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(D) number is divisible by both $2$ and $3$ if and only if it is divisible by $6$. In the first $100$ natural numbers,the numbers divisible by $6$ are

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$4/1155$

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(D) number is divisible by both $2$ and $3$ if and only if it is divisible by $6$. In the first $100$ natural numbers,the numbers divisible by $6$ are $6, 12, 18, \dots, 96$. The number of such values is $\lfloor 100/6 floor = 16$. The total number of ways to select $3$ distinct numbers from $100$ is $^{100}C_3$. The number of ways to select $3$ distinct numbers from the $16$ numbers divisible by $6$ is $^{16}C_3$. The required probability is $P = \frac{^{16}C_3}{^{100}C_3}$. $P = \frac{16 	imes 15 	imes 14}{100 	imes 99 	imes 98} = \frac{16 	imes 15 	imes 14}{100 	imes 99 	imes 98} = \frac{3360}{970200} = \frac{4}{1155}$.

If three distinct numbers are selected from the first $100$ natural numbers,what is the probability that all selected numbers are divisible by both $2$ and $3$?

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