the total possible outcome of a die is 6
n(T) = 6
the sample space {1,2,3,4,5,6}
the odd numbers are {1,3,5}
thus n(O) = 3
numbers greater than 3 are {4,5,6}
thus n(>3) = 3
the probability of getting an odd number or a number greater than 3
is Pr(O) U Pr(>3)
![\begin{gathered} Pr\text{ (O) = }(n(O))/(n(T))=(3)/(6)=(1)/(2) \\ Pr(>3)\text{ = }(n(>3))/(n(T))=\text{ }(3)/(6)=(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kb3c2mbw93syib4tq9u44du4jn9arqvypc.png)
![\begin{gathered} Pr\text{ (O U >3) = Pr(O) + Pr(>3)} \\ \text{ = }(1)/(2)\text{ + }(1)/(2)\text{ = 1} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4n55ow2tzkl4try7mvc1yj2kwl74gpmhga.png)
the probabilty of that it is an odd number or a number greater than 3 is 1.000 (nearest thousandth)