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.. 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

The possible number of outcomes, n(S) = 12 

(i) Number of favorable outcomes, 

n(E) = 1

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{1}{12}\)

(ii) Number of favorable outcomes, 

n(E) = 6

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\)

(iii) Number of favorable outcomes, 

n(E) = 4

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{12}\) = \(\frac{1}{3}\)

(iv) Number of favorable outcomes, 

n(E) = 6

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\)