Question:

Let $S = \{1, 2, \ldots, 2000\}$ be the set of numbers. The total number of elements in $S$ is $|S| = 2000$. Let $A$ be the event that a chosen number is a multiple of 3. The number of multiples of 3 in $S$ is given by: $$|A| = \left\lfloor \frac{2000}{3} \right\rfloor = 666$$ Let $B$ be the event that a chosen number is a multiple of 7. The number of multiples of 7 in $S$ is given by: $$|B| = \left\lfloor \frac{2000}{7} \right\rfloor = 285$$ Let $A \cap B$ be the event that a chosen number is a multiple of both 3 and 7. This means the number is a multiple of their least common multiple, which is 21. The number of multiples of 21 in $S$ is given by: $$|A \cap B| = \left\lfloor \frac{2000}{21} \right\rfloor = 95$$ The probability $p$ that the chosen number is a multiple of 3 or a multiple of 7 is given by the Principle of Inclusion-Exclusion: $$|A \cup B| = |A| + |B| - |A \cap B|$$ $$|A \cup B| = 666 + 285 - 95 = 951 - 95 = 856$$ The probability $p$ is the ratio of the number of favorable outcomes to the total number of outcomes: $$p = \frac{|A \cup B|}{|S|} = \frac{856}{2000}$$ This fraction can be simplified: $$p = \frac{856 \div 8}{2000 \div 8} = \frac{107}{250}$$ Finally, we need to calculate $7000p$: $$7000p = 7000 \times \frac{107}{250}$$ $$7000p = \frac{7000}{250} \times 107$$ $$7000p = 28 \times 107$$ $$7000p = 2996$$