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Sagot :
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)
[tex]\begin{gathered} Pr\text{ (O) = }\frac{n(O)}{n(T)}=\frac{3}{6}=\frac{1}{2} \\ Pr(>3)\text{ = }\frac{n(>3)}{n(T)}=\text{ }\frac{3}{6}=\frac{1}{2} \end{gathered}[/tex][tex]\begin{gathered} Pr\text{ (O U >3) = Pr(O) + Pr(>3)} \\ \text{ = }\frac{1}{2}\text{ + }\frac{1}{2}\text{ = 1} \end{gathered}[/tex]the probabilty of that it is an odd number or a number greater than 3 is 1.000 (nearest thousandth)
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