If three distinct numbers are chosen randomly from the first $100$ natural numbers, the the probability that all three of them are divisible by both $2$ and $3$ is $p/q$ . Then the value of $p+q$ is ?????

{p and q are coprime numbers}

The answer is 1159.

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If a number is divisible by 2 and 3, it is divisible by 6. The number chosen must be divisible by 6. There are 16 numbers between 1 and 100 that have 6 as a factor, because 6(16) = 96. Therefore, the probability of choosing one of these numbers is 16/100. Since we must pick 3 distinct numbers, the probability for the next two picks are 15/99 and 14/98. Multiplying these probabilities and simplifying gives 4/1155 = p/q. Thus, p + q = 1159.