Sarthaks Test
0 votes
124 views
in Mathematics by (7.9k points)

A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number, 1, 2, 3, ..., 12 as shown in Fig. below. What is the probability that it will point to:

(i) 10?

(ii) an odd number?

(iii) a number which is multiple of 3?

(iv) an even number?

1 Answer

0 votes
by (13.1k points)
selected by
 
Best answer

Total no. of possible outcomes = 12 {1, 2, 3,…., 12}

(i) Let E ⟶ event of pointing 10

No. favourable outcomes = 1 {10}

P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 1/12

(ii) Let E ⟶ event of pointing at an odd no.

No. favourable outcomes = 6 {1, 3, 5, 7, 9, 11}

P(E) = 6/12 = 1/2

(iii) Let E ⟶ event of pointing at a no. multiple of 3

No. favourable outcomes = 4 {3, 6, 9, 12}

P(E) = 4/12 = 1/3

(iv) Let E ⟶ event of pointing at an even no.

No. favourable outcomes = 6 {2, 4, 6, 8, 10, 12}

P(E) = 6/12 = 1/2

Related questions

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.

Categories

...