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(D) The total number of ways to choose $3$ numbers from $15$ is given by $^{15}C_3 = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 455$.To find

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$\frac{22}{35}$

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(D) The total number of ways to choose $3$ numbers from $15$ is given by $^{15}C_3 = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 455$.To find the number of ways to choose $3$ numbers such that no two are consecutive,we use the formula for choosing $r$ non-consecutive items from $n$ items,which is $^{n-r+1}C_r$.Here,$n = 15$ and $r = 3$. So,the number of favorable ways is $^{15-3+1}C_3 = ^{13}C_3$.$^{13}C_3 = \frac{13 	imes 12 	imes 11}{3 	imes 2 	imes 1} = 13 	imes 2 	imes 11 = 286$.The required probability is $\frac{286}{455}$.Dividing both numerator and denominator by $13$,we get $\frac{286 \div 13}{455 \div 13} = \frac{22}{35}$.

Three numbers are chosen at random from $1$ to $15$. The probability that no two numbers are consecutive is:

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