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?
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?

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

to find: Probability of following

So, Total numbers on the spin is 12

(i) Now, Favorable outcomes i.e. to get $10$ is $1$

Therefore, total number of favorable outcomes i.e. to get $10$ is $1$

We know that the Probability = Number of favorable outcomes/ Total number of outcomes

Hence, the probability of getting a $10=1/12$

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

Thus, total number of favorable outcomes i.e. to get a prime number is $6$

We know that the Probability = Number of favorable outcomes/ Total number of outcomes

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