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(A) regular $n$-sided polygon has $N = \frac{n(n-3)}{2}$ diagonals.For $n = 15$,the total number of diagonals is $N = \frac{15(15-3)}{2} = \frac{15

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$\frac{2}{3}$

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(A) regular $n$-sided polygon has $N = \frac{n(n-3)}{2}$ diagonals.For $n = 15$,the total number of diagonals is $N = \frac{15(15-3)}{2} = \frac{15 	imes 12}{2} = 90$.Shortest diagonals connect vertices separated by one vertex (e.g.,$V_1$ to $V_3$). There are $n = 15$ such diagonals.Longest diagonals connect vertices separated by $\frac{n-1}{2}$ vertices (e.g.,$V_1$ to $V_8$). There are $n = 15$ such diagonals.The number of diagonals that are neither shortest nor longest is $90 - (15 + 15) = 90 - 30 = 60$.The probability is $\frac{60}{90} = \frac{2}{3}$.

In a regular $15$-sided polygon with all its diagonals drawn,a diagonal is chosen at random. The probability that it is neither a shortest diagonal nor a longest diagonal is

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