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(D) number is divisible by both $2$ and $3$ if and only if it is divisible by their least common multiple,which is $6$. In the first $100$ natural num

Vedclass · Accepted Answer

$4/1155$

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(D) number is divisible by both $2$ and $3$ if and only if it is divisible by their least common multiple,which is $6$. In the first $100$ natural numbers,the numbers divisible by $6$ are $6, 12, 18, \dots, 96$. Using the formula $a_n = a + (n-1)d$,we have $96 = 6 + (n-1)6$,which gives $n = 16$. There are $16$ such numbers. The total number of ways to select $3$ distinct numbers from $100$ is $^{100}C_3$. The number of ways to select $3$ numbers divisible by $6$ is $^{16}C_3$. The required probability is $\frac{^{16}C_3}{^{100}C_3} = \frac{16 	imes 15 	imes 14}{100 	imes 99 	imes 98} = \frac{3360}{970200} = \frac{4}{1155}$.

Three distinct numbers are selected from the first $100$ natural numbers. The probability that all three numbers are divisible by both $2$ and $3$ is:

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