Examsnet
Unconfined exams practice
Home
Exams
Banking Entrance Exams
CUET Exam Papers
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exam Practice
Insurance Exams
International Exams
JEE Exams
LAW Entrance Exams
MBA Entrance Exams
MCA Entrance Exams
Medical Entrance Exams
Other Entrance Exams
Police Exams
Public Service Commission (PSC)
RRB Entrance Exams
SSC Exams
State Govt Exams
Subjectwise Practice
Teacher Exams
SET Exams(State Eligibility Test)
UPSC Entrance Exams
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
CBSE
CBSE Question Papers
NCERT Books
NCERT Exemplar Books
NCERT Study Notes
CBSE Study Concepts
CBSE Class 10 Solutions
CBSE Class 12 Solutions
NCERT Text Book Class 11 Solutions
NCERT Text Book Class 12 Solutions
ICSE Class 10 Papers
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Careers
Contact Us
Our Apps
Privacy
Test Index
JEE Advanced 2020 Paper 2
Show Para
Hide Para
Share question:
© examsnet.com
Question : 51
Total: 54
Two fair dice, each with faces numbered 1,2,3,4,5 and 6, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If p is the probability that this perfect square is an odd number, then the value of 14p is _____
[JEE Adv 2020 P2]
Your Answer:
Validate
Solution:
Output
Ways to Occur
Perfect squares → 1, 4, 9
7 (1 will never occur)
primes → 2, 3, 5, 7, 11
15
Remaining → 6, 8, 10, 12
14
Given process stopped by a perfect square = E (Let event name)
and Process stopped by a odd square = O (Let event name)
To find
P
(
O
E
)
=
P
(
O
∩
E
)
P
(
E
)
P
(
O
∩
E
)
= Probability of occuring 9 before 1, 4, 2, 3, 5, 7, 11
=
(
4
36
)
+
(
4
36
)
(
14
36
)
+
(
14
36
)
2
(
4
36
)
+
.
.
.
.
.
.
∞
P(E) = Probability of occuring 4 or 9 before 2, 3, 5, 7, 11
=
(
7
36
)
+
(
14
36
)
(
7
36
)
+
(
14
36
)
2
(
7
36
)
+
.
.
.
.
.
∞
⇒
P
(
O
E
)
=
4
36
7
36
=
4
7
=
p
∴
14
p
=
8
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Prev Question
Next Question
More Free Exams:
AIEEE Previous Papers
BITSAT Exam Previous Papers
JEE Advanced Chapters Wise Questions
JEE Advanced Model Papers
JEE Mains Chapters Wise Previous Papers
JEE Mains Model Papers
JEE Mains Previous Papers
VITEEE Previous Papers