A bag contains $12$ marbles numbered $1$ to $12$. If a marble is chosen at random, what is the probability that it is either shaded (suppose marbles numbered $2$, $4$, $6$, $8$, $10$, and $12$ are shaded) or is labeled with a multiple of $3$?

• A. $\frac{8}{12}$
• B. $\frac{6}{12}$
• C. $\frac{7}{12}$
• D. $\frac{9}{12}$

Answer: (D)

Verified step by step guidance

1. Identify the total number of marbles, which is 12, so the sample space size is \(N = 12\).
2. Determine the set of shaded marbles: these are marbles numbered 2, 4, 6, 8, 10, and 12. Count them to find \(|S| = 6\).
3. Determine the set of marbles labeled with multiples of 3: these are 3, 6, 9, and 12. Count them to find \(|M| = 4\).
4. Find the intersection of the two sets (marbles that are both shaded and multiples of 3). These are marbles 6 and 12, so \(|S \cap M| = 2\).
5. Use the formula for the probability of the union of two events: \(P(S \cup M) = \frac{|S| + |M| - |S \cap M|}{N} = \frac{6 + 4 - 2}{12}\). This fraction represents the probability that a randomly chosen marble is either shaded or labeled with a multiple of 3.