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 figure. 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

Given: A game of chance consists of spinning an arrow which is equally likely to come to rest pointing number 1, 2, 3 ….12

Required to find: Probability of following

Total numbers on the spin is 12

(i) Favourable outcomes i.e. to get 10 is 1

So, total number of favourable outcomes i.e. to get 10 is 1

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a 10 = 1/12

(ii) Favourable outcomes i.e. to get an odd number are 1, 3, 5, 7, 9, and 11

So, total number of favourable outcomes i.e. to get a prime number is 6

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a prime number = 6/12 = 1/2

(iii) Favourable outcomes i.e. to get a multiple of 3 are 3, 6, 9, and 12

So, total number of favourable outcomes i.e. to get a multiple of 3 is 4

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting multiple of 3 = 4/12 = 1/3

(iv) Favourable outcomes i.e. to get an even number are 2, 4, 6, 8, 10, and 12

So, total number of favourable outcomes i.e. to get an even number is 6

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an even number = 6/12 = 1/2