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Page 1
Q u e s t i o n : 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?
S o l u t i o n :
Total number of tosses = 500
Number of heads = 285
Number of tails = 215
i
Let E be the event of getting a head.
Pgettingahead
= P (E) =
Number of heads coming up
Total number of trials
=
285
500
= 0. 57
ii
Let F be the event of getting a tail.
Pgettingatail
= P (F) =
Number of tails coming up
Total number of trials
=
215
500
= 0. 43
Q u e s t i o n : 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?
S o l u t i o n :
T
otal 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 E
1
, E
2
, E
3
be the events of getting 2 heads, 1 head and 0 head, respectively.
Then,
i
Pgetting2heads
= P(E
1
) =
Number of times 2 heads appear
Total number of trials
=
112
400
= 0. 28
ii
Pgetting1head
= P(E
2
) =
Number of times 1 head appears
Total number of trials
=
160
400
= 0. 4
Page 2
Q u e s t i o n : 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?
S o l u t i o n :
Total number of tosses = 500
Number of heads = 285
Number of tails = 215
i
Let E be the event of getting a head.
Pgettingahead
= P (E) =
Number of heads coming up
Total number of trials
=
285
500
= 0. 57
ii
Let F be the event of getting a tail.
Pgettingatail
= P (F) =
Number of tails coming up
Total number of trials
=
215
500
= 0. 43
Q u e s t i o n : 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?
S o l u t i o n :
T
otal 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 E
1
, E
2
, E
3
be the events of getting 2 heads, 1 head and 0 head, respectively.
Then,
i
Pgetting2heads
= P(E
1
) =
Number of times 2 heads appear
Total number of trials
=
112
400
= 0. 28
ii
Pgetting1head
= P(E
2
) =
Number of times 1 head appears
Total number of trials
=
160
400
= 0. 4
iii
Pgetting0head
= P(E
3
) =
Number of times 0 head appears
Total number of trials
=
128
400
= 0. 32
Remark: Clearly, when two coins are tossed, the only possible outcomes are E
1
, E
2
and E
3
and P(E
1
) + P(E
2
) +
P(E
3
) = 0.28 +0.4 +0.32
= 1
Q u e s t i o n : 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?
S o l u t i o n :
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 E
1
, E
2
, E
3
and E
4
be the events of getting 3 heads, 2 heads, 1 head and 0
head, respectively. Then;
i
Pgetting3heads
= P(E
1
) =
Number of times 3 heads appear
Total number of trials
=
39
200
= 0. 195
ii
Pgetting1head
= P(E
2
) =
Number of times 1 head appears
Total number of trials
=
67
200
= 0. 335
iii
Pgetting0head
= P(E
3
) =
Number of times 0 head appears
Total number of trials
=
36
200
= 0. 18
iv
Pgetting2heads
= P(E
4
) =
Number of times 2 heads appear
Total number of trials
=
58
200
= 0. 29
Remark: Clearly, when three coins are tossed, the only possible outcomes are E
1
, E
2
, E
3 and
E
4
and P(E
1
) + P(E
2
) +
P(E
3
) + P(E
4
) = 0.195 +0.335 +0.18 +0.29
Page 3
Q u e s t i o n : 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?
S o l u t i o n :
Total number of tosses = 500
Number of heads = 285
Number of tails = 215
i
Let E be the event of getting a head.
Pgettingahead
= P (E) =
Number of heads coming up
Total number of trials
=
285
500
= 0. 57
ii
Let F be the event of getting a tail.
Pgettingatail
= P (F) =
Number of tails coming up
Total number of trials
=
215
500
= 0. 43
Q u e s t i o n : 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?
S o l u t i o n :
T
otal 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 E
1
, E
2
, E
3
be the events of getting 2 heads, 1 head and 0 head, respectively.
Then,
i
Pgetting2heads
= P(E
1
) =
Number of times 2 heads appear
Total number of trials
=
112
400
= 0. 28
ii
Pgetting1head
= P(E
2
) =
Number of times 1 head appears
Total number of trials
=
160
400
= 0. 4
iii
Pgetting0head
= P(E
3
) =
Number of times 0 head appears
Total number of trials
=
128
400
= 0. 32
Remark: Clearly, when two coins are tossed, the only possible outcomes are E
1
, E
2
and E
3
and P(E
1
) + P(E
2
) +
P(E
3
) = 0.28 +0.4 +0.32
= 1
Q u e s t i o n : 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?
S o l u t i o n :
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 E
1
, E
2
, E
3
and E
4
be the events of getting 3 heads, 2 heads, 1 head and 0
head, respectively. Then;
i
Pgetting3heads
= P(E
1
) =
Number of times 3 heads appear
Total number of trials
=
39
200
= 0. 195
ii
Pgetting1head
= P(E
2
) =
Number of times 1 head appears
Total number of trials
=
67
200
= 0. 335
iii
Pgetting0head
= P(E
3
) =
Number of times 0 head appears
Total number of trials
=
36
200
= 0. 18
iv
Pgetting2heads
= P(E
4
) =
Number of times 2 heads appear
Total number of trials
=
58
200
= 0. 29
Remark: Clearly, when three coins are tossed, the only possible outcomes are E
1
, E
2
, E
3 and
E
4
and P(E
1
) + P(E
2
) +
P(E
3
) + P(E
4
) = 0.195 +0.335 +0.18 +0.29
= 1
Q u e s t i o n : 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?
S o l u t i o n :
Total number of throws = 300
In a random throw of a dice, let E
1
, E
2
, E
3
, E
4
, be the events of getting 3, 6, 5 and 1, respectively. Then,
i
Pgetting3
= P(E
1
) =
Number of times 3 appears
Total number of trials
=
54
300
= 0. 18
ii
Pgetting6
= P(E
2
) =
Number of times 6 appears
Total number of trials
=
33
300
= 0. 11
iii
Pgetting5
= P(E
3
) =
Number of times 5 appears
Total number of trials
=
39
300
= 0. 13
iv
Pgetting1
= P(E
4
) =
Number of times 1 appears
Total number of trials
=
60
300
= 0. 20
Q u e s t i o n : 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
S o l u t i o n :
Total number of ladies = 200
Number of ladies who like coffee = 142
Number of ladies who dislike coffee = 58
Let E
1
and E
2
be the events that the selected lady likes and dislikes coffee, respectively.Then,
i
Pselectedladylikescoffee
Page 4
Q u e s t i o n : 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?
S o l u t i o n :
Total number of tosses = 500
Number of heads = 285
Number of tails = 215
i
Let E be the event of getting a head.
Pgettingahead
= P (E) =
Number of heads coming up
Total number of trials
=
285
500
= 0. 57
ii
Let F be the event of getting a tail.
Pgettingatail
= P (F) =
Number of tails coming up
Total number of trials
=
215
500
= 0. 43
Q u e s t i o n : 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?
S o l u t i o n :
T
otal 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 E
1
, E
2
, E
3
be the events of getting 2 heads, 1 head and 0 head, respectively.
Then,
i
Pgetting2heads
= P(E
1
) =
Number of times 2 heads appear
Total number of trials
=
112
400
= 0. 28
ii
Pgetting1head
= P(E
2
) =
Number of times 1 head appears
Total number of trials
=
160
400
= 0. 4
iii
Pgetting0head
= P(E
3
) =
Number of times 0 head appears
Total number of trials
=
128
400
= 0. 32
Remark: Clearly, when two coins are tossed, the only possible outcomes are E
1
, E
2
and E
3
and P(E
1
) + P(E
2
) +
P(E
3
) = 0.28 +0.4 +0.32
= 1
Q u e s t i o n : 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?
S o l u t i o n :
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 E
1
, E
2
, E
3
and E
4
be the events of getting 3 heads, 2 heads, 1 head and 0
head, respectively. Then;
i
Pgetting3heads
= P(E
1
) =
Number of times 3 heads appear
Total number of trials
=
39
200
= 0. 195
ii
Pgetting1head
= P(E
2
) =
Number of times 1 head appears
Total number of trials
=
67
200
= 0. 335
iii
Pgetting0head
= P(E
3
) =
Number of times 0 head appears
Total number of trials
=
36
200
= 0. 18
iv
Pgetting2heads
= P(E
4
) =
Number of times 2 heads appear
Total number of trials
=
58
200
= 0. 29
Remark: Clearly, when three coins are tossed, the only possible outcomes are E
1
, E
2
, E
3 and
E
4
and P(E
1
) + P(E
2
) +
P(E
3
) + P(E
4
) = 0.195 +0.335 +0.18 +0.29
= 1
Q u e s t i o n : 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?
S o l u t i o n :
Total number of throws = 300
In a random throw of a dice, let E
1
, E
2
, E
3
, E
4
, be the events of getting 3, 6, 5 and 1, respectively. Then,
i
Pgetting3
= P(E
1
) =
Number of times 3 appears
Total number of trials
=
54
300
= 0. 18
ii
Pgetting6
= P(E
2
) =
Number of times 6 appears
Total number of trials
=
33
300
= 0. 11
iii
Pgetting5
= P(E
3
) =
Number of times 5 appears
Total number of trials
=
39
300
= 0. 13
iv
Pgetting1
= P(E
4
) =
Number of times 1 appears
Total number of trials
=
60
300
= 0. 20
Q u e s t i o n : 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
S o l u t i o n :
Total number of ladies = 200
Number of ladies who like coffee = 142
Number of ladies who dislike coffee = 58
Let E
1
and E
2
be the events that the selected lady likes and dislikes coffee, respectively.Then,
i
Pselectedladylikescoffee
= P(E
1
) =
Number of ladies who like coffee
Total number of ladies
=
142
200
= 0. 71
ii
? P selectedladydislikescoffee
= P(E
2
) = ? Number of ladies who dislike coffee
Total number of ladies
=
58
200
= 0. 29
REMARK: In the given survey, the only possible outcomes are E
1
and E
2
and ?P(E
1
) + ?P(E
2
) = 0.71 +0.29
= 1
Q u e s t i o n : 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?
S o l u t i o n :
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% marks
Total number of unit tests
=
2
6
=
1
3
Q u e s t i o n : 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?
S o l u t i o n :
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) =
84
240
= 0.35
Q u e s t i o n : 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?
S o l u t i o n :
Total phone numbers on the directory page = 200
= =
Total number of unit tests 6 3
Page 5
Q u e s t i o n : 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?
S o l u t i o n :
Total number of tosses = 500
Number of heads = 285
Number of tails = 215
i
Let E be the event of getting a head.
Pgettingahead
= P (E) =
Number of heads coming up
Total number of trials
=
285
500
= 0. 57
ii
Let F be the event of getting a tail.
Pgettingatail
= P (F) =
Number of tails coming up
Total number of trials
=
215
500
= 0. 43
Q u e s t i o n : 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?
S o l u t i o n :
T
otal 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 E
1
, E
2
, E
3
be the events of getting 2 heads, 1 head and 0 head, respectively.
Then,
i
Pgetting2heads
= P(E
1
) =
Number of times 2 heads appear
Total number of trials
=
112
400
= 0. 28
ii
Pgetting1head
= P(E
2
) =
Number of times 1 head appears
Total number of trials
=
160
400
= 0. 4
iii
Pgetting0head
= P(E
3
) =
Number of times 0 head appears
Total number of trials
=
128
400
= 0. 32
Remark: Clearly, when two coins are tossed, the only possible outcomes are E
1
, E
2
and E
3
and P(E
1
) + P(E
2
) +
P(E
3
) = 0.28 +0.4 +0.32
= 1
Q u e s t i o n : 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?
S o l u t i o n :
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 E
1
, E
2
, E
3
and E
4
be the events of getting 3 heads, 2 heads, 1 head and 0
head, respectively. Then;
i
Pgetting3heads
= P(E
1
) =
Number of times 3 heads appear
Total number of trials
=
39
200
= 0. 195
ii
Pgetting1head
= P(E
2
) =
Number of times 1 head appears
Total number of trials
=
67
200
= 0. 335
iii
Pgetting0head
= P(E
3
) =
Number of times 0 head appears
Total number of trials
=
36
200
= 0. 18
iv
Pgetting2heads
= P(E
4
) =
Number of times 2 heads appear
Total number of trials
=
58
200
= 0. 29
Remark: Clearly, when three coins are tossed, the only possible outcomes are E
1
, E
2
, E
3 and
E
4
and P(E
1
) + P(E
2
) +
P(E
3
) + P(E
4
) = 0.195 +0.335 +0.18 +0.29
= 1
Q u e s t i o n : 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?
S o l u t i o n :
Total number of throws = 300
In a random throw of a dice, let E
1
, E
2
, E
3
, E
4
, be the events of getting 3, 6, 5 and 1, respectively. Then,
i
Pgetting3
= P(E
1
) =
Number of times 3 appears
Total number of trials
=
54
300
= 0. 18
ii
Pgetting6
= P(E
2
) =
Number of times 6 appears
Total number of trials
=
33
300
= 0. 11
iii
Pgetting5
= P(E
3
) =
Number of times 5 appears
Total number of trials
=
39
300
= 0. 13
iv
Pgetting1
= P(E
4
) =
Number of times 1 appears
Total number of trials
=
60
300
= 0. 20
Q u e s t i o n : 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
S o l u t i o n :
Total number of ladies = 200
Number of ladies who like coffee = 142
Number of ladies who dislike coffee = 58
Let E
1
and E
2
be the events that the selected lady likes and dislikes coffee, respectively.Then,
i
Pselectedladylikescoffee
= P(E
1
) =
Number of ladies who like coffee
Total number of ladies
=
142
200
= 0. 71
ii
? P selectedladydislikescoffee
= P(E
2
) = ? Number of ladies who dislike coffee
Total number of ladies
=
58
200
= 0. 29
REMARK: In the given survey, the only possible outcomes are E
1
and E
2
and ?P(E
1
) + ?P(E
2
) = 0.71 +0.29
= 1
Q u e s t i o n : 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?
S o l u t i o n :
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% marks
Total number of unit tests
=
2
6
=
1
3
Q u e s t i o n : 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?
S o l u t i o n :
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) =
84
240
= 0.35
Q u e s t i o n : 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?
S o l u t i o n :
Total phone numbers on the directory page = 200
= =
Total number of unit tests 6 3
i Number of numbers with units digit 5 = 24
Let E
1
be the event that the units digit of selected number is 5.
? Required probability = P(E
1
) =
24
200
= 0. 12
ii Number of numbers with units digit 8 = 16
Let E
2
be the event that the units digit of selected number is 8.
? Required probability = P(E
2
) =
16
200
= 0. 08
Q u e s t i o n : 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?
S o l u t i o n :
Total number of students = 40
i Number of students with blood group O = 14
Let E
1
be the event that the selected student's blood group is O.
? Required probability = P(E
1
) =
14
40
= 0. 35
ii Number of students with blood group AB = 6
Let E
2
be the event that the selected student's blood group is AB.
? Required probability = P(E
2
) =
6
40
= 0. 15
Q u e s t i o n : 1 0
12 Packets of salt, each marked 2 kg, actually contained the following weights inkg
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?
S o l u t i o n :
Total number of salt packets = 12
Number of packets which contains more than 2 kg of salt = 5
? PChosenpacketcontainsmorethan2kgofsalt
=
Number of packets which contains more than 2 kg of salt
Total number of salt packets
=
5
12
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FAQs on RS Aggarwal Solutions: Probability - Mathematics (Maths) Class 9
| 1. What is the importance of studying probability in Class 9? |  |
Ans. Studying probability in Class 9 is important as it helps students understand the concepts of uncertainty and randomness in real-life situations. It equips them with the skills to analyze and predict outcomes, which is essential in fields such as statistics, economics, and science.
| 2. How is probability calculated in Class 9 mathematics? | |
Ans. Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. This gives a value between 0 and 1, where 0 represents impossibility and 1 represents certainty. By understanding the concept of probability, students can determine the likelihood of an event occurring.
| 3. Can you provide an example of probability in Class 9 mathematics? | |
Ans. Sure! Let's consider the probability of rolling a fair die and getting an even number. The favorable outcomes are 2, 4, and 6, while the total number of possible outcomes is 6 (1, 2, 3, 4, 5, and 6). Therefore, the probability of getting an even number is 3/6, which simplifies to 1/2 or 0.5.
| 4. How does probability help in decision making? | |
Ans. Probability helps in decision making by providing a quantitative measure of uncertainty. By calculating probabilities, individuals can assess the likelihood of different outcomes and make more informed choices. For example, in business, probability analysis can help in assessing the risks and rewards associated with various investment options.
| 5. What are the practical applications of probability? | |
Ans. Probability has various practical applications in our daily lives. It is used in weather forecasting to predict the chance of rain, in insurance to calculate premiums based on risk assessment, in sports to determine the likelihood of a team winning, and in genetics to understand the probability of inheriting certain traits. Probability is also utilized in gambling and games of chance.
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