A die has two faces each with number $^{\prime}1^{\prime}$ , three faces each with number $^{\prime}2^{\prime}$ and one face with number $^{\prime}3^{\prime}$. If die is rolled once, determine $P($ not $3)$
A die has two faces each with number $^{\prime}1^{\prime}$ , three faces each with number $^{\prime}2^{\prime}$ and one face with number $^{\prime}3^{\prime}$. If die is rolled once, determine $P($ not $3)$
Total number of faces $=6$
Number of faces with number $^{\prime}3^{\prime}=1$
$\therefore $ $P(3)=\frac{1}{6}$
Thus, $P($ not $3)=1-P(3)=\frac{1}{6}=\frac{5}{6}$
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