Use the information about student enrollment in two classes and the Tormuld sriown to \( 35 \% \) of students take spanish or Engineering 305 of students take Spanish \( 25 \% \) of students take Engineering \( P(A \) or \( B)=P(A)+P(B)-P(A \) and \( B) \) What is the probability that a student takes Spanish and Engineering? A. \( 15 \% \) B. \( 20 \% \) C. \( 25 \% \) D. \( 30 \% \) E. \( 40 \% \)

To find the probability that a student takes both Spanish and Engineering, we can use the formula you've provided: \[ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B), \] where \( P(A) \) is the probability of taking Spanish and \( P(B) \) is the probability of taking Engineering. From the problem, we know that: - \( P(A \text{ or } B) = 35\% \) - \( P(A) = \frac{305}{Total\ Students} \) - \( P(B) = 25\% \) Let's denote the total number of students as \( x \). We can express \( P(A) \): \[ P(A) = \frac{305}{x} \] Now substitute the known probabilities into the formula: \[ 35\% = \frac{305}{x} + 25\% - P(A \text{ and } B) \] Rearranging gives: \[ P(A \text{ and } B) = \frac{305}{x} + 25\% - 35\% \] \[ P(A \text{ and } B) = \frac{305}{x} - 10\% \] Now we need to find \( x \). Since there's no total number indicated, based on \( P(A) = \frac{305}{x} \) and if we considered \( x \) such that the probabilities make sense, let’s solve for \( P(A \text{ and } B) \) under the assumption of some values for \( x \) or using the known percentage. Assuming \( x \) is large enough, the simplest values of \( P \) will fall close together. If we use common total student numbers like \( 1000 \) (just for calculation's sake): \[ P(A) = \frac{305}{1000} = 30.5\% \] So substituting in: \[ P(A \text{ and } B) = 30.5\% - 10\% = 20.5\% \] Rounding appropriately suggests likely categories likely lead us back to option B: B. \( 20\% \) would be closestly reasonable based on our aggregate terms regarding the students set. Thus, the correct answer is **B. \( 20\% \)**.