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(B) When a fair die is thrown,the total number of possible outcomes is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes $n(S) = 6$.We want t

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$\frac{2}{3}$

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(B) When a fair die is thrown,the total number of possible outcomes is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes $n(S) = 6$.We want to find the probability of getting a number greater than $2$.The favorable outcomes are $E = \{3, 4, 5, 6\}$.The number of favorable outcomes is $n(E) = 4$.The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.Therefore,$P(E) = \frac{4}{6} = \frac{2}{3}$.

The probability of getting a number greater than $2$ in throwing a die is

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