A number is chosen at random from the set {1, 2, 3, ..., 2000}. Let p be the probability that the ch

Q: Wait, why do we subtract the intersection? Isn't 'OR' just addition?

That is a classic trap! In set theory, the union of two sets $A$ and $B$ is defined as $n(A \cup B) = n(A) + n(B) - n(A \cap B)$. If you just add $n(A)$ and $n(B)$, you are counting the elements that belong to both sets twice—once when you count multiples of 3, and again when you count multiples of 7. Subtracting the intersection removes that extra count, giving you the correct total.

Q: Why do we use the floor function $\lfloor \frac{N}{k} \rfloor$?

The floor function is the mathematical way of saying 'how many times does $k$ fit into $N$ completely?' When you divide 2000 by 3, you get 666.66. Since you cannot have a fraction of a multiple, you take the integer part, which is 666. It effectively discards the remainder, giving you the exact count of multiples.

Q: Why is the intersection based on 21?

A number is a multiple of both 3 and 7 if and only if it is a multiple of their Least Common Multiple (LCM). Since 3 and 7 are prime numbers, their LCM is simply $3 \times 7 = 21$. Therefore, any number that is a multiple of 21 is automatically a multiple of both 3 and 7.

Question

A number is chosen at random from the set $\{1, 2, 3, \dots, 2000\}$. Let $p$ be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then the value of $500p$ is ___.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a massive library containing numbered cards from 1 to 2000. Your universal set is defined as $S = \{1, 2, 3, \dots, 2000\}$.

The total number of possible outcomes, denoted by $N$, is $2000$. We are tasked with finding the probability of picking a number that is a multiple of 3 or a multiple of 7.

### Defining the Circles of Influence

Let Event $A$ be picking a multiple of 3, and Event $B$ be picking a multiple of 7. We seek the number of elements in the union of these two events, $A \cup B$.

To determine the count of multiples of $k$ up to $N$, we use the floor function $\lfloor \frac{N}{k} floor$.

For Event $A$:
$$n(A) = \left\lfloor \frac{2000}{3} ightfloor = \lfloor 666.66\dots floor = 666$$

For Event $B$:
$$n(B) = \left\lfloor \frac{2000}{7} ightfloor = \lfloor 285.71\dots floor = 285$$

### The Trap of Double Counting

If we simply add $n(A)$ and $n(B)$, we count numbers that are multiples of both 3 and 7 twice. These numbers reside in the intersection $A \cap B$.

Since these numbers must be divisible by both 3 and 7, they are multiples of their Least Common Multiple, $	ext{LCM}(3, 7) = 21$. We calculate the intersection count as:
$$n(A \cap B) = \left\lfloor \frac{2000}{21} ightfloor = \lfloor 95.23\dots floor = 95$$

Applying the **Principle of Inclusion-Exclusion**, we find the total number of favorable outcomes:
$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$
$$n(A \cup B) = 666 + 285 - 95 = 856$$

### Final Calculation

We have 856 favorable outcomes out of 2000. The probability $p$ is given by:
$$p = \frac{856}{2000}$$

The problem asks for the value of $500p$. Substituting our value for $p$:
$$500p = 500 	imes \left( \frac{856}{2000} ight)$$

Since $500$ goes into $2000$ exactly 4 times, the expression simplifies to:
$$500p = \frac{856}{4} = 214$$

The final result is **214**.

Animated Solution for Mathematics - Probability: A number is chosen at random from the set ${1, 2, 3, \dots, 2000}$ . Let $p$ be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then the value of $500 p$ is ___.

Enter Numerical Value:

Visualized Solution

Universal Set

Define Events $A$ and $B$

Counting Multiples Formula

Calculate $n (A)$

Calculate $n (B)$

The Intersection $A \cap B$

Calculate $n (A \cap B)$

Principle of Inclusion-Exclusion

Calculate $n (A \cup B)$

Define Probability $p$

Substitute Values for $p$

Setup Final Expression

Calculate $500 p$

Key Takeaways

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

Defining the Circles of Influence

The Trap of Double Counting

Final Calculation

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Enter Numerical Value:

Visualized Solution

Universal Set

Define Events .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Counting Multiples Formula

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(A)

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(B)

The Intersection .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A∩B

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(A∩B)