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You roll a six-sided die. What is the probability that it is an odd number or greater than three? Round your answer to the nearest thousandth. The probability is about

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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}


\begin{gathered} Pr\text{ (O U >3) = Pr(O) + Pr(>3)} \\ \text{ = }(1)/(2)\text{ + }(1)/(2)\text{ = 1} \end{gathered}

the probabilty of that it is an odd number or a number greater than 3 is 1.000 (nearest thousandth)

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