A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1,2,3,4,5,6,1,8 (see Figure), and these are equally likely outcomes. What is the probability that it will point at(i)         8?(ii)        an odd number?(iii)       a number greater than 2?(iv)       a number less than 9?

Solution: Here, sample space, $S=\left\{1,2,3,4,5,6,7,8\right\}$
So, $n\left(S\right)=8$
As, pointing a number are equally likely outcomes, so, pointing a number will have a probability of $\frac{1}{8}$.

(i) From above, we can say that arrow pointing at number $8$ will have a probability of $\frac{1}{8}$.

(ii)Odd numbers in given sample space, $E_o=\left\{1,3,5,7\right\}$
So,$n\left(E_o\right)=4$
Probability of arrow pointing at odd number, $P\left(E_o\right)=\frac{4}{8}=\frac{1}{2}$

(iii)Number greater than 2,$E_g=\left\{3,4,5,6,7,8\right\}$
So,$n\left(E_g\right)=6$
Probability of arrow pointing a number greater than 2, $P\left(E_g\right)=\frac{6}{8}=\frac{3}{4}$

(iii)Number less than 9,$E_l=\left\{1,2,3,4,5,6,7,8\right\}$
So,$n\left(E_l\right)=6$
Probability of arrow pointing a number less than 9, $P\left(E_l\right)=\frac{8}{8}=1$