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?
Total number times coin tossed: 500
Number of times head occurred: 285
Number of times tail occurred: 215
Probability P() =
(i). Let p(H) be probability of head
P (H) =
P (H) = = 0.57
(ii). Let p(T) be probability of tale
P (T) =
P (T) =
Two coins are tossed 400 times and we get
Two heads:112 times; one head 160 times ;0 times; 0 head :128 times.
When two coins are tossed at random, what is the probability of getting
(i) 2 heads? (ii) 1 heads? (iii) 0 head?
Total number times coin tossed: 400
Number of times Two heads occurred: 112
Number of times One head occurred: 160
Number of times No head (zero head) occurred: 128
Probability P()=
(i). Let p(H1) be probability of getting Two heads
P (H1) =
P (H1) = = 0.28
(ii). Let p(H2) be probability of getting One head
P (H2) =
P (H2) = = 0.4
(iii). Let p(H3) be probability of getting No head
P (H3) =
P (H3) = = 0.32
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 there coins are tossed at random, what is the probability of getting
(i) 3 heads? (ii) 1 head? (iii) 0 head (iv) 2 heads?
Total number times coins tossed: 200
Number of times Three heads occurred: 39
Number of times Two head occurred: 58
Number of times One heads occurred: 67
Number of times No(Zero) head occurred: 36
Probability P() =
(i). Let P(H1) be probability of getting Three heads
P (H1) =
P (H1) = = 0.195
(ii). Let P(H2) be probability of getting One head
P (H2) =
P(H2) = = 0.335
(iii). Let P(H3) be probability of getting No heads (zero heads)
P (H3) =
P (H3) = = 0.18
(iv). Let P(H4) be probability of getting Two heads
P (H4) =
P (H4) = = 0.29
A die is thrown 300 times and the outcomes are noted as given below.
When a die is thrown at random, what is the probability of getting a
(i) 3? (ii) 6? (iii) 5? (iv)1?
Total number times a Die Rolled: 300
Number of times 3 occurred on die: 54
Number of times 6 occurred on die: 33
Number of times 5 occurred on die: 39
Number of times 1 occurred in die: 60
Probability =
(i). Let p(3) be probability of getting 3 on die
P (3) =
P (3) = = 0.18
(ii). Let p(6) be probability of getting 6 on die
P (6) =
P (6) = = 0.11
(iii). Let p(5) be probability of getting 5 on die
P (5) =
P (5) = = 0.13
(iv). Let p(1) be probability of getting 1 on die
P (1) =
P (1) = = 0.2
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.
Total number of ladies: 200
Number of ladies who like coffee: 142
Number of ladies who dislike coffee: 58
Probability P() =
(i). Let p(Coffee) be probability of ladies who like coffee
P (Coffee) =
P (Coffee) = = 0.71
(ii). Let p(No Coffee) be probability of ladies who dislikes coffee
P (No Coffee) =
P (No Coffee) = = 0.29
The percentages of marks obtained by a student in six unit tests are given below:
A unit test is selected at random. What is the probability that the student gets more than 60% marks in the test ?
Total number of unit tests: 6
Number of Unit tests in which, the student got more than 60%: 2
That includes 1.) unit test II and
2.) unit test V
Probability P() =
1.Let P(U) be probability of Student scoring more than 60% in unit tests
P (U) =
P (U) = =
On a particular day, at a crossing in a city, the various types of 240 vehicles going past during a time interval were observed as under:
Out of these vehicles, one is chosen at random, What is the probability that the chosen vehicle is a two wheeler?
Total number of vehicles observed: 240
Number of Two wheeler vehicles: 84
Probability P() =
1.Let P(Two) be probability of Two wheelers
P (Two) =
P (Two) = = 0.35
On one page of a telephone directory, there are 200 phone number. The frequency distribution of their units digits is given below:
One of the number is chosen at random from the page. What is the probability that the units digit of the chosen number is (i) 5? (8)?
Total number of Numbers on a page: 200
Number of telephone Numbers which have 5 in its units place: 24
Number of telephone Numbers which have 8 in its units place: 16
Probability P() =
1.Let P(5) be probability of telephone number having 5 in the units place
P (5) =
P (5) = = 0.12
2.Let P(8) be probability of telephone number having 8 in units place P (8) =
P (8) = = 0.08
The following table shows the blood groups of 40 students of a class.
One student of the class is chosen at random. What is the probability that the chosen student has blood group(i) O? (ii) AB?
Total number of Students: 40
Number of Students having blood group O: 14
Number of Students having blood group AB: 6
Probability P() =
1.Let P(O) be probability of selecting a student with Blood group O
P (O) =
P (O) = = 0.35
2.Let P(AB) be probability of selecting a student with Blood group AB
P (AB) =
P (AB) = = 0.15
The table given below shows the marks obtained by 30 students in a test.
Out of these students, one is chosen at random. What is the probability that his marks lie in the interval 21-30?
Total number of Students: 30
Number of Students having marks in the range 21-30: 6
Probability P() =
1.Let P(S) be probability of selecting a student having marks in the range 21-30
P (S) =
P (S) = = 0.2
Following are the ages (in years) of 360 patients, getting medical treatment in a hospital:
One of the patients is selected at random.
Find the probability that his age is
(i) 30 Year or more but less than 40 years.
(ii) 50 year or more but less than 70 years
(iii) Less than 10 years.
(iv) 10 years or more.
Total number of Patients: 360
Number of Patients who are 30 Years or more but less than 40 years: 60
(This include age groups between 30-40)
Number of Patients who are 50 Years or more but less than 70 years: 80
(This include patients of age groups 50-60 and 60-70 therefore 50+30=80)
Number of Patients who are less than 10 years: 0 (No patients below 10 years)
Number of Patients who are 10 years or more: 360 (this include all age -
groups admitted in the hospital)
Probability P() =
(i). Let P(P1) be probability of patients between age groups 30-40
P (P1) =
P (P2) =
(ii). Let P(P2) be probability of patients between age groups 50-70
P (P2) =
P (P2) =
(iii). Let P(P3) be probability of patients who are less than 10 years
P (P3) =
P (P3) =
(iv). Let P(P4) be probability of patients whose age is more than 10 years
P (P4) =
P (P4) =
A coin is tossed 100 times with 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.
B.
C.
D.
Total number times a coin is tossed = 100
Number of times head occurred = 43
Number of times tail occurred = 57
Probability P() =
Probability of getting a head is P(head) = =
A coin is tossed 200 times with following outcomes:
Head 112 times and tail 88 times.
In a single throw of a coin, what is the probability of getting a head?
A.
B.
C.
D.
Total number times a coin is tossed = 200
Number of times head occurred = 112
Number of times tail occurred = 88
Probability P() =
Probability of getting a head is P(head) = = =
A survey of 200 persons of a locality shows the liking and disliking of tea.
Out of these one was chosen at random. What is the probability that the chosen person likes tea?
A.
B.
C.
D.
Total number of persons: 200
Number of persons who like tea: 148
Number of persons who dislike tea: 52
Probability P() =
Let p(Tea) be probability of person who like tea
P (Tea) =
P (Tea) = =
In a locality, 1000 families were chosen at random and the following data was collected:
Out of these families, a family was chosen at random. What is the probability that the chosen family has 2 children?
A.
B.
C.
D.
Total number of families: 1000
Number of families having no children: 6
Number of families having 1 child: 184
Number of families having 2 children: 672
Number of families having 3 children: 127
Number of families having 4 or more children: 11
Probability P() =
Let p(2) be probability of families having 2 children
P (2) =
P (2) = =
The table given below shows the month of birth of 36 students of a class:
A students is chosen at random from the class. What is the probability that the chosen student was born in October?
A.
B.
C.
D.
Total number of students: 36
Number of students born in October: 3
Probability P() =
Let p(Oct.) be probability of students born in October.
P (Oct.) =
P (Oct.) = =
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.
B.
C.
D.
Total number times a coin is tossed = 50
Number of times tail occurred = 32
Number of times tail occurred = 50 – 32 = 18
Probability P() =
Probability of getting a head is P(head) = = =
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 throw, the ball does not hit the boundary?
A.
B.
C.
D.
Total number balls batsman played = 30
Number of batsman hits a boundary = 6
Number of batsman doesn’t hit a boundary = 30 – 6 = 24
Probability P() =
Probability of batsman not hitting a boundary is
P(no boundary) = = =
A die is thrown 40 times and each time the number on the uppermost face is noted. It was recorded as under:
A die is thrown at random, What is the probability of getting a 5?
A.
B.
C.
D.
Total number of times die is thrown: 40
Number of times 5 noted on the uppermost face: 7
Probability P() =
Probability of getting 5 on uppermost face die is
P(5) = =
In 50 throws of a die, the outcomes were noted as under:
A die is the at random. What is the probability of getting an even number?
A.
B.
C.
D.
Total number of times die is thrown: 50
Number of times 2 noted on the uppermost face: 9
Number of times 4 noted on the uppermost face: 7
Number of times 6 noted on the uppermost face: 8
Number of times even number noted on the uppermost face: 9+7+8 = 24
Probability P() =
Probability of getting even number on uppermost face of die is
P(even) = = =
In 65 thrown of a die, the outcomes were noted as under:
A die is thrown at random. What is the probability of getting a prime number?
A.
B.
C.
D.
Total number of times die is thrown: 65
Number of times 2 noted on the uppermost face: 10
Number of times 3 noted on the uppermost face: 12
Number of times 5 noted on the uppermost face: 9
Number of times prime number noted on the uppermost face: 10+12+9 = 31
Probability P() =
Probability of getting even number on uppermost face of die is
P(prime) = =
On one page of a directory, there are 160 telephone numbers. The frequency distribution of the unit place digit is given as under:
From this page, one of the numbers is chosen at random. What is the probability that the unit place digit in the chosen number is 6?
A.
B.
C.
D.
Total number of telephone numbers on the page: 160
Number of telephone numbers which have 6 in units place is: 15
Probability P() =
Probability of getting 6 in unit place of a telephone number.
P(6) = = =
Two coins are tossed 1000 times and the outcomes are recorded as under:
A coin is thrown at random. What is the probability of getting at most one head?
A.
B.
C.
D.
Total number times a coin is tossed = 1000
Number of times no heads occurred = 194
Number of times one head occurred = 540
Number of times two head occurred = 266
Probability P() =
Probability of getting at most one head is
P(at most one head) = = = =
80 bulbs are selected at random from a lot and their lifetime is recorded in the from of a frequency table given below:
A bulb is chosen at random from the lot. What is the probability that the bulb chosen has lifetime less than 900 hours?
A.
B.
C.
D.
Total number bulbs= 80
Number of bulbs, which have lifetime of 300 hrs = 10
Number of bulbs, which have lifetime of 500 hrs = 15
Number of bulbs, which have lifetime of 700 hrs = 23
Number of bulbs, which have lifetime of 900 hrs = 25
Number of bulbs, which have lifetime of 1100 hrs = 7
Number of bulbs , having lifetime less than 900 hrs = 10 + 15 +23 = 48
Probability P() =
Probability of selecting a bulb which have a lifetime of 900 hrs
P(900) = = =
In a medical examination of 40 students of a class, the following blood, groups are recorded:
From this class, a student is chosen at random, What is the probability that the chosen student has blood group B?
A.
B.
C.
D.
Total number students= 40
Number of students with blood group ‘A’ = 11
Number of students with blood group ‘B’ = 15
Number of students with blood group ‘AB’ = 9
Number of students with blood group ‘O’ = 5
Probability P() =
Probability of selecting a student with blood group ‘B’
P(B) = = =
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.
B.
C.
D.
Total number of members in a group: 60
Number of members who like coffee in the group: 35
Number of members who dislike coffee in the group: 60 – 35 = 25
Probability P() =
Let p(No coffee) be probability of a member who dislikes the coffee
P (No coffee) =
P (No coffee) = =
A die is thrown 50 times and the outcomes are recorded as under.
If a die is thrown at random, what is the probability of getting 8?
A.
B.
C.
D. 0
Total number times a die is thrown: 50
We know that a die contains only 6 sides (1,2,3,4,5,6) and there is no 8 on any side of the die
Therefore, probability of getting 8 when a die is thrown is : 0
That is P(8) = 0
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
Given: P(winning) = 0.7
We know that if P(favorable) is the probability of favorable outcomes, then
P(unfavorable) , is probability of unfavorable outcomes Is given by
P(unfavorable) = 1 – P(favorable)
Therefore,
P(losing) = 1 – P(winning) = 1 – 0.7 = 0.3
A coin is tossed 60 times and the tail appears 35 times. What is the probability of getting a head?
A.
B.
C.
D.
Total number times a coin is tossed = 60
Number of times tail occurred = 35
Number of times tail occurred = 60 – 35 = 25
Probability P() =
Probability of getting a head is P(head) = = =
Each question consists of two statements, namely, Assertion(A) and Reason (R). Choose the correct options.
A. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion(A).
B. Both Assertion (A) and Reason (R) are true and Reason (R) is a Not explanation of Assertion(A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (A) is false and Reason (R) is false.
Here P(boundary) = =
P(no boundary) = 1 – =
Each question consists of two statements, namely, Assertion(A) and Reason (R) . Choose the correct options.
A. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion(A).
B. Both Assertion (A) and Reason (R) are true and Reason (R) is a Not explanation of Assertion(A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (A) is false and Reason (R) is false.
Here always the probability of a sure event is 1
and
P(E), where E is some event, always ranges between 0P (E ) 1.
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 ____________
(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(event)
(iv) p (E) +P (not E) = 1
(V) P (E) lies between 0 and 1
The marks obtained by 90 students of a school in Mathematics out of 100 are given as under:
From these students, a students is chosen at random.
What is the probability that the chosen student
(i) gets 20% or less marks?
(ii) gets 60% or more marks?
Total number of students : 90
Number of students who scored less than 20% : 7
Number of students who scored more than 60% : 19
Probability P() =
(i) Let P(20) be probability of students who scored less than 20%
P(20) = =
(ii) Let P(60) be probability of students who scored more than 60%
P(60) = =
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 nondefective?
Total number of electric bulbs = 800
Number of Non-defective bulbs = 800-36 = 764
Probability P() =
Probability a Non- defective bulbs P(bulbs) = = =
The table given below shows the ages of 75 teachers in a school.
Note Here 18-29 means from 18 to 29 including both. A teacher from the school is chosen at ramdom. What is the probability that the teacher chosen is
(i) 40 years or more then 40 years old?
(ii) 49 years or less then 49 years old?
(iii) 60 years or more then 60 years old?
Total number of teachers: 75
Number of teachers having age 40 years or more than 40 years old: 45
Number of teachers having age 49 years or less than 40 years old: 65 (5+25+35)
Number of teachers having age 60 years or more than 60 years old: 0
Probability P() =
(i) Let P(more than 40) be probability of teachers having age 40 years or more than 40 years old
P(more than 40) =
(ii) Let P(less than 49) of teachers having age 49 years or less than 40 years old
P(less than 49) =
(iii) Let P(more than 60) be probability 60 years or more then 60 years old
P(more than 60) =
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 the chosen bulb is defective?
A.
B.
C.
D.
Total number of electric bulbs = 600
Number of bulbs which are defective: 20
Number of Non-defective bulbs = 600-20 = 580
Probability P() =
Probability of getting a Non-defective bulbs P(bulbs) =
A bag contains 5 red, 8 black and 7 white balls. Is chosen at random. What is the probability that the chosen ball is black?
A.
B.
C.
D.
Total number of balls bag containing is: 5 red + 8 black + 7 white = 20 balls
Number of red balls = 8
Probability P() =
Probability of getting a black ball P(black) =
A bag contains 16 cards bearing number1,2,3…, 16 respectively. one card is chosen at random. What is the probability that the chosen card bears a number which is divisible by 3?
A.
B.
C.
D.
Total number of cards = 16
Chances of drawing a numbered card which is divisible by 3 = 5 (They are 3,6,9,12,15)
Probability P() =
Probability of drawing a numbered cards which is divisible by 3
P(card divisible by 3) =
In a cricket match a batsmen hits a boundary 4 times out of the 32 balls he plays. In a given ball. What is the probability that he does not hit the ball?
A.
B.
C.
D.
Total number balls batsman played = 32
Number of batsman hits a boundary = 4
Number of batsman doesn’t hit a boundary = 32 – 4 = 28
Probability P() =
Probability of batsman not hitting a boundary is
P(no boundary) =
Define the probability of an event E.
The probability of event E is defined as number of outcomes favorable to E divided by total numbers of equally likely outcomes in the sample space S of the experiment.
That is
P(E) =
A coin is tossed 60 times with the following outcomes:
Head 28 times and tail 32 times.
In a single throw of a coin, find the probability of getting a head.
Total number times a coin is tossed = 60
Number of times head occurred = 28
Number of times tail occurred = 32
Probability P() =
Probability of getting a head is P(head) =
When a die is thrown, write down all possible outcomes.
A die has total of 6 sides.
Therefore,
Total possible outcomes area : 6
They are {1,2,3,4,5,6}
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)< _______
(i) Probability of a sure event = 1
(ii) probability of an impossible event = 0
(iii) If E be an event, then p (E) +p (not E) = 1
(iv) If E is an event, then 0 <p (E)< 1
A die is thrown 80 times and the outcomes were noted as under:
If the die is thrown at random, What is the probability of getting a 2?
Total number of outcomes: 80
Number of times 2 occurred : 11
Probability P() =
Probability of getting a 2 on die is P(2) =
The probability of losing a game is 0.6 What is the probability of winning the game?
Given: P(winning) = 0.6
We know that if P(favorable) is the probability of favorable outcomes, then
P(unfavorable), is probability of unfavorable outcomes It given by
P(unfavorable) = 1 – P(favorable)
Therefore,
P(losing) = 1 – P(winning) = 1 – 0.6 = 0.4
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?
Total number times a coin is tossed = 50
Number of times tail occurred = 28
Number of times tail occurred = 50 – 28 = 22
Probability P() =
Probability of getting a head is P(head) =
In a one day cricket match, a batsman hits the boundary 8 times out of 48 balls he plays. Find the probability that he does not hit the boundary.
Total number balls batsman played = 48
Number of batsman hits a boundary = 8
Number of batsman doesn’t hit a boundary = 48 – 8 = 40
Probability P() =
Probability of batsman not hitting a boundary is
P(no boundary) =
Two coins are tossed simultaneously 80 times and the outcomes were recorded as under:
If two coins are tossed at random,find the probability of getting
(i) Two heads (ii) one head (iii) no head
Total number times a coin is tossed = 80
Number of times two heads occurred = 24
Number of times one head occurred = 36
Number of times no head occurred = 20
Probability P() =
(i) Probability of getting two heads is
P(Two heads) =
(ii) Probability of getting one head is
P(one heads) =
(iii) Probability of getting no head is
P(no head) =
Marks obtained by 90 students of Class IX in a test are given below:
Out of these students one is chosen at random. Find the probability that the chosen student obtains
(i) less than 20% marks (ii) 80% of more marks
(iii) 60% or more marks
Total number of students : 90
Number of students who scored less than 20% : 8
Number of students who scored more than 80% : 9
Number of students who scored more than 60% : 35
Probability P() =
(i) Let P(20) be probability of students who scored less than 20%
P(20) =
(ii) Let P(80) be probability of students who scored more than 80%
P(80) =
(iii) Let P(60) be probability of students who scored more than 60%
P(60) =
The blood group of 30 students of a class were recorded as under:
If a students of this class is chosen at random, What is the probability that the chosen students has blood group
(i)0? (ii) A? (iii) AB?
(i) O
Total number students= 30
Number of students with blood group ‘A’ = 9
Number of students with blood group ‘B’ = 11
Number of students with blood group ‘AB’ = 4
Number of students with blood group ‘O’ = 6
Probability P() =
(i) Probability of selecting a student with blood group ‘o’
P(B) =
(ii) Probability of selecting a student with blood group ‘A’
P(B) =
(iii) Probability of selecting a student with blood group ‘AB’
P(B) =
In a survey of 100 families having 2 or less boys. The following data was obtained:
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 boy?
Total number families surveyed = 100
Number of families having no boys = 18
Number of families having one boy = 46
Number of families having two boys = 36
Probability P() =
(i) Probability of family having one boy
P(one boy) =
(ii) Probability of family having 2 boys
P(two boy) =
(iii) Probability of family having no boy
P(no boy) =
A die is thrown 100 times and following observations were recorded:
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
Total number of outcomes : 100
(i) Chances of getting a number less than 3 on the die = 2 (They are 1,2)
Its frequency is: 12+18 = 30
Probability P() =
Probability of getting a number less than 3 on die P(less than 3)
=
(ii) Chances of getting a even number on the die = 3 (They are 2,4,6)
Frequency : 18+26+16 = 60
Probability P() =
Probability of getting a composite number on die the P(even number)
(iii) Chances of getting a number greater than 4 on the die = 2 (They are 5,6)
Frequence : 14+16 = 30
Probability P() =
Probability of getting a number not less than 4 on die P(greater than 4)
A survey of 600 students about their liking on coffee was conducted and recoded as under:
Out of these students one is chosen at random. What is the probability that the chosen student
(i) likes coffee? (ii) dislikes coffee?
Total number of students: 600
Number of students who like coffee: 360
Number of students who dislike coffee: 140
Probability P() =
(i) Let p( like ) be probability of ladies who dislike coffee
P (like) =
P (like) =
(ii) Let p( dislike ) be probability of ladies who dislike coffee
P (dislike) =
P (dislike) = =
Two coins are tossed 1000 times and the outcomes are recorded as given below:
In a simple throw of two coins, what is the probability of getting
(i) at most one head? (ii) At least one head?
Total number times a coin is tossed = 1000
Number of times no heads occurred = 240
Number of times one head occurred = 450
Number of times two head occurred = 310
Probability P() =
(i) Probability of getting at most one head is (that is Probability of no head and one head)
P( at most one head) =
(ii) Probability of getting at least one head is (that is Probability of one head and two head)
P( at least one head) =
A coin is tossed 80 times with 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.
Total number times a coin is tossed = 80
Number of times head occurred = 35
Number of times tail occurred = 45
Probability P() =
Probability of getting a head is P(head) =
Probability of getting a tails is P(tail) =
Here P(head) = and P(tail) =
Now, P(head) + P(tail)
Hence proved.