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(C) The total number of possible outcomes when a balanced die is rolled is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes $n(S) = 6$.Let $

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$1/2$

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(C) The total number of possible outcomes when a balanced die is rolled is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes $n(S) = 6$.Let $E$ be the event that the number on the die is greater than $3$.The favorable outcomes are $E = \{4, 5, 6\}$,so the number of favorable outcomes $n(E) = 3$.The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.Substituting the values,we get $P(E) = \frac{3}{6} = \frac{1}{2}$.

$A$ balanced die is rolled once. Then,the probability that the number on the die is greater than $3$ is ........

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