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(C) The first $100$ natural numbers are $\{1, 2, 3, \ldots, 100\}$.Numbers divisible by both $2$ and $3$ are multiples of $	ext{lcm}(2, 3) = 6$.These

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$\frac{4}{1155}$

Vedclass · Answer

(C) The first $100$ natural numbers are $\{1, 2, 3, \ldots, 100\}$.Numbers divisible by both $2$ and $3$ are multiples of $	ext{lcm}(2, 3) = 6$.These numbers are $\{6, 12, 18, \ldots, 96\}$.The number of such values is $\lfloor \frac{100}{6} floor = 16$.The total number of ways to choose $3$ distinct numbers from $100$ is $^{100}C_3$.The number of ways to choose $3$ numbers divisible by $6$ from the $16$ available is $^{16}C_3$.The required probability is $P = \frac{^{16}C_3}{^{100}C_3}$.$P = \frac{16 	imes 15 	imes 14}{3 	imes 2 	imes 1} 	imes \frac{3 	imes 2 	imes 1}{100 	imes 99 	imes 98} = \frac{16 	imes 15 	imes 14}{100 	imes 99 	imes 98}$.$P = \frac{3360}{970200} = \frac{4}{1155}$.

If three distinct numbers are chosen randomly from the first $100$ natural numbers,then the probability that all three of them are divisible by both $2$ and $3$ is

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