Question

A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number $1,2,3.....12$,then the probability that it will point to an odd number is

Answer 1

Let's find the probability that it will point to an odd number is

The total number of possible outcomes here is $1,2,3,4,5,6,7,8,9,10,11,12$

So, $n\left(S\right)=12$

Let $A$ be the event that the arrow will point to an odd number.

Thus, the favourable outcomes = $1,3,5,7,9,11.$

I.e., $n\left(A\right)=6$

Hence, the probability that the arrow will point to an odd number,

$$\begin{gathered} P\left(A\right)=\frac{6}{12} \\ P\left(A\right)=\frac{1}{2} \end{gathered}$$

Hence, the answer is $\frac{1}{2}$