Answer: B. $30$

Explanation:1. Given Information: Total number of visitors = $375$. Each visitor is in room A, room B, or room C. Probability of selecting a visitor from room A, $P(A) = 0.68$. Probability of selecting a visitor from room B, $P(B) = 0.24$. We need to find the number of visitors in room C.

2. Property of Probabilities: Since each visitor must be in one of the three rooms (A, B, or C), these are mutually exclusive and exhaustive events. The sum of their probabilities must be $1$. $P(A) + P(B) + P(C) = 1$

3. Calculate the Probability of Selecting a Visitor from Room C, $P(C)$: $0.68 + 0.24 + P(C) = 1$ $0.92 + P(C) = 1$ $P(C) = 1 - 0.92$ $P(C) = 0.08$

4. Calculate the Number of Visitors in Room C: Number of visitors in room C = Total number of visitors $\times P(C)$ Number of visitors in room C = $375 \times 0.08$

5. Perform the Multiplication: $375 \times 0.08$: We can do $375 \times 8$ and then adjust decimal places. $375$ $\times \ 8$ ------- $3000$ ($375 \times 8 = (300 \times 8) + (70 \times 8) + (5 \times 8) = 2400 + 560 + 40 = 3000$) Since $0.08$ has two decimal places, $375 \times 0.08 = 30.00$, or $30$. Alternatively: $375 \times \dfrac{8}{100} = \dfrac{375 \times 2}{25} = \dfrac{750}{25} = \dfrac{1500}{50} = \dfrac{3000}{100} = 30$. Another way: $375 \times 0.08 = 375 \times (2 \times 0.04) = (375 \times 2) \times 0.04 = 750 \times 0.04 = 750 \times \dfrac{4}{100} = \dfrac{3000}{100} = 30$.

6. Compare with Options: [A] $8$ [B] $30$ - This matches our calculation. [C] $61$ [D] $165$

7. Conclusion: There are $30$ visitors located in room C.