Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
647 views
in Statistics and probability by (54.8k points)

Let n = 10k + r when k, r ∈ N, 0 ≤ r ≤ 9. A number a is chosen at random from the set {1, 2, ...., n} and let pn denote the probability that a2 – 1 is divisible by 10.

If r = 9, then pn equals

(A) 2k/n

(B) 2(k + 1)/n

(C) (2k + 1)/n

(D) k/n 

1 Answer

+1 vote
by (52.5k points)
selected by
 
Best answer

Answer is (B) 2(k + 1)/n

n = 10k + r, k, r ∈ N, 0 ≤ r ≤ 9

Unit place of a2 will contain 0, 1, 4, 5, 6, 9 only.

Hence, a2 – 1 is divisible by 10 only if unit place of a2 contain 1.

If unit place of a2 is 1, then unit place of a will be 1 or 9.

n = 10k + 9

Number of a whose unit place is 1 or 9 = 2(k + 1)

Therefore,

pn = 2(k + 1)/n

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...