) Number of families having two boy children = 300

Probability that a family chosen at random has two boy children = 300/1000 = 0.3

(iv) Number of families having at least one boy child = 560 + 300 = 860

Probability that a family chosen at random has at least one boy child = 860/1000 = 0.86

(v) Number of families having at most one boy child = 140 + 560 = 700

Probability that a family chosen at random has at most one boy child = 700/1000 = 0.7

Q10. The percentage of marks obtained by a student in the monthly unit tests are given below:

Unit:	I	II	III	IV	V
Percentage of marks obtained	58	64	76	62	85

Find the probability that the student gets:

(i) a first division i.e. at least 60% marks

(ii) marks between 70% and 80%

(iii) a distinction i.e. 75% or above.

(iv) less than 65% marks.

Solution 10: Student percentage marks

Total number of unit tests held = 5

(i) Number of unit tests in which the student gets a first class i.e. at least 60% marks = 4

Probability that the student gets a first class=4/5 = 0.8

(ii) Number of unit tests in which the student gets marks between 70% and 80% = 1

Probability that a student gets marks between 70% and 80% = 1/5 = 0.2

(iii) Number of unit tests in which the student gets distinction = 2

Probability that the student gets distinction = 2/5 = 0.4

(iv) Number of unit tests in which the student gets less than 65% marks = 3

Probability that a student gets less than 65% marks = 3/5 = 0.6

Q11. On the page of a telephone directory, there were 200 telephone numbers. The frequency distribution of their unit place digit is given in the table below:

Digit:	0	1	2	3	4	5	6	7	8	9
Frequency:	22	26	22	22	20	10	14	28	16	20

A number is chosen at random, find the probability that the digit at its unit's place is:

(i) 6

(ii) a non-zero multiple of 3

(iii) a non-zero even number

(iv) an odd number

Solution 11: Telephone directory problem

We have, Total number of selected telephone numbers = 200

(i) It is given that the digit 6 occurs 14 times at unit's place.

Probability that the digit at unit's place is 6 = 14/200 = 0.07

(ii) A non-zero multiple of 3 i.e. 3, 6, and 9.

Number of telephone number in which unit's digit is either 3 or 6 or 9 = 22 + 14 + 20 = 56

Probability of getting a telephone number having a multiple of 3 at unit's place = 56/200 = 0.28

(iii) Number of telephone numbers having an even number (2 or 4 or 6 or 8) at unit's place

= 22 + 20 + 14 + 16 = 72

Probability of getting a telephone number having an even number at unit's place = 72/200 = 0.36

(iv) Number of telephone numbers having an odd digit (1 or 3 or 5 or 7 or 9) at unit's place

= 26 + 22 + 10 + 28 + 20 = 106

Probability of getting a telephone number having an odd number at unit's place = 106/200 = 0.53

Q12. Two similar coins were tossed simultaneously 1000 times and the frequency distribution of heads obtained on each toss is as below:

Number of heads	0	1	2
Frequency	200	500	300

Find the probabilities of the following:

(i) Probability of getting one head.

(ii) Probability of getting two heads.

(iii) Probability of getting at least one head.

(iv) Probability of getting less than two heads.

(v) Probability of getting three heads.

(vi) Probability of getting not more than two heads.

Solution 12: Two coins tossed 1000 times

Total number of cases = 1000

(i) Total number of cases favouring the event of getting one head = 500.

so, P(one head) = 500/1000 = 0.5

(ii) P(two head) = 300/1000 = 0.3

(iii) P(at least one head) = P(1 head or 2 heads) = (500 + 300)/1000 = 800/1000 = 0.8

(iv) P(less than two head) = P(0 head or 1 head) = (200 + 500)/1000 = 700/1000 = 0.7

(v) P(three heads) = 0/1000 = 0

(vi) P(not more than two heads) = (200 + 500 + 300)/1000 = 1000/1000 = 1

Q13. A factory manufacturing car batteries made a survey in the field about the life of these batteries. The data obtained are given in the following table:

Life time (in months)	Less than 24	24 to 36	36 to 48	More than 48	Total number of batteries
Frequency or the number of batteries	40	220	540	200	1000

A battery of this company is put in a car, what is the probability that

(i) the battery will last for more than 36 months?

(ii) the battery will last for less than 48 months?

(iii) the battery will last for 36 to 48 months?

Solution 13: Car battery life problem

Total frequency or the total number of cases made = 1000

(i) The total number of batteries which last for more than 36 months = 540 + 200 = 740

Now, P(battery will last for more than 36 months) = 740/1000 = 0.74

(ii) The total number of batteries which last for less than 48 months = 40 + 220 + 540 = 800

So, P(battery will last for less than 48 months) = 800/1000 = 0.80

(iii) The total number of batteries which last for 36 to 48 months = 540

So, P(battery will last for 36 to 48 months) = 540/1000 = 0.54

Q14. Fifty seeds were selected at random from each of 5 bags A, B, C, D, E of seeds, and were kept under standardized conditions equally favourable to germination. After 20 days, the number of seeds which had germinated in each collection were counted and recorded as follow:

Bag	A	B	C	D	E
Number of seeds germinated	40	48	42	39	41

What is the probability of germination of

(i) more than 40 seeds in a bag?

(ii) 49 seeds in a bag?

(iii) more than 35 seeds in bag?

Solution 14: Seed germination problem

(i) Number of bags in which more than 40 seeds out of 50 seeds germinated = 3 (these are B, C and E)

Total number of bags = 5

So, P(more than 40 seeds in a bag germinated) = 3/5 = 0.60

(ii) Number of bags in which 49 seeds germinated = 0

So, P(49 seeds germinated in a bag) = 0/5 = 0

(iii) Number of bags in which more than 35 seeds out 50 seeds germinated = 5

Total number of bags = 5

So, P(more than 35 seeds in a bag germinated) = 5/5 = 1

Q15. An insurance company selected 2000 drivers at random in a particular city to find a relationship between age and accidents. The data obtained are given in the following table:

Age group of drivers (in years)	Number of accidents in one year
0	1	2	3	more than 3
18 – 29	440	160	110	61	35
30 – 50	505	125	60	22	18
Above 50	360	45	35	15	9

Find the probability of the following events for a driver selected at random from the city:

(i) being 18 – 29 years of age and having exactly 3 accidents in one year

(ii) being 30 – 50 years of age and having one or more accidents in a year.

(iii) having no accident in one year.

Solution 15: Insurance and accidents problem

(i) The number of drivers in the age group 18–29 having exactly 3 accidents = 61

Total number of cases = 2000

So, P(driver in age group 18–29 having exactly 3 accidents in one year) = 61/2000 = 0.0305

(ii) The number of drivers having in the age group 30–50 and having one or more than one accidents in one year = 125 + 60 + 22 + 18 = 225

P(driver in age group 30–50 having one or more accidents in one year) = 225/2000 = 0.1125

(iii) The number of drivers having no accident in one year = 440 + 505 + 360 = 1305

So, P(driver having no accident) = 1305/2000 = 0.6525

Q16. 100 plants each were planted in 100 schools during Van Mahotsava. After one month, the number of plants that survived was recorded as in data below:

Number of plants survived	Less than 25	26 – 50	51 – 60	61 – 70	More than 70	Total number of schools
Number of schools (frequency)	15	20	30	30	5	100

When a school is selected at random for inspection, what is the probability that:

(i) More than 25 plants survived in the school?

(ii) Less than 61 plants survived in the school?

(iii) 61 to 70 plants survived in the school?

Solution 16: Plant survival problem

Total frequency or the total number of school in which plants were planted = 100

(i) Number of schools in which more than 25 plants survived

= 20 + 30 + 30 + 5 = 85

P(more than 25 plants survived in the school) = 85/100 = 0.85

(ii) Number of schools in which less than 61 plants survived) = 15 + 20 + 30 = 65

P(Less than 61 plants survived) = 65/100 = 0.65

(iii) P(61 to 70 plants survived) = 30/100 = 0.30

Exercise 1 (Multiple Choice Questions)

Q1. If P(E) = 0.37, then P(not E) will be

(a) 0.37

(b) 0.63

(c) 0.57

(d) None of these

Q2. Probability of getting even number in a single throw of dice is

(a) 1/2

(b) 1/6

(c) 5/6

(d) None of these

Q3. There are 5 prizes on 1000 tickets of a lottery of company. Probability of winning prize is

(a) 199/200

(b) 1/200

(c) 198/200

(d) None of these

Q4. Probability of drawing an ace from a deck of 52 cards is

(a) 1/52

(b) 1/26

(c) 1/13

(d) 3/52

Q5. There are 5 red and 3 black balls in a bag. Probability of drawing a black ball is

(a) 5/8

(b) 1/2

(c) 3/8

(d) None of these

Q6. If the probability of winning a game is 0.3, then probability of losing it is

(a) 0.6

(b) 0.7

(c) 0.5

(d) None of these

Q7. Probability of drawing '10' of a black suit from a deck of 52 cards is

(a) 1/52

(b) 1/26

(c) 1/13

(d) None of these

Q8. A dice is thrown once. Probability of getting a number 3 or 4 is

(a) 1/6

(b) 2/3

(c) 1/2

(d) 1/3

Q9. There are 50 tickets numbered from 1 to 50 in a box. Probability of drawing a ticket bearing prime number is

(a) 13/50

(b) 3/10

(c) 17/50

(d) None of these

Q10. Probability of a leap year having 53 Sundays is
(a) 1/7

(b) 3/7

(c) 2/7

(d) None of these

Q11. 3 Coins are tossed simultaneously. The probability of getting at least 2 heads is

(a) 3/10

(b) 3/4

(c) 3/8

(d) 1/2

Q12. Two cards are drawn successively with replacement from a pack of 52 cards. The probability of drawing two aces is

(a) 1/169

(b) 1/221

(c) 1/265

(d) 4/663

Q13. In a single throw of two dice, the probability of getting more than 7 is

(a) 7/36

(b) 7/12

(c) 5/12

(d) 5/36

Q14. Two cards are drawn at random from a pack of 52 cards. The probability that both are the cards of spade is

(a) 1/26

(b) 1/4

(c) 1/17

(d) None of these

Q15. Two dice are thrown together. The probability that sum of the two numbers will be a multiple of 4 is

(a) 1/9

(b) 1/3

(c) 1/4

(d) 5/9

Q16. If the odds in favour of an event be 3 : 5 then the probability of non-happening of the event is

(a) 3/5

(b) 5/3

(c) 3/8

(d) 5/8

Q17. In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

(a) 0.8

(b) 0.6

(c) 0.5

(d) 0.2

Q18. If the three coins are simultaneously tossed again compute the probability of 2 heads coming up.

(a) 3/8

(b) 1/4

(c) 5/8

(d) 3/4

Q19. A coin is tossed successively three times. The probability of getting one head or two heads is:

(a) 2/3

(b) 3/4

(c) 4/9

(d) 1/9

Q20. One card is drawn from a pack of 52 cards. What is the probability that the drawn card is either red or king:

(a) 15/26

(b) 1/2

(c) 7/13

(d) 17/32

Answers to Exercise 1

1. (b)	2. (a)	3. (b)	4. (c)	5. (c)
6. (b)	7. (b)	8. (d)	9. (b)	10. (c)
11. (d)	12. (a)	13. (c)	14. (c)	15. (c)
16. (d)	17. (a)	18. (a)	19. (b)	20. (c)

Exercise 2 (Descriptive Questions)

Q1. Two dice are thrown together. Find the probability of getting a total of 9.

Q2. A coin and a dice are tossed simultaneously find the sample space.

Q3. A dice is thrown repeatedly until a six comes up. What is the sample space for this experiment.

Q4. On a simultaneous toss of three coins, find the probability of getting
(i) at least 2 heads
(ii) at most 2 heads
(iii) exactly 2 heads

Q5. Two dice are thrown simultaneously. Find the probability of getting

(i) an even number as the sum

(ii) the sum as a prime number

(iii) a doubled of even number

Q6. Three dice are thrown together. Find the probability of getting a total of at least 6.

Q7. Find the probability that a leap year selected at random will contain 53 Tuesday.

Q8. A coin is tossed 80 times with the following outcomes:

(i) head: 35

(ii) tail: 45

Find the probability of each event.

Q9. Two coins are tossed simultaneously 150 times and we get the following outcomes.

(a) No tail = 45

(b) One tail = 55

(c) Two tails = 50

Find the probability of each event.

Q10. In a cricket match a batsman hits a boundary 10 times out of 36 balls he play. Find the probability that he did not hit the boundary.

Q11. In a cricket match a batsman hits a boundary 3 times in 3 over he play. Find the probability that he did not hit the boundary.

Q12. A bag which contains 7 blue marbles, 4 black marbles and 9 white marbles. A marble drawn at random from the bag then what is the probability that the drawn marble is
(i) blue
(ii) white or black

Q13. The odds in favour of an event are 3 : 5 find the probability of occurrence of this event.

Q14. Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box, find the probability that the number on card is

(i) An even number

(ii) A number less than 14

(iii) A number which is a perfect square.

(iv) A prime number less than 20

Q15. An urn contains 6 oranges, 7 apples & 11 mango. A fruit is drawn at random, what is the probability of drawing.

(i) An orange

(ii) Not apple

(iii) An apple or a mango

Q16. A card is drawn at random from a well shuffled deck of playing cards. Find the probability that the card drawn is

(i) A card of spade or an ace

(ii) A red king

(iii) Neither a king nor a queen

(iv) Either a king or a queen

Q17. A box contains 19 balls bearing numbers 1,2,3..... 19. A ball is drawn at random from the box. Find the probability that the number on the ball is

(i) A prime number

(ii) Divisible by 3 or 5

(iii) Neither divisible by 5 nor by 10

(iv) An even number

Q18. There are 30 cards of same size in a bag containing numbers 1 to 30. One card is taken out from the bag at random. Find the probability that the number on the selected card is not divisible by 3.

Answers to Exercise 2

1. 1/9

2. (H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6) (T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6)

3. {6, (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (1, 1, 6) (1, 2, 6).... }

4. (1/2, 7/8, 3/8)

5. (1/2, 5/12, 1/12)

6. 103/108

7. 2/7

8. (i) 7/16 (ii) 9/16

9. (a) 3/10 (b) 11/30 (c) 1/3

10. 13/18

11. 3/8

12. (i) 7/20 (ii) 13/20

13. 3/8

14. (i) 1/2 (ii) 3/25 (iii) 2/25 or 9/100 (iv) 2/25

15. (i) 1/4 (ii) 17/24 (iii) 3/4

16. (i) 4/13 (ii) 1/26 (iii) 11/13 (iv) 2/13

17. 8/19, 8/19, 16/19, 9/19

18. 2/3