Question

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?

Answer 1

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