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(D) Let $S$ be the set of the first $200$ positive integers,so $n(S) = 200$.Let $A$ be the set of integers divisible by $6$. Then $n(A) = \lfloor 200/

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$1/4$

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(D) Let $S$ be the set of the first $200$ positive integers,so $n(S) = 200$.Let $A$ be the set of integers divisible by $6$. Then $n(A) = \lfloor 200/6 floor = 33$.Let $B$ be the set of integers divisible by $8$. Then $n(B) = \lfloor 200/8 floor = 25$.The intersection $A \cap B$ contains integers divisible by $	ext{lcm}(6, 8) = 24$. Then $n(A \cap B) = \lfloor 200/24 floor = 8$.Using the inclusion-exclusion principle,the number of integers divisible by $6$ or $8$ is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 33 + 25 - 8 = 50$.The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{50}{200} = \frac{1}{4}$.

If a number is chosen at random from the first $200$ positive integers,what is the probability that it is divisible by $6$ or $8$?

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