Let S be the sample space
If two dice are rolled then n(S) = 6
2
= 36
A be the event of getting the number in each die is even then
n(A) = { (2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6) } = 9
P
(
A
)
=
n
(
A
)
n
(
S
)
=
9
36
=
1
4
B be the event of getting sum of the numbers appearing on two dice is 5 then
n(B) = { (1,4),(2,3),(3,2),(4,1) } = 4
P
(
B
)
=
n
(
B
)
n
(
S
)
=
4
36
=
1
9
Q.
A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the numbers
1
,
2
,
3
,
.
.
.
,
12
as shown in the figure. What is the probability that it will point to
(i)
6
?
(ii) an even number?
(iii) a prime number ?
(iv) a number which is a multiple of
5
?
Solution:
Total number of outcomes
=
12
(i) Favourable outcome:
6
No. of favourable outcomes
=
1
∴
P
(Getting
6
)
=
1
12
(ii) Favourable outcomes:
2
,
4
,
6
,
8
,
10
,
12
No. of favourable outcomes
=
6
∴
P
(Getting even number)
=
6
12
=
1
2
(iii) Favourable outcome:
2
,
3
,
5
,
7
,
11
No. of favourable outcomes
=
11
∴
P
(Getting a prime number)
=
5
12
(iv) Favourable outcome:
5
,
10
No. of favourable outcomes
=
2
∴
P
(Getting a multiple of
5
)
=
2
12
=
1
6
Q.
A lot consists of 144 ballpoint pens of which 20 are defective and others good. Tanvy will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her.
What is the probability that (i) she will buy it, (ii) she will not buy it?
Solution:
Given that total number of pens n(S) = 144.
Given that total number of defective pens = 20.
That means total number of non-defective pens = 144 - 20
= 124.
(1) She will buy:
Let A be event that she will buy a non-defective pen
n(A) = .124.
Therefore the required probability P(A) = n(A)/n(S)
= 124/144
= 31/36.
(2) She will not buy it.
Let B be the event that of getting a defective pen.
n(B) = 20.
Therefore the required probability P(B) = n(B)/n(S)
= 20/144
= 5/36.
Q.
A box contains 90 disces which are numbered from 1 to 90. If one disc is drawn at random from the box, find rthe probability that it bears (i) a two-digit number, (ii) a perfect square number, (iii) a number divisible by 5.
Solution:
1. number of favourable outcomes = 90 - 9 =81
Probability=
81
90
=
9
10
2. number of favourable outcomes = {1,4,9,16,25,36,49,64,81} = 9
Probability =
9
90
=
1
10
3. number of favourable outcomes ={5,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90}
P =
18
90
=
1
5
Q.
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this blb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and not replaced Now, bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
(i) Total number of bulbs = 20
Total number of defective bulbs = 4
P (getting a defective bulb) =
n
u
m
b
e
r
o
f
f
a
v
o
u
r
a
b
l
e
o
u
t
c
o
m
e
s
t
o
t
a
l
n
u
m
b
e
r
o
f
o
u
t
c
o
m
e
s
=
4
20
=
1
5
(ii) Remaining total number of bulbs = 19
Remaining total number of non-defective bulbs = 16 − 1 = 15
P (getting a not defective bulb)=
15
19
Q.
One card is drawn from a well- shuffled deck of 52 cards. Find the probability of drawing
(i) an ace (ii) a '4' of spades
(iii) a '9' of a black suit (iv) a red king.
Solution:
One card is drawn from a well- shuffled deck of 52 cards
i) favourable outcomes = 4
probability =
4
52
=
1
13
ii) favorable outcome = 1
probability =
1
52
iii) favourable outcomes = 2
probability =
2
52
= probability =
1
26
iv)favourable outcomes = 2
probability =
2
52
= probability =
1
26
Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 15 Exercise 15.1
-
Clear Understanding:
The solutions provide detailed step-by-step explanations that help students grasp the fundamental concepts of probability making it easier to understand how to calculate the likelihood of different events.
-
Enhanced Problem-Solving Skills:
By working through the exercises and solutions students improve their ability to solve various probability problems which strengthens their overall mathematical skills.
-
Effective Exam Preparation:
The solutions cover important probability concepts and types of questions that are commonly found in exams, helping students prepare thoroughly and perform better in their board exams.
-
Concept Reinforcement:
The exercise helps reinforce key probability principles, ensuring that students have a solid understanding of the subject matter and are able to apply these principles effectively.
-
Expert Guidance:
Prepared by subject experts, these solutions ensure accuracy and reliability, providing students with trusted resources for learning and practicing probability.
-
Confidence Building:
With clear explanations and practice problems, students can build their confidence in handling probability questions, reducing exam-related anxiety.
(2) possible outcomes = 17
Favourable outcomes are 5,10,15
Number of favourable outcomes = 3
P(. getting a number divisible by 5) =
