Probability of Randomly Choosing a Number Divisible by Three or Seven

To find the value of $500p$, we first calculate the number of integers in the set $S = \{1, 2, 3, \dots, 2000\}$ that are divisible by $3$, $7$, or both.

Let:

• $A$ be the set of integers in $S$ divisible by $3$.
• $B$ be the set of integers in $S$ divisible by $7$.

The total number of elements in the sample space $S$ is $|S| = 2000$.

The number of elements in $A$ is given by: $|A| = \left\lfloor \frac{2000}{3} \right\rfloor = 666$

The number of elements in $B$ is given by: $|B| = \left\lfloor \frac{2000}{7} \right\rfloor = 285$

The intersection set $A \cap B$ contains the integers in $S$ that are divisible by both $3$ and $7$, which means they are divisible by $\text{lcm}(3, 7) = 21$.

The number of elements in $A \cap B$ is given by: $|A \cap B| = \left\lfloor \frac{2000}{21} \right\rfloor = 95$

By the Principle of Inclusion-Exclusion, the number of elements in $A \cup B$ (i.e., integers divisible by $3$ or $7$) is: $|A \cup B| = |A| + |B| - |A \cap B|$ $|A \cup B| = 666 + 285 - 95 = 856$

Thus, the probability $p$ of selecting an integer divisible by $3$ or $7$ is: $p = \frac{|A \cup B|}{|S|} = \frac{856}{2000}$

Now, we calculate the required value $500p$: $500p = 500 \times \frac{856}{2000} = \frac{856}{4} = 214$