Three numbers are chosen from 1 to 30. The pr | vedclass

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(D) The total number of ways to choose $3$ numbers from $30$ is given by $C(30, 3) = \frac{30 	imes 29 	imes 28}{3 	imes 2 	imes 1} = 4060$. Next,

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$\frac{144}{145}$

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(D) The total number of ways to choose $3$ numbers from $30$ is given by $C(30, 3) = \frac{30 	imes 29 	imes 28}{3 	imes 2 	imes 1} = 4060$. Next,we find the number of ways to choose $3$ consecutive numbers. These are $(1, 2, 3), (2, 3, 4), \dots, (28, 29, 30)$. There are $28$ such sets of consecutive numbers. The probability of choosing $3$ consecutive numbers is $P(E) = \frac{28}{4060} = \frac{1}{145}$. The probability that they are not three consecutive numbers is $1 - P(E) = 1 - \frac{1}{145} = \frac{144}{145}$.

Three numbers are chosen from $1$ to $30$. The probability that they are not three consecutive numbers is

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